# Miscellaneous Exercise on Chapter 4 Solutions

This page contains the NCERT mathematics class 12 chapter Determinants Miscellaneous Exercise on Chapter 4 Solutions. You can find the numerical questions solutions for the chapter 4/Miscellaneous Exercise on Chapter 4 of NCERT class 12 mathematics in this page. So is the case if you are looking for NCERT class 12 Maths related topic Determinants Miscellaneous Exercise on Chapter 4 solutions. This page contains Miscellaneous Exercise on Chapter 4 solutions. If you’re looking for summary of the chapter or other exercise solutions, they’re available at
Miscellaneous Exercise on Chapter 4 Solutions
1. Prove that the determinant {\begin{vmatrix} x & \sin \text{θ} & \cos \text{θ} \\[5pt] -\sin \text{θ} & -x & 1 \\[5pt] \cos \text{θ} & 1 & x \end{vmatrix}} is independent of θ
Expanding along {\text{C}_1,} we have,
{\begin{vmatrix} x & \sin \text{θ} & \cos \text{θ} \\[5pt] -\sin \text{θ} & -x & 1 \\[5pt] \cos \text{θ} & 1 & x \end{vmatrix}}
{= x \begin{vmatrix} -x & 1 \\[5pt] 1 & x \end{vmatrix} - (-\sin \text{θ}) \begin{vmatrix} \sin \text{θ} & \cos \text{θ} \\[5pt] 1 & x \end{vmatrix} + \cos \text{θ} \begin{vmatrix} \sin \text{θ} & \cos \text{θ} \\[5pt] -x & 1 \end{vmatrix}}
{= x [(-x)x - 1(1)] + \sin \text{θ}\left[(\sin \text{θ})x - (\cos \text{θ})1)\right] + \cos \text{θ} \left[(\sin \text{θ})1 - (\cos \text{θ})(-x)\right]}
{= x(-x^2 - 1) + \sin \text{θ}(x\sin \text{θ} - \cos \text{θ}) + \cos \text{θ}(\sin \text{θ} + x\cos \text{θ})}
{= -x^3 - x + x\sin^2 \text{θ} - \sin \text{θ} \cos \text{θ} + \sin \text{θ} \cos \text{θ} + x\cos^2 \text{θ}}
{= -x^3 - x + x(\sin^2 \text{θ} + \cos^2 \text{θ})}
{= -x^3 - x + x(1)}
{= -x^3 - x + x}
{= -x^3}
As the final value of the determinant does not have any term that contains θ, it is proved that the determinant {\begin{vmatrix} x & \sin \text{θ} & \cos \text{θ} \\[5pt] -\sin \text{θ} & -x & 1 \\[5pt] \cos \text{θ} & 1 & x \end{vmatrix}} is independent of θ
2. Without expanding the determinant prove that {\begin{vmatrix} a & a^2 & bc \\[5pt] b & b^2 & ca \\[5pt] c & c^2 & ab \end{vmatrix} = \begin{vmatrix} 1 & a^2 & a^3 \\[5pt] 1 & b^2 & b^3 \\[5pt] 1 & c^2 & c^3 \end{vmatrix}}
Let {Δ = \begin{vmatrix} a & a^2 & bc \\[5pt] b & b^2 & ca \\[5pt] c & c^2 & ab \end{vmatrix}}
Multiplying {\text{R}_1} with {a,} {\text{R}_2} with {b} and {\text{R}_3} with c and dividing the determinant value with {abc} (so that the determinant value does not change), we have
{Δ = \dfrac{1}{abc}\begin{vmatrix} a^2 & a^3 & abc \\[5pt] b^2 & b^3 & abc \\[5pt] c^2 & c^3 & abc \end{vmatrix}}
Taking out common factor {abc} from {\text{C}_3,} we have,
{Δ = \dfrac{1}{abc} × abc \begin{vmatrix} a^2 & a^3 & 1 \\[5pt] b^2 & b^3 & 1 \\[5pt] c^2 & c^3 & 1 \end{vmatrix}}
Interchanging {\text{C}_2 ↔ \text{C}_3} (and changing the sign of the determinant,) we have,
{Δ = -\begin{vmatrix} a^2 & 1 & a^3 \\[5pt] b^2 & 1 & b^3 \\[5pt] c^2 & 1 & c^3 \end{vmatrix}}
Interchanging {\text{C}_1 ↔ \text{C}_2} (and changing the sign of the determinant,) we have,
{Δ = \begin{vmatrix} 1 & a^2 & a^3 \\[5pt] 1 & b^2 & b^3 \\[5pt] 1 & c^2 & c^3 \end{vmatrix}}
∴ Without expanding the determinant, it is proved that {\begin{vmatrix} a & a^2 & bc \\[5pt] b & b^2 & ca \\[5pt] c & c^2 & ab \end{vmatrix} = \begin{vmatrix} 1 & a^2 & a^3 \\[5pt] 1 & b^2 & b^3 \\[5pt] 1 & c^2 & c^3 \end{vmatrix}}
3. Evaluate {\begin{vmatrix} \cos \text{α} \cos \text{β} & \cos \text{α} \sin \text{β} & -\sin \text{α} \\[5pt] -\sin \text{β} & \cos \text{β} & 0 \\[5pt] \sin \text{α} \cos \text{β} & \sin \text{α} \sin \text{β} & \cos \text{α} \end{vmatrix}}
This problem can be solved in two ways:
Method 1:
Using column transformation
Method 2:
Expanding directly.
(i) Method 1: Using column transformation.
Let {Δ = \begin{vmatrix} \cos \text{α} \cos \text{β} & \cos \text{α} \sin \text{β} & -\sin \text{α} \\[5pt] -\sin \text{β} & \cos \text{β} & 0 \\[5pt] \sin \text{α} \cos \text{β} & \sin \text{α} \sin \text{β} & \cos \text{α} \end{vmatrix}}
Multiplying {\text{C}_1} with {\cos \text{β}} and multiplying {\text{C}_2} with {\sin \text{β}} and dividing the determinant with {\cos \text{β} \sin \text{β}} (so that the value of the determinant won’t change), we have,
Δ
{= \begin{vmatrix} \cos \text{α} \cos \text{β} \cos \text{β} & \cos \text{α} \sin \text{β} \sin \text{β} & -\sin \text{α} \\[5pt] -\sin \text{β} \cos \text{β} & \cos \text{β} \sin \text{β} & 0 \\[5pt] \sin \text{α} \cos \text{β} \cos \text{β} & \sin \text{α} \sin \text{β} \sin \text{β} & \cos \text{α} \end{vmatrix}}
{= \dfrac{1}{\cos \text{β} \sin \text{β}} \begin{vmatrix} \cos \text{α} \cos^2 \text{β} & \cos \text{α} \sin^2 \text{β} & -\sin \text{α} \\[5pt] -\sin \text{β} \cos \text{β} & \cos \text{β} \sin \text{β} & 0 \\[5pt] \sin \text{α} \cos^2 \text{β} & \sin \text{α} \sin^2 \text{β} & \cos \text{α} \end{vmatrix}}
Applying {\text{C}_1 → \text{C}_1 + \text{C}_2}
Δ
{= \dfrac{1}{\cos \text{β} \sin \text{β}} \begin{vmatrix} \cos \text{α} \cos^2 \text{β} + \cos \text{α} \sin^2 \text{β} & \cos \text{α} \sin^2 \text{β} & -\sin \text{α} \\[5pt] -\sin \text{β} \cos \text{β} + \cos \text{β} \sin \text{β} & \cos \text{β} \sin \text{β} & 0 \\[5pt] \sin \text{α} \cos^2 \text{β} + \sin \text{α} \sin^2 \text{β} & \sin \text{α} \sin^2 \text{β} & \cos \text{α} \end{vmatrix}}
{= \dfrac{1}{\cos \text{β} \sin \text{β}} \begin{vmatrix} \cos \text{α}(\cos^2 \text{β} + \sin^2 \text{β}) & \cos \text{α} \sin^2 \text{β} & -\sin \text{α} \\[5pt] 0 & \cos \text{β} \sin \text{β} & 0 \\[5pt] \sin \text{α}(\cos^2 \text{β} + \sin^2 \text{β}) & \sin \text{α} \sin^2 \text{β} & \cos \text{α} \end{vmatrix}}
{= \dfrac{1}{\cos \text{β} \sin \text{β}} \begin{vmatrix} (\cos \text{α})1 & \cos \text{α} \sin^2 \text{β} & -\sin \text{α} \\[5pt] 0 & \cos \text{β} \sin \text{β} & 0 \\[5pt] (\sin \text{α})1 & \sin \text{α} \sin^2 \text{β} & \cos \text{α} \end{vmatrix}}
{= \dfrac{1}{\cos \text{β} \sin \text{β}} \begin{vmatrix} \cos \text{α} & \cos \text{α} \sin^2 \text{β} & -\sin \text{α} \\[5pt] 0 & \cos \text{β} \sin \text{β} & 0 \\[5pt] \sin \text{α} & \sin \text{α} \sin^2 \text{β} & \cos \text{α} \end{vmatrix}}
Expanding along {\text{R}_2,} we have,
Δ
{= \dfrac{1}{\cos \text{β} \sin \text{β}} × \left(- 0\begin{vmatrix} \cos \text{α} \sin^2 \text{β} & -\sin \text{α} \\[5pt] \sin \text{α} \sin^2 \text{β} & \cos \text{α} \end{vmatrix} + \cos \text{β} \sin \text{β} \begin{vmatrix} \cos \text{α} & -\sin \text{α} \\[5pt] \sin \text{α} & \cos \text{α} \end{vmatrix} - 0\begin{vmatrix} \cos \text{α} & \cos \text{α} \sin^2 \text{β} \\[5pt] \sin \text{α} & \sin \text{α} \sin^2 \text{β} \end{vmatrix}\right)}
{= \dfrac{1}{\cos \text{β} \sin \text{β}} × \left[- 0 + \cos \text{β} \sin \text{β} \left(\cos \text{α} \cos \text{α} - (-\sin \text{α}) \sin \text{α}\right) - 0\right]}
{= \dfrac{1}{\cos \text{β} \sin \text{β}} × \cos \text{β} \sin \text{β} (\cos^2 \text{α} +\sin^2 \text{α})}
= 1
{\begin{vmatrix} \cos \text{α} \cos \text{β} & \cos \text{α} \sin \text{β} & -\sin \text{α} \\[5pt] -\sin \text{β} & \cos \text{β} & 0 \\[5pt] \sin \text{α} \cos \text{β} & \sin \text{α} \sin \text{β} & \cos \text{α} \end{vmatrix} = 1}
(ii) Method 2: Direct expansion.
Expanding along {\text{C}_1,} we have,
Let {Δ = \begin{vmatrix} \cos \text{α} \cos \text{β} & \cos \text{α} \sin \text{β} & -\sin \text{α} \\[5pt] -\sin \text{β} & \cos \text{β} & 0 \\[5pt] \sin \text{α} \cos \text{β} & \sin \text{α} \sin \text{β} & \cos \text{α} \end{vmatrix}}
Expanding along {\text{C}_3,} we have,
Δ
{= -\sin \text{α} \begin{vmatrix} -\sin \text{β} & \cos \text{β} \\[5pt] \sin \text{α} \cos \text{β} & \sin \text{α} \sin \text{β} \end{vmatrix} - 0 \begin{vmatrix} \cos \text{α} \cos \text{β} & \cos \text{α} \sin \text{β} \\[5pt] \sin \text{α} \cos \text{β} & \sin \text{α} \sin \text{β} \end{vmatrix} + \cos \text{α} \begin{vmatrix} \cos \text{α} \cos \text{β} & \cos \text{α} \sin \text{β} \\[5pt] -\sin \text{β} & \cos \text{β} \end{vmatrix}}
{= -\sin \text{α} \left[(-\sin \text{β}) \sin \text{α} \sin \text{β} - \cos \text{β} \sin \text{α} \cos \text{β}\right] - 0 + \cos \text{α} \left[\cos \text{α} \cos \text{β} \cos \text{β} - \cos \text{α} \sin \text{β} (-\sin \text{β})\right]}
{= -\sin \text{α} \left(-\sin \text{α} \sin^2 \text{β} - \sin \text{α} \cos^2 \text{β}\right) + \cos \text{α} \left(\cos \text{α} \cos^2 \text{β} + \cos \text{α} \sin^2 \text{β}\right)}
{= (-\sin \text{α})(-\sin \text{α})(\sin^2 \text{β} + \cos^2 \text{β}) + \cos \text{α} \cos \text{α}(\cos^2 \text{β} + \sin^2 \text{β})}
{= (\sin^2 \text{α})1 + (\cos^2 \text{α})1}
{= \sin^2 \text{α} + \cos^2 \text{α}}
= 1
{\begin{vmatrix} \cos \text{α} \cos \text{β} & \cos \text{α} \sin \text{β} & -\sin \text{α} \\[5pt] -\sin \text{β} & \cos \text{β} & 0 \\[5pt] \sin \text{α} \cos \text{β} & \sin \text{α} \sin \text{β} & \cos \text{α} \end{vmatrix} = 1}
4. If {a,} {b} and c are real numbers, and {Δ = \begin{vmatrix} b + c & c + a & a + b \\[5pt] c + a & a + b & b + c \\[5pt] a + b & b + c & c + a \end{vmatrix} = 0,} show that either {a + b + c = 0} or {a = b = c.}
It is given in the problem that
{Δ = \begin{vmatrix} b + c & c + a & a + b \\[5pt] c + a & a + b & b + c \\[5pt] a + b & b + c & c + a \end{vmatrix} = 0}
Applying the row transformation {\text{R}_1 → \text{R}_1 + \text{R}_2 + \text{R}_3} to the given determinant, we have,
Δ
{= \begin{vmatrix} b + c + (c + a) + (a + b) & c + a + (a + b) + (b + c) & a + b + (b + c) + (c + a) \\[5pt] c + a & a + b & b + c \\[5pt] a + b & b + c & c + a \end{vmatrix}}
{= \begin{vmatrix} 2(a + b + c) & 2(a + b + c) & 2(a + b + c) \\[5pt] c + a & a + b & b + c \\[5pt] a + b & b + c & c + a \end{vmatrix}}
Taking out common factor {2(a + b + c)} out of {\text{R}_1,} we have,
{Δ = 2(a + b + c)\begin{vmatrix} 1 & 1 & 1 \\[5pt] c + a & a + b & b + c \\[5pt] a + b & b + c & c + a \end{vmatrix}}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_1,} and {\text{C}_3 → \text{C}_3 - \text{C}_1} we have,
Δ
{= 2(a + b + c)\begin{vmatrix} 1 & 1 - 1 & 1 - 1 \\[5pt] c + a & (a + b) - (c + a) & (b + c) - (c + a) \\[5pt] a + b & (b + c) - (a + b) & (c + a) - (a + b) \end{vmatrix}}
{= 2(a + b + c)\begin{vmatrix} 1 & 0 & 0 \\[5pt] c + a & b - c & b - a \\[5pt] a + b & c - a & c - b \end{vmatrix}}
Expanding along {\text{R}_1,} we have,
Δ
{= 2(a + b + c)\left[1 \begin{vmatrix} b - c & b - a \\[5pt] c - a & c - b \end{vmatrix} - 0 \begin{vmatrix} c + a & b - a \\[5pt] a + b & c - b \end{vmatrix} + 0 \begin{vmatrix} c + a & b - c \\[5pt] a + b & c - a \end{vmatrix}\right]}
{= 2(a + b + c)\left[(b - c)(c - b) - (b - a)(c - a) - 0 + 0\right]}
{= 2(a + b + c)\left[(bc - b^2 - c^2 + bc) - (bc - ab - ac + a^2)\right]}
{= 2(a + b + c)(bc - b^2 - c^2 + bc - bc + ab + ac - a^2)}
{= 2(a + b + c) × -(a^2 + b^2 + c^2 - ab - bc - ca)}
{= -(a + b + c)(2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca)}
{= -(a + b + c)\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right]}
It is given in the problem that Δ = 0
{⇒ (a + b + c)\left[(a - b)^2 + (b - c)^2 + (c - a)^2\right] = 0}
⇒ either {(a + b + c) = 0} or {(a - b)^2 + (b - c)^2 + (c - a)^2 = 0}
As {(a - b)^2 \ge 0,} {(b - c)^2 \ge 0} and {(a - b)^2 \ge 0} (as {a,} {b} and c are real numbers and as each of the terms in this expression is a square, they can never be negative).
However, as their sum is zero, it is possible only when {a - b = 0,} {b - c = 0} and {c - a = 0}
{⇒ a = b,} {b = c} and {c = a}
{⇒ a = b = c}
∴ It is showed that either {a + b + c = 0} or {a = b = c.}
5. Solve the equation {\begin{vmatrix} x + a & x & x \\[5pt] x & x + a & x \\[5pt] x & x & x + a \end{vmatrix} = 0,} {a} ≠ 0
The given equation is {\begin{vmatrix} x + a & x & x \\[5pt] x & x + a & x \\[5pt] x & x & x + a \end{vmatrix} = 0,}
Applying row transformation {\text{R}_1 → \text{R}_1 - \text{R}_2,} we have,
{\begin{vmatrix} x + a - x & x - (x + a) & x - x \\[5pt] x & x + a & x \\[5pt] x & x & x + a \end{vmatrix} = 0,}
{⇒ \begin{vmatrix} a & -a & 0 \\[5pt] x & x + a & x \\[5pt] x & x & x + a \end{vmatrix} = 0,}
Applying row transformation {\text{R}_2 → \text{R}_2 - \text{R}_3,} we have,
{\begin{vmatrix} a & -a & 0 \\[5pt] x - x & x + a - x & x - (x + a) \\[5pt] x & x & x + a \end{vmatrix} = 0,}
{⇒ \begin{vmatrix} a & -a & 0 \\[5pt] 0 & a & -a \\[5pt] x & x & x + a \end{vmatrix} = 0,}
Applying the row transformation {\text{C}_2 → \text{C}_2 + \text{C}_1,} we have,
{\begin{vmatrix} a & -a + a & 0 \\[5pt] 0 & a + 0 & -a \\[5pt] x & x + x & x + a \end{vmatrix} = 0,}
{⇒ \begin{vmatrix} a & 0 & 0 \\[5pt] 0 & a & -a \\[5pt] x & 2x & x + a \end{vmatrix} = 0,}
Expanding along {\text{R}_1,} we have,
{a \begin{vmatrix} a & -a \\[5pt] 2x & x + a \end{vmatrix} - 0 \begin{vmatrix} 0 & -a \\[5pt] x & x + a \end{vmatrix} + 0 \begin{vmatrix} 0 & a \\[5pt] x & 2x \end{vmatrix} = 0}
{⇒ a \left[(a × (x + a) - (-a) × 2x\right] - 0 + 0 = 0}
{⇒ (ax + a^2 + 2ax) = 0}
{⇒ 3ax + a^2 = 0}
{⇒ 3ax = -a^2}
{⇒ x = \dfrac{-a^2}{3a}}
{⇒ x = \dfrac{-a}{3}}
6. Prove that {\begin{vmatrix} a^2 & bc & ac + c^2 \\[5pt] a^2 + ab & b^2 & ac \\[5pt] ab & b^2 + bc & c^2 \end{vmatrix} = 4a^2b^2c^2}
Let the given determinant be
{Δ = \begin{vmatrix} a^2 & bc & ac + c^2 \\[5pt] a^2 + ab & b^2 & ac \\[5pt] ab & b^2 + bc & c^2 \end{vmatrix}}
Taking out common factors a from {\text{C}_1,} b from {\text{C}_2} and c from {\text{C}_3,} we have
{Δ = abc\begin{vmatrix} a & c & a + c \\[5pt] a + b & b & a \\[5pt] b & b + c & c \end{vmatrix}}
Applying row transformation {\text{R}_1 → \text{R}_1 + \text{R}_2 + \text{R}_3,} we have,
Δ
{abc\begin{vmatrix} a + (a + b) + b & c + b + (b + c) & a + c + a + c \\[5pt] a + b & b & a \\[5pt] b & b + c & c \end{vmatrix}}
{= abc\begin{vmatrix} 2(a + b) & 2(b + c) & 2(a + c) \\[5pt] a + b & b & a \\[5pt] b & b + c & c \end{vmatrix}}
Taking out common factor 2 from {\text{R}_1,} we have,
{Δ = 2abc\begin{vmatrix} a + b & b + c & a + c \\[5pt] a + b & b & a \\[5pt] b & b + c & c \end{vmatrix}}
Applying row transformations {\text{R}_2 → \text{R}_2 - \text{R}_1} and {\text{R}_3 → \text{R}_3 - \text{R}_1,} we have,
Δ
{= 2abc\begin{vmatrix} a + b & b + c & a + c \\[5pt] a + b - (a + b) & b - (b + c) & a - (a + c) \\[5pt] b - (a + b) & b + c - (b + c) & c - (a + c) \end{vmatrix}}
{= 2abc\begin{vmatrix} a + b & b + c & a + c \\[5pt] 0 & -c & -c \\[5pt] -a & 0 & -a \end{vmatrix}}
Taking out common factor {-c} from {\text{R}_2} and {-a} from {\text{R}_3,} we have,
Δ
{= 2abc(-c)(-a)\begin{vmatrix} a + b & b + c & a + c \\[5pt] 0 & 1 & 1 \\[5pt] 1 & 0 & 1 \end{vmatrix}}
{= 2a^2bc^2\begin{vmatrix} a + b & b + c & a + c \\[5pt] 0 & 1 & 1 \\[5pt] 1 & 0 & 1 \end{vmatrix}}
Expanding along {\text{R}_3,} we have,
Δ
{= 2a^2bc^2 × \left(1 × \begin{vmatrix} b + c & a + c \\[5pt] 1 & 1 \end{vmatrix} - 0 \begin{vmatrix} a + b & a + c \\[5pt] 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} a + b & b + c \\[5pt] 0 & 1 \end{vmatrix}\right)}
{= 2a^2bc^2 \Big(1\big((b + c)1 - (a + c)1\big) - 0 + 1\big((a + b)1 - (b + c)0\big)\Big)}
{= 2a^2bc^2\Big(\big((b + c) - (a + c)\big) + \big((a + b) - 0\big)\Big)}
{= 2a^2bc^2(b + c - a - c + a + b)}
{= 2a^2bc^2(2b)}
{= 4a^2b^2c^2}
∴ It is proved that {\begin{vmatrix} a^2 & bc & ac + c^2 \\[5pt] a^2 + ab & b^2 & ac \\[5pt] ab & b^2 + bc & c^2 \end{vmatrix} = 4a^2b^2c^2}
7. If {\text{A}^{-1} = \begin{bmatrix} 3 & -1 & 1 \\[5pt] -15 & 6 & 5 \\[5pt] 5 & -2 & 2 \end{bmatrix}} and {\text{B} = \begin{bmatrix} 1 & 2 & -2 \\[5pt] -1 & 3 & 0 \\[5pt] 0 & -2 & 1 \end{bmatrix},} find {\text{(AB)}^{-1}}
We know that
{\text{(AB)}^{-1} = \text{B}^{-1}\text{A}^{-1}}
So, let’s first find {\text{B}^{-1}}
Now,
Expanding the determinant of B along {\text{C}_1,} we have,
|B|
{= 1 \begin{vmatrix} 3 & 0 \\[5pt] -2 & 1 \end{vmatrix} - (-1) \begin{vmatrix} 2 & -2 \\[5pt] -2 & 1 \end{vmatrix} + 0 \begin{vmatrix} 2 & -2 \\[5pt] 3 & 0 \end{vmatrix}}
{= 1\big(3(1) - 0(-2)\big) + 1\big(2(1) - (-2)(-2)\big) + 0}
= 1(3 – 0) + 1(2 – 4)
= 1(3) + 1(-2)
= 3 – 2
= 1 ———-❶
≠ 0
As |B| ≠ 0, B is a nonsingular matrix and hence B is invertible.
As {\text{B}^{-1} = \dfrac{1}{\text{|B|}}(adj \text{ B)},} lets’s first find {adj \text{ B},} by first finding the minors and then the Cofactors of the elements of B.
Element
{b_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{b_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{red}{\cancel{-1}} & \textcolor{green}{3} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{-2} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 3 & 0 \\[5pt] -2 & 1 \end{vmatrix} = 3(1) - 0(-2) = 3 - 0 = 3}
{b_{12} = 2}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{-1} & \textcolor{red}{\cancel{3}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{-2}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} -1 & 0 \\[5pt] 0 & 1 \end{vmatrix} = (-1)1 - 0(0) = -1 - 0 = -1}
{b_{13} = -2}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{-1} & \textcolor{green}{3} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{-2} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} -1 & 3 \\[5pt] 0 & -2 \end{vmatrix} = (-1)(-2) - 3(0) = 2 - 0 = 2}
{b_{21} = -1}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{2} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{-2} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 2 & -2 \\[5pt] -2 & 1 \end{vmatrix} = 2(1) - (-2)(-2) = 2 - 4 = -2}
{b_{22} = 3}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{-2}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & -2 \\[5pt] 0 & 1 \end{vmatrix} = 1(1) - (-2)(0) = 1 - 0 = 1}
{b_{23} = 0}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{2} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{red}{\cancel{-1}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{-2} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] 0 & -2 \end{vmatrix} = 1(-2) - 2(0) = -2 - 0 = -2}
{b_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{2} & \textcolor{green}{-2} \\[5pt] \textcolor{red}{\cancel{-1}} & \textcolor{green}{3} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 2 & -2 \\[5pt] 3 & 0 \end{vmatrix} = 2(0) - (-2)3 = 0 + 6 = 6}
{b_{32} = -2}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{2}} & \textcolor{green}{-2} \\[5pt] \textcolor{green}{-1} & \textcolor{red}{\cancel{3}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -2 \\[5pt] -1 & 0 \end{vmatrix} = 1(0) - (-2)(-1) = 0 - 2 = -2}
{b_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{2} & \textcolor{red}{\cancel{-2}} \\[5pt] \textcolor{green}{-1} & \textcolor{green}{3} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 2 \\[5pt] -1 & 3 \end{vmatrix} = 1(3) - 2(-1) = 3 + 2 = 5}
Element
{b_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{B}_{ij}}
{b_{11} = 1}
{\text{M}_{11} = 3}
{\text{B}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × 3 = 3}
{b_{12} = 2}
{\text{M}_{12} = -1}
{\text{B}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × -1 = 1}
{b_{13} = -2}
{\text{M}_{13} = 2}
{\text{B}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × 2 = 2}
{b_{21} = -1}
{\text{M}_{21} = -2}
{\text{B}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × -2 = 2}
{b_{22} = 3}
{\text{M}_{22} = 1}
{\text{B}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × 1 = 1}
{b_{23} = 0}
{\text{M}_{23} = -2}
{\text{B}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × -2 = 2}
{b_{31} = 0}
{\text{M}_{31} = 6}
{\text{B}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × 6 = 6}
{b_{32} = -2}
{\text{M}_{32} = -2}
{\text{B}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × -2 = 2}
{b_{33} = 1}
{\text{M}_{33} = 5}
{\text{B}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × 5 = 5}
So,
{= \begin{bmatrix} \text{B}_{11} & \text{B}_{21} & \text{B}_{31} \\[5pt] \text{B}_{12} & \text{B}_{22} & \text{B}_{32} \\[5pt] \text{B}_{13} & \text{B}_{23} & \text{B}_{33} \end{bmatrix}}
{= \begin{bmatrix} 3 & 2 & 6 \\[5pt] 1 & 1 & 2 \\[5pt] 2 & 2 & 5 \end{bmatrix}} ———-❷
{\text{ B}^{-1}}
{= \dfrac{1}{1}\begin{bmatrix} 3 & 2 & 6 \\[5pt] 1 & 1 & 2 \\[5pt] 2 & 2 & 5 \end{bmatrix}} (from ❶ and ❷)
{= \begin{bmatrix} 3 & 2 & 6 \\[5pt] 1 & 1 & 2 \\[5pt] 2 & 2 & 5 \end{bmatrix}} ———-❸
Now,
{\text{(AB)}^{-1}}
{= \text{B}^{-1}.\text{A}^{-1}}
{= \begin{bmatrix} 3 & 2 & 6 \\[5pt] 1 & 1 & 2 \\[5pt] 2 & 2 & 5 \end{bmatrix} \begin{bmatrix} 3 & -1 & 1 \\[5pt] -15 & 6 & 5 \\[5pt] 5 & -2 & 2 \end{bmatrix}} (from ❸)
{= \begin{bmatrix} 3(3) + 2(-15) + 6(5) & 3(-1) + 2(6) + 6(-2) & 3(1) + 2(-5) + 6(2) \\[5pt] 1(3) + 1(-15) + 2(5) & 1(-1) + 1(6) + 2(-2) & 1(1) + 1(-5) + 2(2) \\[5pt] 2(3) + 2(-15) + 5(5) & 2(-1) + 2(6) + 5(-2) & 2(1) + 2(-5) + 5(2) \end{bmatrix}}
{= \begin{bmatrix} 9 - 30 + 30 & -3 + 12 - 12 & 3 - 10 + 12 \\[5pt] 3 - 15 + 10 & -1 + 6 - 4 & 1 - 5 + 4 \\[5pt] 6 - 30 + 25 & -2 + 12 - 10 & 2 - 10 + 10 \end{bmatrix}}
{= \begin{bmatrix} 9 & -3 & 5 \\[5pt] -2 & 1 & 0 \\[5pt] 1 & 0 & 2 \end{bmatrix}}
{\text{(AB)}^{-1} = \begin{bmatrix} 9 & -3 & 5 \\[5pt] -2 & 1 & 0 \\[5pt] 1 & 0 & 2 \end{bmatrix}}
8. Let {\text{A} = \begin{bmatrix} 1 & -2 & 1 \\[5pt] -2 & 3 & 1 \\[5pt] 1 & 1 & 5 \end{bmatrix}.} Verify that
(ii) {\left(\text{A}^{-1}\right)^{-1} = \text{A}}
⚠️Note: In some versions of the pdf book, {a_{12}} and {a_{21}} values have a typo and given as ‘2’. Consider it as ‘-2’ (as given in the question above).
To solve this problem, we need to find,
Step I:
To Find {\text{A}^{-1}}
Step II:
Step III:
Step IV:
To Find {\left(\text{A}^{-1}\right)^{-1}}
Given that {\text{A} = \begin{bmatrix} 1 & -2 & 1 \\[5pt] -2 & 3 & 1 \\[5pt] 1 & 1 & 5 \end{bmatrix}}
I.(i) Find |A|
Exanding the determinant along {\text{R}_1,} we have,
|A|
{= 1 \begin{vmatrix} 3 & 1 \\[5pt] 1 & 5 \end{vmatrix} - (-2) \begin{vmatrix} -2 & 1 \\[5pt] 1 & 5 \end{vmatrix} + 1 \begin{vmatrix} -2 & 3 \\[5pt] 1 & 1 \end{vmatrix}}
{= 1 \left[3(5) - 1(1)\right] - (-2) \left[(-2)5 - 1(1)\right] + 1 \left[(-2)1 - 3(1)\right]}
= 1(15 – 1) + 2(-10 – 1) + 1(-2 – 3)
= 1(14) + 2(-11) + 1(-5)
= 14 – 22 – 5
= -13 ———-❶
≠ 0
As |A| ≠ 0, A is nonsingular and hence invertible.
I.(ii) To find {adj \text{ A}}
To find {adj \text{ A},} lets’s first find {adj \text{ A},} by first finding the minors and then the Cofactors of the elements of A.
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{green}{3} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{1} & \textcolor{green}{5} \end{vmatrix}}
{= \begin{vmatrix} 3 & 1 \\[5pt] 1 & 5 \end{vmatrix} = (3 × 5 - 1 × 1) = 15 - 1 = 14}
{a_{12} = -2}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{-2} & \textcolor{red}{\cancel{3}} & \textcolor{green}{1} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{1}} & \textcolor{green}{5} \end{vmatrix}}
{= \begin{vmatrix} -2 & 1 \\[5pt] 1 & 5 \end{vmatrix} = \big((-2) × 5 - 1 × 1\big) = -10 - 1 = -11}
{a_{13} = 1}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{-2} & \textcolor{green}{3} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{1} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} -2 & 3 \\[5pt] 1 & 1 \end{vmatrix} = \big((-2) × 1 - 3 × 1\big) = -2 - 3 = -5}
{a_{21} = -2}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-2} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{1} & \textcolor{green}{5} \end{vmatrix}}
{= \begin{vmatrix} -2 & 1 \\[5pt] 1 & 5 \end{vmatrix} = \big((-2) × 5 - 1 × 1\big) = -10 - 1 = -11}
{a_{22} = 3}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-2}} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{1}} & \textcolor{green}{5} \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 \\[5pt] 1 & 5 \end{vmatrix} = (1 × 5 - 1 × 1) = 5 - 1 = 4}
{a_{23} = 1}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-2} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{1} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -2 \\[5pt] 1 & 1 \end{vmatrix} = \big(1 × 1 - (-2) × 1\big) = 1 + 2 = 3}
{a_{31} = 1}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{-2} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{-2}} & \textcolor{green}{3} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} -2 & 1 \\[5pt] 3 & 1 \end{vmatrix} = \big((-2) × 1 - 1 × 3\big) = -2 - 3 = -5}
{a_{32} = 1}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{-2}} & \textcolor{green}{1} \\[5pt] \textcolor{green}{-2} & \textcolor{red}{\cancel{3}} & \textcolor{green}{1} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 \\[5pt] -2 & 1 \end{vmatrix} = \big(1 × 1 - 1 × (-2)\big) = 1 + 2 = 3}
{a_{33} = 5}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{-2} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{-2} & \textcolor{green}{3} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{5}} \end{vmatrix}}
{= \begin{vmatrix} 1 & -2 \\[5pt] -2 & 3 \end{vmatrix} = \big(1 × 3 - (-2) × (-2)\big) = 3 - 4 = -1}
Element
{b_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{B}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 14}
{\text{A}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × 14 = 14}
{a_{12} = -2}
{\text{M}_{12} = -11}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × -11 = 11}
{a_{13} = 1}
{\text{M}_{13} = -5}
{\text{a}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × -5 = -5}
{a_{21} = -2}
{\text{M}_{21} = -11}
{\text{a}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × -11 = 11}
{a_{22} = 3}
{\text{M}_{22} = 4}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × 4 = 4}
{a_{23} = 1}
{\text{M}_{23} = 3}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 3 = -3}
{a_{31} = 1}
{\text{M}_{31} = -5}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × -5 = -5}
{a_{32} = 1}
{\text{M}_{32} = 3}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × 3 = -3}
{a_{33} = 5}
{\text{M}_{33} = -1}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × -1 = -1}
So,
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} 14 & 11 & -5 \\[5pt] 11 & 4 & -3 \\[5pt] -5 & -3 & -1 \end{bmatrix}} ———-❷
{\text{A}^{-1}}
{= \dfrac{1}{-13}\begin{bmatrix} 14 & 11 & -5 \\[5pt] 11 & 4 & -3 \\[5pt] -5 & -3 & -1 \end{bmatrix}} (from ❶ and ❷)
{= \dfrac{1}{13}\begin{bmatrix} -14 & -11 & 5 \\[5pt] -11 & -4 & 3 \\[5pt] 5 & 3 & 1 \end{bmatrix}} ———-❸
II.(i) To find {|adj \text{ A}|}
Expanding the determinant of {adj \text{ A}} along {\text{R}_1,} we have,
{= \begin{vmatrix} 14 & 11 & -5 \\[5pt] 11 & 4 & -3 \\[5pt] -5 & -3 & -1 \end{vmatrix}}
{= 14 \begin{vmatrix} 4 & -3 \\[5pt] -3 & -1 \end{vmatrix} - 11 \begin{vmatrix} 11 & -3 \\[5pt] -5 & -1 \end{vmatrix} + (-5) \begin{vmatrix} 11 & 4 \\[5pt] -5 & -3 \end{vmatrix}}
{= 14\left[4(-1) - (-3)(-3)\right] - 11\left[11(-1) - (-3)(-5)\right] + (-5)\left[11(-3) - 4(-5)\right]}
= 14(-4 – 9) – 11(-11 – 15) – 5(-33 + 20)
= 14(-13) – 11(-26) – 5(-13)
= -182 + 286 + 65
= 169 ———-❹
≠ 0
Alternatively, we know that if A is a square matrix of order {n,} then {\left|adj \text{ A}\right| = \left|\text{A}\right|^{n - 1}.}
So, in our case as A is a square matrix of order 3, we have,
{= \left|\text{A}\right|^{3 - 1}}
{= 13^2}
= 169 ———-❹
≠ 0
Thus, {adj \text{ A}} is also a nonsingular matrix and is invertible.
For the sake of simplicity, let’s assume that

{⇒ \text{B} = \begin{bmatrix} 14 & 11 & -5 \\[5pt] 11 & 4 & -3 \\[5pt] -5 & -3 & -1 \end{bmatrix}}
To find {\left(adj \text{ B}\right)^{-1},} lets’s first find {adj \text{ B},} by first finding the Minors and then the Cofactors of the elements of B.
Element
{b_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
{b_{11} = 14}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{14}} & \textcolor{red}{\cancel{11}} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{red}{\cancel{11}} & \textcolor{green}{4} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{-5}} & \textcolor{green}{-3} & \textcolor{green}{-1} \end{vmatrix}}
{= \begin{vmatrix} 4 & -3 \\[5pt] -3 & -1 \end{vmatrix} = \big(4 × (-1) - (-3) × (-3)\big) = -4 - 9 = -13}
{b_{12} = 11}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{14}} & \textcolor{red}{\cancel{11}} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{green}{11} & \textcolor{red}{\cancel{4}} & \textcolor{green}{-3} \\[5pt] \textcolor{green}{-5} & \textcolor{red}{\cancel{-3}} & \textcolor{green}{-1} \end{vmatrix}}
{= \begin{vmatrix} 11 & -3 \\[5pt] -5 & -1 \end{vmatrix} = \big(11 × (-1) - (-3) × (-5)\big) = -11 - 15 = -26}
{b_{13} = -5}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{14}} & \textcolor{red}{\cancel{11}} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{green}{11} & \textcolor{green}{4} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{-5} & \textcolor{green}{-3} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 11 & 4 \\[5pt] -5 & -3 \end{vmatrix} = \big(11 × (-3) - 4 × (-5)\big) = -33 + 20 = -13}
{b_{21} = 11}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{14}} & \textcolor{green}{11} & \textcolor{green}{-5} \\[5pt] \textcolor{red}{\cancel{11}} & \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{red}{\cancel{-5}} & \textcolor{green}{-3} & \textcolor{green}{-1} \end{vmatrix}}
{= \begin{vmatrix} 11 & -5 \\[5pt] -3 & -1 \end{vmatrix} = \big(11 × (-1) - (-5) × (-3)\big) = -11 - 15 = -26}
{b_{22} = 4}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{14} & \textcolor{red}{\cancel{11}} & \textcolor{green}{-5} \\[5pt] \textcolor{red}{\cancel{11}} & \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{-5} & \textcolor{red}{\cancel{-3}} & \textcolor{green}{-1} \end{vmatrix}}
{= \begin{vmatrix} 14 & -5 \\[5pt] -5 & -1 \end{vmatrix} = \big(14 × (-1) - (-5) × (-5)\big) = -14 - 25 = -39}
{b_{23} = -3}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{14} & \textcolor{green}{11} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{red}{\cancel{11}} & \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{green}{-5} & \textcolor{green}{-3} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 14 & 11 \\[5pt] -5 & -3 \end{vmatrix} = \big(14 × (-3) - 11 × (-5)\big) = -42 + 55 = 13}
{b_{31} = -5}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{14}} & \textcolor{green}{11} & \textcolor{green}{-5} \\[5pt] \textcolor{red}{\cancel{11}} & \textcolor{green}{4} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{-5}} & \textcolor{red}{\cancel{-3}} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 11 & -5 \\[5pt] 4 & -3 \end{vmatrix} = \big(11 × (-3) - (-5) × 4\big) = -33 + 20 = -13}
{b_{32} = -3}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{14} & \textcolor{red}{\cancel{11}} & \textcolor{green}{-5} \\[5pt] \textcolor{green}{11} & \textcolor{red}{\cancel{4}} & \textcolor{green}{-3} \\[5pt] \textcolor{red}{\cancel{-5}} & \textcolor{red}{\cancel{-3}} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 14 & -5 \\[5pt] 11 & -3 \end{vmatrix} = \big(14 × (-3) - (-5) × 11\big) = -42 + 55 = 13}
{b_{33} = -1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{14} & \textcolor{green}{11} & \textcolor{red}{\cancel{-5}} \\[5pt] \textcolor{green}{11} & \textcolor{green}{4} & \textcolor{red}{\cancel{-3}} \\[5pt] \textcolor{red}{\cancel{-5}} & \textcolor{red}{\cancel{-3}} & \textcolor{red}{\cancel{-1}} \end{vmatrix}}
{= \begin{vmatrix} 14 & 11 \\[5pt] 11 & 4 \end{vmatrix} = (14 × 4 - 11 × 11) = 56 - 121 = -65}
Element
{b_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{B}_{ij}}
{b_{11} = 14}
{\text{M}_{11} = -13}
{\text{B}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × -13 = -13}
{b_{12} = 11}
{\text{M}_{12} = -26}
{\text{B}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × -26 = 26}
{b_{13} = -5}
{\text{M}_{13} = -13}
{\text{B}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × -13 = -13}
{b_{21} = 11}
{\text{M}_{21} = -26}
{\text{B}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × -26 = 26}
{b_{22} = 4}
{\text{M}_{22} = -39}
{\text{B}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × -39 = -39}
{b_{23} = -3}
{\text{M}_{23} = 13}
{\text{B}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 13 = -13}
{b_{31} = -5}
{\text{M}_{31} = -13}
{\text{B}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × -13 = -13}
{b_{32} = -3}
{\text{M}_{32} = 13}
{\text{B}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × 13 = -13}
{b_{33} = -1}
{\text{M}_{33} = -65}
{\text{B}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × -65 = -65}
So,
{= \begin{bmatrix} \text{B}_{11} & \text{B}_{21} & \text{B}_{31} \\[5pt] \text{B}_{12} & \text{B}_{22} & \text{B}_{32} \\[5pt] \text{B}_{13} & \text{B}_{23} & \text{B}_{33} \end{bmatrix}}
{= \begin{bmatrix} -13 & 26 & -13 \\[5pt] 26 & -39 & -13 \\[5pt] -13 & -13 & -65 \end{bmatrix}} ———-❺
{\text{B}^{-1} =}
=
{\dfrac{1}{169}\begin{bmatrix} -13 & 26 & -13 \\[5pt] 26 & -39 & -13 \\[5pt] -13 & -13 & -65 \end{bmatrix}}
(from ❹ and ❺)
=
{\dfrac{1}{169} × 13 \begin{bmatrix} -1 & 2 & -1 \\[5pt] 2 & -3 & -1 \\[5pt] -1 & -1 & -5 \end{bmatrix}}
(taking 13 common)
=
{\dfrac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\[5pt] 2 & -3 & -1 \\[5pt] -1 & -1 & -5 \end{bmatrix}}
As we assumed that {\text{B} = adj \text{ A}}
{\left(adj \text{ A}\right)^{-1} = \dfrac{1}{13} \begin{bmatrix} -1 & 2 & -1 \\[5pt] 2 & -3 & -1 \\[5pt] -1 & -1 & -5 \end{bmatrix}} ———-❻
III. To find {adj \left(\text{ A}^{-1}\right)}
From ❸, we have,
{\text{A}^{-1} = \dfrac{1}{13}\begin{bmatrix} -14 & -11 & 5 \\[5pt] -11 & -4 & 3 \\[5pt] 5 & 3 & 1 \end{bmatrix}}
For the sake of simplicity, let’s assume,
C
{= \text{A}^{-1}}
{= \dfrac{1}{13}\begin{bmatrix} -14 & -11 & 5 \\[5pt] -11 & -4 & 3 \\[5pt] 5 & 3 & 1 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{-14}{13} & \dfrac{-11}{13} & \dfrac{5}{13} \\[10pt] \dfrac{-11}{13} & \dfrac{-4}{13} & \dfrac{3}{13} \\[10pt] \dfrac{5}{13} & \dfrac{3}{13} & \dfrac{1}{13} \end{bmatrix}}
To find {adj \text{ C},} lets’s first find the Minors and then the Cofactors of the elements of C.
Element
{c_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{c_{11} = \dfrac{-14}{13}}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{\dfrac{-14}{13}}} & \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{red}{\cancel{\dfrac{5}{13}}} \\[10pt] \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{green}{\dfrac{-4}{13}} & \textcolor{green}{\dfrac{3}{13}} \\[10pt] \textcolor{red}{\cancel{\dfrac{5}{13}}} & \textcolor{green}{\dfrac{3}{13}} & \textcolor{green}{\dfrac{1}{13}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-4}{13} & \dfrac{3}{13} \\[10pt] \dfrac{3}{13} & \dfrac{1}{13} \end{vmatrix}}
{= \dfrac{-4}{13} × \dfrac{1}{13} - \dfrac{3}{13} × \dfrac{3}{13}}
{= \dfrac{-4}{169} - \dfrac{9}{169}}
{= \dfrac{-13}{169}}
{= \dfrac{-1}{13}}
{c_{12} = \dfrac{-11}{13}}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{\dfrac{-14}{13}}} & \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{red}{\cancel{\dfrac{5}{13}}} \\[10pt] \textcolor{green}{\dfrac{-11}{13}} & \textcolor{red}{\cancel{\dfrac{-4}{13}}} & \textcolor{green}{\dfrac{3}{13}} \\[10pt] \textcolor{green}{\dfrac{5}{13}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} & \textcolor{green}{\dfrac{1}{13}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-11}{13} & \dfrac{3}{13} \\[10pt] \dfrac{5}{13} & \dfrac{1}{13} \end{vmatrix}}
{= \dfrac{-11}{13} × \dfrac{1}{13} - \dfrac{3}{13} × \dfrac{5}{13}}
{= \dfrac{-11}{169} - \dfrac{15}{169}}
{= \dfrac{-26}{169}}
{= \dfrac{-2}{13}}
{c_{13} = \dfrac{5}{13}}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{\dfrac{-14}{13}}} & \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{red}{\cancel{\dfrac{5}{13}}} \\[10pt] \textcolor{green}{\dfrac{-11}{13}} & \textcolor{green}{\dfrac{-4}{13}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} \\[10pt] \textcolor{green}{\dfrac{5}{13}} & \textcolor{green}{\dfrac{3}{13}} & \textcolor{red}{\cancel{\dfrac{1}{13}}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-11}{13} & \dfrac{-4}{13} \\[10pt] \dfrac{5}{13} & \dfrac{3}{13} \end{vmatrix}}
{ = \dfrac{-11}{13} × \dfrac{3}{13} - \dfrac{-4}{13} × \dfrac{5}{13}}
{= \dfrac{-33}{169} + \dfrac{20}{169}}
{= \dfrac{-13}{169}}
{= \dfrac{-1}{13}}
{c_{21} = \dfrac{-11}{13}}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{\dfrac{-14}{13}}} & \textcolor{green}{\dfrac{-11}{13}} & \textcolor{green}{\dfrac{5}{13}} \\[10pt] \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{red}{\cancel{\dfrac{-4}{13}}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} \\[10pt] \textcolor{red}{\cancel{\dfrac{5}{13}}} & \textcolor{green}{\dfrac{3}{13}} & \textcolor{green}{\dfrac{1}{13}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-11}{13} & \dfrac{5}{13} \\[10pt] \dfrac{3}{13} & \dfrac{1}{13} \end{vmatrix}}
{= \dfrac{-11}{13} × \dfrac{1}{13} - \dfrac{5}{13} × \dfrac{3}{13}}
{= \dfrac{-11}{169} - \dfrac{15}{169}}
{= \dfrac{-26}{169}}
{= \dfrac{-2}{13}}
{c_{22} = \dfrac{-4}{13}}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{\dfrac{-14}{13}} & \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{green}{\dfrac{5}{13}} \\[10pt] \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{red}{\cancel{\dfrac{-4}{13}}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} \\[10pt] \textcolor{green}{\dfrac{5}{13}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} & \textcolor{green}{\dfrac{1}{13}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-14}{13} & \dfrac{5}{13} \\[10pt] \dfrac{5}{13} & \dfrac{1}{13} \end{vmatrix}}
{= \dfrac{-14}{13} × \dfrac{1}{13} - \dfrac{5}{13} × \dfrac{5}{13}}
{= \dfrac{-14}{169} - \dfrac{25}{169}}
{= \dfrac{-39}{169}}
{= \dfrac{-3}{13}}
{c_{23} = \dfrac{3}{13}}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{\dfrac{-14}{13}} & \textcolor{green}{\dfrac{-11}{13}} & \textcolor{red}{\cancel{\dfrac{5}{13}}} \\[10pt] \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{red}{\cancel{\dfrac{-4}{13}}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} \\[10pt] \textcolor{green}{\dfrac{5}{13}} & \textcolor{green}{\dfrac{3}{13}} & \textcolor{red}{\cancel{\dfrac{1}{13}}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-14}{13} & \dfrac{-11}{13} \\[10pt] \dfrac{5}{13} & \dfrac{3}{13} \end{vmatrix}}
{= \dfrac{-14}{13} × \dfrac{3}{13} - \dfrac{-11}{13} × \dfrac{5}{13}}
{= \dfrac{-42}{169} + \dfrac{55}{169}}
{= \dfrac{13}{169}}
{= \dfrac{1}{13}}
{c_{31} = \dfrac{5}{13}}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{\dfrac{-14}{13}}} & \textcolor{green}{\dfrac{-11}{13}} & \textcolor{green}{\dfrac{5}{13}} \\[10pt] \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{green}{\dfrac{-4}{13}} & \textcolor{green}{\dfrac{3}{13}} \\[10pt] \textcolor{red}{\cancel{\dfrac{5}{13}}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} & \textcolor{red}{\cancel{\dfrac{1}{13}}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-11}{13} & \dfrac{5}{13} \\[10pt] \dfrac{-4}{13} & \dfrac{3}{13} \end{vmatrix}}
{= \dfrac{-11}{13} × \dfrac{3}{13} - \dfrac{5}{13} × \dfrac{-4}{13}}
{= \dfrac{-33}{169} + \dfrac{20}{169}}
{= \dfrac{-13}{169}}
{= \dfrac{-1}{13}}
{c_{32} = \dfrac{3}{13}}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{\dfrac{-14}{13}} & \textcolor{red}{\cancel{\dfrac{-11}{13}}} & \textcolor{green}{\dfrac{5}{13}} \\[10pt] \textcolor{green}{\dfrac{-11}{13}} & \textcolor{red}{\cancel{\dfrac{-4}{13}}} & \textcolor{green}{\dfrac{3}{13}} \\[10pt] \textcolor{red}{\cancel{\dfrac{5}{13}}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} & \textcolor{red}{\cancel{\dfrac{1}{13}}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-14}{13} & \dfrac{5}{13} \\[10pt] \dfrac{-11}{13} & \dfrac{3}{13} \end{vmatrix}}
{= \dfrac{-14}{13} × \dfrac{3}{13} - \dfrac{5}{13} × \dfrac{-11}{13}}
{= \dfrac{-42}{169} + \dfrac{55}{169}}
{= \dfrac{13}{169}}
{= \dfrac{1}{13}}
{c_{33} = \dfrac{1}{13}}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{\dfrac{-14}{13}} & \textcolor{green}{\dfrac{-11}{13}} & \textcolor{red}{\cancel{\dfrac{5}{13}}} \\[10pt] \textcolor{green}{\dfrac{-11}{13}} & \textcolor{green}{\dfrac{-4}{13}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} \\[10pt] \textcolor{red}{\cancel{\dfrac{5}{13}}} & \textcolor{red}{\cancel{\dfrac{3}{13}}} & \textcolor{red}{\cancel{\dfrac{1}{13}}} \end{vmatrix}}
{= \begin{vmatrix} \dfrac{-14}{13} & \dfrac{-11}{13} \\[10pt] \dfrac{-11}{13} & \dfrac{-4}{13} \end{vmatrix}}
{= \dfrac{-14}{13} × \dfrac{-4}{13} - \dfrac{-11}{13} × \dfrac{-11}{13}}
{= \dfrac{1}{56} - \dfrac{121}{13}}
{= \dfrac{-65}{13}}
{= \dfrac{-5}{13}}
Element
{c_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{C}_{ij}}
{c_{11} = \dfrac{-14}{13}}
{\text{M}_{11} = \dfrac{-1}{13}}
{\text{C}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × \dfrac{-1}{13} = \dfrac{-1}{13}}
{c_{12} = \dfrac{-11}{13}}
{\text{M}_{12} = \dfrac{-2}{13}}
{\text{C}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × \dfrac{-2}{13} = \dfrac{2}{13}}
{c_{13} = \dfrac{5}{13}}
{\text{M}_{13} = \dfrac{-1}{13}}
{\text{C}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × \dfrac{-1}{13} = \dfrac{-1}{13}}
{c_{21} = \dfrac{-11}{13}}
{\text{M}_{21} = \dfrac{-2}{13}}
{\text{C}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × \dfrac{-2}{13} = \dfrac{2}{13}}
{c_{22} = \dfrac{-4}{13}}
{\text{M}_{22} = \dfrac{-3}{13}}
{\text{C}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × \dfrac{-3}{13} = \dfrac{-3}{13}}
{c_{23} = \dfrac{3}{13}}
{\text{M}_{23} = \dfrac{1}{13}}
{\text{C}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × \dfrac{1}{13} = \dfrac{-1}{13}}
{c_{31} = \dfrac{5}{13}}
{\text{M}_{31} = \dfrac{-1}{13}}
{\text{C}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × \dfrac{-1}{13} = \dfrac{-1}{13}}
{c_{32} = \dfrac{3}{13}}
{\text{M}_{32} = \dfrac{1}{13}}
{\text{C}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × \dfrac{1}{13} = \dfrac{-1}{13}}
{c_{33} = \dfrac{1}{13}}
{\text{M}_{33} = \dfrac{-5}{13}}
{\text{C}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × \dfrac{-5}{13} = \dfrac{-5}{13}}
{= \begin{bmatrix} \text{C}_{11} & \text{C}_{21} & \text{C}_{31} \\[5pt] \text{C}_{12} & \text{C}_{22} & \text{C}_{32} \\[5pt] \text{C}_{13} & \text{C}_{23} & \text{C}_{33} \end{bmatrix}}
{= \begin{bmatrix} \dfrac{-1}{13} & \dfrac{2}{13} & \dfrac{-1}{13} \\[10pt] \dfrac{2}{13} & \dfrac{-3}{13} & \dfrac{-1}{13} \\[10pt] \dfrac{-1}{13} & \dfrac{-1}{13} & \dfrac{-5}{13} \end{bmatrix}}
{= \dfrac{1}{13}\begin{bmatrix} -1 & 2 & -1 \\[5pt] 2 & -3 & -1 \\[5pt] -1 & -1 & -5 \end{bmatrix}}
As we assumed that {\text{C = A}^{-1},} we have,
{adj \text{ A}^{-1}= \begin{bmatrix} \dfrac{-1}{13} & \dfrac{2}{13} & \dfrac{-1}{13} \\[10pt] \dfrac{2}{13} & \dfrac{-3}{13} & \dfrac{-1}{13} \\[10pt] \dfrac{-1}{13} & \dfrac{-1}{13} & \dfrac{-5}{13} \end{bmatrix}} ———-❼
∴ from ❻ and ❼, we have, {\left[adj \text{ A}\right]^{-1} = adj \left(\text{A}^{-1}\right)}
8.ii To prove that {\left(\text{A}^{-1}\right)^{-1} = \text{A}}
We’ve assumed in step III above that {\text{C = A}^{-1},} we have,
So, {\text{C} = \begin{bmatrix} \dfrac{-14}{13} & \dfrac{-11}{13} & \dfrac{5}{13} \\[10pt] \dfrac{-11}{13} & \dfrac{-4}{13} & \dfrac{3}{13} \\[10pt] \dfrac{5}{13} & \dfrac{3}{13} & \dfrac{1}{13} \end{bmatrix}}
{\left(\text{A}^{-1}\right)^{-1} = \text{C}^{-1}}
Now,
{\text{|C|} = \begin{vmatrix} \dfrac{-14}{13} & \dfrac{-11}{13} & \dfrac{5}{13} \\[10pt] \dfrac{-11}{13} & \dfrac{-4}{13} & \dfrac{3}{13} \\[10pt] \dfrac{5}{13} & \dfrac{3}{13} & \dfrac{1}{13} \end{vmatrix}}
Taking {\dfrac{1}{13}} common from {\text{R}_1,} {\text{R}_2} and {\text{R}_3,} we have,
{\text{|C|} = \dfrac{1}{13} × \dfrac{1}{13} × \dfrac{1}{13} × \begin{vmatrix} -14 & -11 & 5 \\[5pt] -11 & -4 & 3 \\[5pt] 5 & 3 & 1 \end{vmatrix}}
Expanding along {\text{R}_1,} we have,
|C|
{= \dfrac{1}{13} × \dfrac{1}{13} × \dfrac{1}{13} × \left(-14 \begin{vmatrix} -4 & 3 \\[5pt] 3 & 1 \end{vmatrix} - (-11) \begin{vmatrix} -11 & 3 \\[5pt] 5 & 1 \end{vmatrix} + 5 \begin{vmatrix} -11 & -4 \\[5pt] 5 & 3 \end{vmatrix}\right)}
{= \dfrac{1}{13} × \dfrac{1}{13} × \dfrac{1}{13} × \left[(-14)\big((-4)1 - 3(3)\big) - (-11)\big((-11)1 - 3(5)\big) + 5\big((-11)3 - (-4)5\big)\right]}
{= \dfrac{1}{13} × \dfrac{1}{13} × \dfrac{1}{13} × \left[(-14)(-4 - 9) - (-11)(-11 - 15) + 5(-33 + 20)\right]}
{= \dfrac{1}{13} × \dfrac{1}{13} × \dfrac{1}{13} × \left[(-14)(-13) - (-11)(-26) + 5(-13)\right]}
Taking 13 common,
|C|
{= \dfrac{1}{13} × \dfrac{1}{13} × \dfrac{1}{13} × 13 × (14 - 22 - 5)}
{= \dfrac{1}{13} × \dfrac{1}{13} × (-13)}
{= -\dfrac{1}{13}} ———-❽
Now,
{\text{C}^{-1}}
{= \dfrac{1}{\left(\dfrac{-1}{13}\right)} × \dfrac{1}{13}\begin{bmatrix} -1 & 2 & -1 \\[5pt] 2 & -3 & -1 \\[5pt] -1 & -1 & -5 \end{bmatrix}} (from ❼ and ❽)
{= -13 × \dfrac{1}{13}\begin{bmatrix} -1 & 2 & -1 \\[5pt] 2 & -3 & -1 \\[5pt] -1 & -1 & -5 \end{bmatrix}}
{= -\begin{bmatrix} -1 & 2 & -1 \\[5pt] 2 & -3 & -1 \\[5pt] -1 & -1 & -5 \end{bmatrix}}
{= \begin{bmatrix} 1 & -2 & 1 \\[5pt] -2 & 3 & 1 \\[5pt] 1 & 1 & 5 \end{bmatrix}}
= A
Thus we’ve proved that {\text{C}^{-1} = \text{A}}
{\left(\text{A}^{-1}\right)^{-1} = \text{A}}
9. Evaluate {\begin{vmatrix} x & y & x + y \\[5pt] y & x + y & x \\[5pt] x + y & x & y \end{vmatrix}}
Let {Δ = \begin{vmatrix} x & y & x + y \\[5pt] y & x + y & x \\[5pt] x + y & x & y \end{vmatrix}}
Applying {\text{R}_1 → \text{R}_1 + \text{R}_2 + \text{R}_3,} we have,
Δ
{= \begin{vmatrix} x + y + (x + y) & y + (x + y) + y & x + y + x + y \\[5pt] y & x + y & x \\[5pt] x + y & x & y \end{vmatrix}}
{= \begin{vmatrix} 2(x + y) & 2(x + y) & 2(x + y) \\[5pt] y & x + y & x \\[5pt] x + y & x & y \end{vmatrix}}
Taking common factor {2(x + y)} from {\text{R}_1,} we have,
{Δ = 2(x + y) \begin{vmatrix} 1 & 1 & 1 \\[5pt] y & x + y & x \\[5pt] x + y & x & y \end{vmatrix}}
Applying {\text{C}_3 → \text{C}_3 - \text{C}_2,} we have,
Δ
{= 2(x + y) \begin{vmatrix} 1 & 1 & 1 - 1 \\[5pt] y & x + y & x - (x + y) \\[5pt] x + y & x & y - x \end{vmatrix}}
{= 2(x + y) \begin{vmatrix} 1 & 1 & 0 \\[5pt] y & x + y & -y \\[5pt] x + y & x & y - x \end{vmatrix}}
Applying {\text{C}_2 → \text{C}_2 - \text{C}_1,} we have,
Δ
{= 2(x + y) \begin{vmatrix} 1 & 1 - 1 & 0 \\[5pt] y & x + y - y & -y \\[5pt] x + y & x - (x + y) & y - x \end{vmatrix}}
{= 2(x + y) \begin{vmatrix} 1 & 0 & 0 \\[5pt] y & x & -y \\[5pt] x + y & -y & y - x \end{vmatrix}}
Expanding along {\text{R}_1,} we have,
Δ
{= 2(x + y) × \left(1 \begin{vmatrix} x & -y \\[5pt] -y & y - x \end{vmatrix} - 0 \begin{vmatrix} y & -y \\[5pt] x + y & y - x \end{vmatrix} + 0 \begin{vmatrix} y & x \\[5pt] x + y & -y \end{vmatrix}\right)}
{= 2(x + y)\left[1\big(x(y - x) - (-y)(-y)\big) - 0 + 0\right]}
{= 2(x + y)(xy -x^2 - y^2)}
{= -2(x + y)(x^2 - xy + y^2)}
{= -2(x^3 + y^3)}
{\begin{vmatrix} x & y & x + y \\[5pt] y & x + y & x \\[5pt] x + y & x & y \end{vmatrix} = -2(x^3 + y^3)}
10. Evaluate {\begin{vmatrix} 1 & x & y \\[5pt] 1 & x + y & y \\[5pt] 1 & x & x + y \end{vmatrix}}
Let {Δ = \begin{vmatrix} 1 & x & y \\[5pt] 1 & x + y & y \\[5pt] 1 & x & x + y \end{vmatrix}}
Applying row transformations {\text{R}_2 → \text{R}_2 - \text{R}_1} and {\text{R}_3 → \text{R}_3 - \text{R}_1,} we have,
Δ
{= \begin{vmatrix} 1 & x & y \\[5pt] 1 - 1 & x + y - x & y - y \\[5pt] 1 - 1 & x - x & x + y - y \end{vmatrix}}
{= \begin{vmatrix} 1 & x & y \\[5pt] 0 & y & 0 \\[5pt] 0 & 0 & x \end{vmatrix}}
Expanding along {\text{C}_1,} we have,
Δ
{= 1 \begin{vmatrix} y & 0 \\[5pt] 0 & x \end{vmatrix} - 0 \begin{vmatrix} x & y \\[5pt] 0 & x \end{vmatrix} + 0 \begin{vmatrix} x & y \\[5pt] y & 0 \end{vmatrix}}
{= 1\big[y(x) - 0(0)\big] - 0 + 0}
{= xy}
{\begin{vmatrix} 1 & x & y \\[5pt] 1 & x + y & y \\[5pt] 1 & x & x + y \end{vmatrix} = xy}
Using properties of determinants in Exercises 11 to 15, prove that:
11. {\begin{vmatrix} \text{α} & \text{α}^2 & \text{β + γ} \\[5pt] \text{β} & \text{β}^2 & \text{γ + α} \\[5pt] \text{γ} & \text{γ}^2 & \text{α + β} \end{vmatrix} = \text{(β - γ)(γ - α)(α - β)(α + β + γ)}}
Let {Δ = \begin{vmatrix} \text{α} & \text{α}^2 & \text{β + γ} \\[5pt] \text{β} & \text{β}^2 & \text{γ + α} \\[5pt] \text{γ} & \text{γ}^2 & \text{α + β} \end{vmatrix}}
Applying column transformation {\text{C}_3 → \text{C}_3 + \text{C}_1,} we have,
{Δ = \begin{vmatrix} \text{α} & \text{α}^2 & \text{β + γ + α} \\[5pt] \text{β} & \text{β}^2 & \text{γ + α + β} \\[5pt] \text{γ} & \text{γ}^2 & \text{α + β + γ} \end{vmatrix}}
Taking common factor {\text{(α + β + γ)}} from {\text{C}_3,} we have,
Let {Δ = \text{(α + β + γ)}\begin{vmatrix} \text{α} & \text{α}^2 & 1 \\[5pt] \text{β} & \text{β}^2 & 1 \\[5pt] \text{γ} & \text{γ}^2 & 1 \end{vmatrix}}
Applying row transformations, {\text{R}_1 → \text{R}_1 - \text{R}_2} and {\text{R}_2 → \text{R}_2 - \text{R}_3,} we have,
Δ
{= \text{(α + β + γ)}\begin{vmatrix} \text{α - β} & \text{α}^2 - \text{β}^2 & 1 - 1 \\[5pt] \text{β - γ} & \text{β}^2 - \text{γ}^2 & 1 - 1 \\[5pt] \text{γ} & \text{γ}^2 & 1 \end{vmatrix}}
{= \text{(α + β + γ)}\begin{vmatrix} \text{α - β} & \text{(α - β)(α + β)} & 0 \\[5pt] \text{β - γ} & \text{(β - γ)(β + γ)} & 0 \\[5pt] \text{γ} & \text{γ}^2 & 1 \end{vmatrix}}
Taking out common factos {\text{(α - β)}} from {\text{R}_1} and {\text{β - γ}} from {\text{R}_2,} we have,
{Δ = \text{(α + β + γ)(α - β)(β - γ)} \begin{vmatrix} 1 & \text{α + β} & 0 \\[5pt] 1 & \text{β + γ} & 0 \\[5pt] \text{γ} & \text{γ}^2 & 1 \end{vmatrix}} ———-❶
Expanding {\begin{vmatrix} 1 & \text{α + β} & 0 \\[5pt] 1 & \text{β + γ} & 0 \\[5pt] \text{γ} & \text{γ}^2 & 1 \end{vmatrix}} along {\text{C}_3,} we have,
{\begin{vmatrix} 1 & \text{β + γ} & 0 \\[5pt] \text{α + β} & 1 \\[5pt] \text{γ} & \text{γ}^2 & 1 \end{vmatrix}}
{= 0 \begin{vmatrix} 1 & \text{β + γ} \\[5pt] \text{γ} & \text{γ}^2 \end{vmatrix} - 0 \begin{vmatrix} 1 & \text{α + β} \\[5pt] \text{γ} & \text{γ}^2 \end{vmatrix} + 1 \begin{vmatrix} 1 & \text{α + β} \\[5pt] 1 & \text{β + γ} \end{vmatrix}}
{= 0 - 0 + 1[1\text{(β + γ) - (α + β)}1]}
{= 1[\text{(β + γ) - (α + β)}]}
{= \text{γ - α}} ———-❷
Substituting ❷ in ❶, we have,
{Δ = \text{(α + β + γ)(α - β)(β - γ)(γ - α)}}
∴ It is proved that {\begin{vmatrix} \text{α} & \text{α}^2 & \text{β + γ} \\[5pt] \text{β} & \text{β}^2 & \text{γ + α} \\[5pt] \text{γ} & \text{γ}^2 & \text{α + β} \end{vmatrix} = \text{(β - γ)(γ - α)(α - β)(α + β + γ)}}
Using properties of determinants in Exercises 11 to 15, prove that:
{12. \begin{vmatrix} x & x^2 & 1 + px^3 \\[5pt] y & y^2 & 1 + py^3 \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix} = (1 + pxyz)(x - y)(y - z)(z - x),} where p is any scalar.
Let {Δ = \begin{vmatrix} x & x^2 & 1 + px^3 \\[5pt] y & y^2 & 1 + py^3 \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
Applying the row transformations, {\text{R}_1 → \text{R}_1 - \text{R}_2,} we have,
Δ
{= \begin{vmatrix} x - y & x^2 - y^2 & (1 + px^3) - (1 + py^3) \\[5pt] y & y^2 & 1 + py^3 \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= \begin{vmatrix} x - y & x^2 - y^2 & p(x^3 - y^3) \\[5pt] y & y^2 & 1 + py^3 \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
Applying the row transformations {\text{R}_2 → \text{R}_2 - \text{R}_3,} we have,
Δ
{= \begin{vmatrix} x - y & x^2 - y^2 & p(x^3 - y^3) \\[5pt] y - z & y^2 - z^2 & (1 + py^3) - (1 + pz^3) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= \begin{vmatrix} x - y & x^2 - y^2 & p(x^3 - y^3) \\[5pt] y - z & y^2 - z^2 & p(y^3 - z^3) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= \begin{vmatrix} x - y & (x - y)(x + y) & p(x - y)(x^2 + xy + y^2) \\[5pt] y - z & (y - z)(y + z) & p(y - z)(y^2 + yz + z^2) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
Taking common factor {(x - y)} from {\text{R}_1} and {(y - z)} from {\text{R}_2,} we have,
{Δ = (x - y)(y - z)\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 1 & y + z & p(y^2 + yz + z^2) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
Applying row transformation {\text{R}_2 → \text{R}_2 - \text{R}_1,} we have,
Δ
{= (x - y)(y - z)\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 1 - 1 & (y + z) - (x + y) & p(y^2 + yz + z^2) - p(x^2 + xy + y^2) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= (x - y)(y - z)\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 0 & z - x & p(y^2 + yz + z^2 - x^2 - xy - y^2) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= (x - y)(y - z)\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 0 & z - x & p(yz + z^2 - x^2 - xy) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= (x - y)(y - z)\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 0 & z - x & p(yz - xy + z^2 - x^2) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= (x - y)(y - z)\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 0 & z - x & p[y(z - x) + (z + x)(z - x)] \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= (x - y)(y - z)\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 0 & z - x & p(z - x)(y + z + x) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
Taking {(z - x)} common from {\text{R}_2,} we have,
{Δ = (x - y)(y - z)(z - x)\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 0 & 1 & p(y + z + x) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}} ———-❶
Expanding {\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 0 & 1 & p(y + z + x) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}} along {\text{C}_1,} we have,
{\begin{vmatrix} 1 & x + y & p(x^2 + xy + y^2) \\[5pt] 0 & 1 & p(y + z + x) \\[5pt] z & z^2 & 1 + pz^3 \end{vmatrix}}
{= 1 \begin{vmatrix} 1 & p(x + y + z) \\[5pt] z^2 & 1 + pz^3 \end{vmatrix}- 0 \begin{vmatrix} x + y & p(x^2 + xy + y^2) \\[5pt] z^2 & 1 + pz^3 \end{vmatrix}+ z \begin{vmatrix} x + y & p(x^2 + xy + y^2) \\[5pt] 1 & p(x + y + z) \end{vmatrix}}
{= 1\left[1(1 + pz^3) - p(x + y + z)(z^2)\right] - 0 + z\left[(x + y)p(x + y + z) - p(x^2 + xy + y^2)(1)\right]}
{= 1 + pz^3 - pxz^2 - pyz^2 - pz^3 + pz(x^2 + 2xy + y^2 + xz + yz - x^2 - xy - y^2)}
{= 1 + pz^3 - pxz^2 - pyz^2 - pz^3 + px^2z + 2pxyz + py^2z + pxz^2 + pyz^2 - px^2z - pxyz - py^2z)}
{= 1 + pxyz)} ———-❷
Substituting ❷ in ❶, we have,
{Δ = (x - y)(y - z)(z - x)(1 + pxyz)} ———-❶
Using properties of determinants in Exercises 11 to 15, prove that:
{13. \begin{vmatrix} 3a & -a + b & 1 + -a + c \\[5pt] -b + a & 3b & -b + c \\[5pt] -c + a & -c + b & 3c \end{vmatrix} = 3(a + b + c)(ab + bc + ca)}
Let {Δ = \begin{vmatrix} 3a & -a + b & -a + c \\[5pt] -b + a & 3b & -b + c \\[5pt] -c + a & -c + b & 3c \end{vmatrix}}
Applying {\text{C}_1 → \text{C}_1 + \text{C}_2 + \text{C}_3,} we have,
Δ
{= \begin{vmatrix} 3a + (-a + b) + (-a + c) & -a + b & -a + c \\[5pt] (-b + a) + 3b + (-b + c) & 3b & -b + c \\[5pt] -c + a + (-c + b) + 3c & -c + b & 3c \end{vmatrix}}
{= \begin{vmatrix} a + b + c & -a + b & -a + c \\[5pt] a + b + c & 3b & -b + c \\[5pt] a + b + c & -c + b & 3c \end{vmatrix}}
Taking common factor {(a + b + c)} from {\text{C}_1,} we have,
{Δ = (a + b + c) \begin{vmatrix} 1 & -a + b & -a + c \\[5pt] 1 & 3b & -b + c \\[5pt] 1 & -c + b & 3c \end{vmatrix}}
Applying row transformation {\text{R}_1 → \text{R}_1 - \text{R}_2} and {\text{R}_2 → \text{R}_2 - \text{R}_3,} we have,
Δ
{= (a + b + c) \begin{vmatrix} 1 - 1 & (-a + b) - 3b & (-a + c) - (-b + c) \\[5pt] 1 - 1 & 3b - (-c + b) & (-b + c) - 3c \\[5pt] 1 & -c + b & 3c \end{vmatrix}}
{= (a + b + c) \begin{vmatrix} 0 & -a - 2b & -a + b \\[5pt] 0 & 2b + c & -b - 2c \\[5pt] 1 & -c + b & 3c \end{vmatrix}}———-❶
Expanding {\begin{vmatrix} 0 & -a - 2b & -a + b \\[5pt] 0 & 2b + c & -b - 2c \\[5pt] 1 & -c + b & 3c \end{vmatrix}} along {\text{C}_1,} we have,
Δ
{= 0 \begin{vmatrix} 2b + c & -b - 2c \\[5pt] -c + b & 3c \end{vmatrix} - 0 \begin{vmatrix} -a - 2b & -a + b \\[5pt] -c + b & 3c \end{vmatrix} + 1 \begin{vmatrix} -a - 2b & -a + b \\[5pt] 2b + c & -b - 2c \end{vmatrix}}
{= 0 - 0 + 1 \left[(-a - 2b)(-b - 2c) - (-a + b)(2b + c)\right]}
{= 1 \left[(ab + 2ca + 2b^2 +4bc) - (-2ab -ac + 2b^2 + bc)\right]}
{= (3ab + 3ca + 3bc)}
{= 3(ab + ca + bc)}
{= 3(ab + bc + ca)} ———-❷
Substituting ❷ in ❶, we have,
{Δ = 3(a + b + c)(ab + bc + ca)}
∴ It is provded that {\begin{vmatrix} 3a & -a + b & 1 + -a + c \\[5pt] -b + a & 3b & -b + c \\[5pt] -c + a & -c + b & 3c \end{vmatrix} = 3(a + b + c)(ab + bc + ca)}
Using properties of determinants in Exercises 11 to 15, prove that:
14. {\begin{vmatrix} 1 & 1 + p & 1 + p + q \\[5pt] 2 & 3 + 2p & 4 + 3p + 2q \\[5pt] 3 & 6 + 3p & 10 + 6p + 3q \end{vmatrix} = 1}
Let {Δ = \begin{vmatrix} 1 & 1 + p & 1 + p + q \\[5pt] 2 & 3 + 2p & 4 + 3p + 2q \\[5pt] 3 & 6 + 3p & 10 + 6p + 3q \end{vmatrix}}
Applying row transformations {\text{R}_2 → \text{R}_2 - 2\text{R}_1} and {\text{R}_3 → \text{R}_3 - 3\text{R}_1,} we have,
Δ
{= \begin{vmatrix} 1 & 1 + p & 1 + p + q \\[5pt] 2 - 2(1) & 3 + 2p - 2(1 + p) & 4 + 3p + 2q - 2(1 + p + q) \\[5pt] 3 - 3(1) & 6 + 3p - 3(1 + p) & 10 + 6p + 3q - 3(1 + p + q) \end{vmatrix}}
{= \begin{vmatrix} 1 & 1 + p & 1 + p + q \\[5pt] 0 & 1 & 2 + p \\[5pt] 0 & 3 & 7 + 3p \end{vmatrix}}
Expanding along {\text{C}_1,} we have,
Δ
{= 1 \begin{vmatrix} 1 & 2 + p \\[5pt] 3 & 7 + 3p \end{vmatrix} - 0 \begin{vmatrix} 1 + p & 1 + p + q \\[5pt] 3 & 7 + 3p \end{vmatrix} + 0 \begin{vmatrix} 1 + p & 1 + p + q \\[5pt] 1 & 2 + p \end{vmatrix}}
{= 1 \left[1(7 + 3p) - (2 + p)3\right] - 0 + 0}
{= 1 (7 + 3p - 6 - 3p)}
= 1
∴ It is proved that {\begin{vmatrix} 1 & 1 + p & 1 + p + q \\[5pt] 2 & 3 + 2p & 4 + 3p + 2q \\[5pt] 3 & 6 + 3p & 10 + 6p + 3q \end{vmatrix} = 1}
Using properties of determinants in Exercises 11 to 15, prove that:
15. {\begin{vmatrix} \sin \text{α} & \cos \text{α} & \cos\text{(α + δ)} \\[5pt] \sin \text{β} & \cos \text{β} & \cos\text{(β + δ)} \\[5pt] \sin \text{γ} & \cos \text{γ} & \cos\text{(γ + δ)} \end{vmatrix} = 0}
Let {Δ = \begin{vmatrix} \sin \text{α} & \cos \text{α} & \cos\text{(α + δ)} \\[5pt] \sin \text{β} & \cos \text{β} & \cos \text{(β + δ)} \\[5pt] \sin \text{γ} & \cos \text{γ} & \cos\text{(γ + δ)} \end{vmatrix}}
Applying column transformation {\text{C}_3 → \text{C}_3 + \text{C}_1 . \sin \text{δ}}
Δ
{= \begin{vmatrix} \sin \text{α} & \cos \text{α} & \cos \text{α} \cos \text{δ} - \sin \text{α} \sin \text{δ} + \sin \text{α} \sin \text{δ} \\[5pt] \sin \text{β} & \cos \text{β} & \cos \text{β} \cos \text{δ} - \sin \text{β} \sin \text{δ} + \sin \text{β} \sin \text{δ} \\[5pt] \sin \text{γ} & \cos \text{γ} & \cos \text{γ} \cos \text{δ} - \sin \text{γ} \sin \text{δ} + \sin \text{γ} \sin \text{δ} \end{vmatrix}}
{= \begin{vmatrix} \sin \text{α} & \cos \text{α} & \cos \text{α} \cos \text{δ} \\[5pt] \sin \text{β} & \cos \text{β} & \cos \text{β} \cos \text{δ} \\[5pt] \sin \text{γ} & \cos \text{γ} & \cos \text{γ} \cos \text{δ} \end{vmatrix}}
Taking common factor {\cos \text{γ}} from {\text{C}_3,} we have,
{Δ = \cos \text{δ}\begin{vmatrix} \sin \text{α} & \cos \text{α} & \cos \text{α} \\[5pt] \sin \text{β} & \cos \text{β} & \cos \text{β} \\[5pt] \sin \text{γ} & \cos \text{γ} & \cos \text{γ} \end{vmatrix}}
As {\text{C}_2} and {\text{C}_3} are identical, using properties of determinants, we can say that, Δ = 0
∴ Using the properties of determinants, it is proved that {\begin{vmatrix} \sin \text{α} & \cos \text{α} & \cos\text{(α + δ)} \\[5pt] \sin \text{β} & \cos \text{β} & \cos\text{(β + δ)} \\[5pt] \sin \text{γ} & \cos \text{γ} & \cos\text{(γ + δ)} \end{vmatrix} = 0}
16. Solve the system of equations
{\dfrac2x + \dfrac3y + \dfrac{10}{z} = 4}
{\dfrac4x - \dfrac6y + \dfrac5z = 1}
{\dfrac6x + \dfrac9y - \dfrac{20}{z} = 2}
The system of equations can be written in the form AX = B, as specified below,
{\begin{matrix} {\begin{bmatrix} 2 & 3 & 10 \\[5pt] 4 & -6 & 5 \\[5pt] 6 & 9 & -20 \end{bmatrix}} & {\begin{bmatrix} \dfrac1x \\[10pt] \dfrac1y \\[10pt] \dfrac1z \end{bmatrix}} & = & {\begin{bmatrix} 4 \\[5pt] 1 \\[5pt] 2 \end{bmatrix}} \\[10pt] \text{A} & \text{X} & = & \text{B} \end{matrix}}
Thus, we have,
{\text{A} = \begin{bmatrix} 2 & 3 & 10 \\[5pt] 4 & -6 & 5 \\[5pt] 6 & 9 & -20 \end{bmatrix},} {\text{X} = \begin{bmatrix} \dfrac1x \\[10pt] \dfrac1y \\[10pt] \dfrac1z \end{bmatrix}} and {\text{B} = \begin{bmatrix} 4 \\[5pt] 1 \\[5pt] 2 \end{bmatrix}}
Now, expanding along {\text{R}_1,} we have,
|A|
{= 2 \begin{vmatrix} -6 & 5 \\[5pt] 9 & -20 \end{vmatrix} - 3 \begin{vmatrix} 4 & 5 \\[5pt] 6 & -20 \end{vmatrix} + 10 \begin{vmatrix} 4 & -6 \\[5pt] 6 & 9 \end{vmatrix}}
{= 2\left[(-6)(-20) - (5)(9)\right] - 3\left[4(-20) - 5(6)\right] +10\left[4(9) - (-6)6\right]}
= 2(120 – 45) – 3(-80 – 30) + 10(36 + 36)
= 2(75) – 3(-110) + 10(72)
= 150 + 330 + 720
= 1200 ———-❶
≠ 0
As |A| ≠ 0, the given system of equations is consistent and has a unique solution.
It also implies that A is invertible.
So, {\text{X = A}^{-1}\text{B}}
Now, let’s find {\text{A}^{-1}}
We have
Now, {adj \text{ A}} can be calculated as below:
Let’s first find the Minors of each of the elements of A and then the Cofactors. We can then find {adj \text{ A}} as below:
{adj \text{ A} = \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{11} = 2}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{10}} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{green}{-6} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{green}{9} & \textcolor{green}{-20} \end{vmatrix}}
{= \begin{vmatrix} -6 & 5 \\[5pt] 9 & -20 \end{vmatrix}}
= (-6)(-20) – 5(9)
= 120 – 45
= 75
{a_{12} = 3}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{10}} \\[5pt] \textcolor{green}{4} & \textcolor{red}{\cancel{-6}} & \textcolor{green}{5} \\[5pt] \textcolor{green}{6} & \textcolor{red}{\cancel{9}} & \textcolor{green}{-20} \end{vmatrix}}
{= \begin{vmatrix} 4 & 5 \\[5pt] 6 & -20 \end{vmatrix}}
= 4(-20) – 5(6)
= -80 – 30
= -110
{a_{13} = 10}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{10}} \\[5pt] \textcolor{green}{4} & \textcolor{green}{-6} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{6} & \textcolor{green}{9} & \textcolor{red}{\cancel{-20}} \end{vmatrix}}
{= \begin{vmatrix} 4 & -6 \\[5pt] 6 & 9 \end{vmatrix}}
= 4(9) – (-6)6
= 36 + 36
= 72
{a_{21} = 4}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{3} & \textcolor{green}{10} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-6}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{green}{9} & \textcolor{green}{-20} \end{vmatrix}}
{= \begin{vmatrix} 3 & 10 \\[5pt] 9 & -20 \end{vmatrix}}
= 3(-20) – 10(9)
= -60 – 90
= -150
{a_{22} = -6}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{3}} & \textcolor{green}{10} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-6}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{6} & \textcolor{red}{\cancel{9}} & \textcolor{green}{-20} \end{vmatrix}}
{= \begin{vmatrix} 2 & 10 \\[5pt] 6 & -20 \end{vmatrix}}
= 2(-20) – 10(6)
= -40 – 60
= -100
{a_{23} = 5}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{green}{3} & \textcolor{red}{\cancel{10}} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{red}{\cancel{-6}} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{green}{6} & \textcolor{green}{9} & \textcolor{red}{\cancel{-20}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 3 \\[5pt] 6 & 9 \end{vmatrix}}
= 2(9) – 3(6)
= 18 – 18
= 0
{a_{31} = 6}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{3} & \textcolor{green}{10} \\[5pt] \textcolor{red}{\cancel{4}} & \textcolor{green}{-6} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{red}{\cancel{9}} & \textcolor{red}{\cancel{-20}} \end{vmatrix}}
{= \begin{vmatrix} 3 & 10 \\[5pt] -6 & 5 \end{vmatrix}}
= 3(5) – 10(-6)
= 15 + 60
= 75
{a_{32} = 9}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{3}} & \textcolor{green}{10} \\[5pt] \textcolor{green}{4} & \textcolor{red}{\cancel{-6}} & \textcolor{green}{5} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{red}{\cancel{9}} & \textcolor{red}{\cancel{-20}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 10 \\[5pt] 4 & 5 \end{vmatrix}}
= 2(5) – 10(4)
= 10 – 40
= -30
{a_{33} = -20}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{green}{3} & \textcolor{red}{\cancel{10}} \\[5pt] \textcolor{green}{4} & \textcolor{green}{-6} & \textcolor{red}{\cancel{5}} \\[5pt] \textcolor{red}{\cancel{6}} & \textcolor{red}{\cancel{9}} & \textcolor{red}{\cancel{-20}} \end{vmatrix}}
{= \begin{vmatrix} 2 & 3 \\[5pt] 4 & -6 \end{vmatrix}}
= 2(-6) – 3(4)
= -12 – 12
= -24
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 2}
{\text{M}_{11} = 75}
{\text{A}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × 75 = 75}
{a_{12} = 3}
{\text{M}_{12} = -110}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × -110 = 110}
{a_{13} = 10}
{\text{M}_{13} = 72}
{\text{A}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × 72 = 72}
{a_{21} = 4}
{\text{M}_{21} = -150}
{\text{A}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × -150 = 150}
{a_{22} = -5}
{\text{M}_{22} = -100}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × -100 = -100}
{a_{23} = 5}
{\text{M}_{23} = 0}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 0 = 0}
{a_{31} = 6}
{\text{M}_{31} = 75}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × 75 = 75}
{a_{32} = 9}
{\text{M}_{32} = -30}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × -30 = 30}
{a_{33} = -20}
{\text{M}_{33} = -24}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × -24 = -24}
So,
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} 75 & 150 & 75 \\[5pt] 110 & -100 & 30 \\[5pt] 72 & 0 & -24 \end{bmatrix}} ———-❷
{\text{A}^{-1} =}
=
{\dfrac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\[5pt] 110 & -100 & 30 \\[5pt] 72 & 0 & -24 \end{bmatrix}}
(from ❶ and ❷)
As we know,
{X = \text{A}^{-1}\text{B}}
{⇒ \begin{bmatrix} \dfrac1x \\[10pt] \dfrac1y \\[10pt] \dfrac1z \end{bmatrix}}
{= \dfrac{1}{1200} \begin{bmatrix} 75 & 150 & 75 \\[5pt] 110 & -100 & 30 \\[5pt] 72 & 0 & -24 \end{bmatrix} \begin{bmatrix} 4 \\[5pt] 1 \\[5pt] 2 \end{bmatrix}}
{= \dfrac{1}{1200} \begin{bmatrix} 75(4) + 150(1) + 75(2) \\[5pt] 110(4) + (-100)1 + 30(2) \\[5pt] 72(4) + 0(1) + (-24)2 \end{bmatrix}}
{= \dfrac{1}{1200} \begin{bmatrix} 300 + 150 + 150 \\[5pt] 440 - 100 + 60 \\[5pt] 288 + 0 - 48 \end{bmatrix}}
{= \dfrac{1}{200} \begin{bmatrix} 600 \\[5pt] 400 \\[5pt] 240 \end{bmatrix}}
{= \begin{bmatrix} \dfrac{1}{1200} × 600 \\[10pt] \dfrac{1}{1200} × 400 \\[10pt] \dfrac{1}{1200} × 240 \end{bmatrix}}
{= \begin{bmatrix} \dfrac12 \\[10pt] \dfrac13 \\[10pt] \dfrac15 \end{bmatrix}}
Comparing the corresponding elements, we have,
{\dfrac1x = \dfrac12,} {\dfrac1y = \dfrac13} and {\dfrac1z = \dfrac15}
∴ The solution of given system of linear equations is {x = 2,} {y = 3} and {z = 5}
Choose the correct answer in Exercise 17 to 19.
17. If {a,} {b} and c are in A.P., then the determinant {\begin{vmatrix} x + 2 & x + 3 & x + 2a \\[5pt] x + 3 & x + 4 & x + 2b \\[5pt] x + 4 & x + 5 & x + 2c \end{vmatrix}} is
(A) 0 ✔
(B) 1
(C) x
(D) {2x}
It is given that {a,} b and c are in A.P.
So, let the common difference of A.P. be
{⇒ (b - a) = (c - b) = d} ———-❶
Now, let the given determinant be
{Δ = \begin{vmatrix} x + 2 & x + 3 & x + 2a \\[5pt] x + 3 & x + 4 & x + 2b \\[5pt] x + 4 & x + 5 & x + 2c \end{vmatrix}}
Applying the row transformations {\text{R}_3 → \text{R}_3 - \text{R}_2,} we have,
Δ
{= \begin{vmatrix} x + 2 & x + 3 & x + 2a \\[5pt] x + 3 & x + 4 & x + 2b \\[5pt] x + 4 - (x + 3) & x + 5 - (x + 4) & x + 2c - (x + 2b) \end{vmatrix}}
{= \begin{vmatrix} x + 2 & x + 3 & x + 2a \\[5pt] x + 3 & x + 4 & x + 2b \\[5pt] 1 & 1 & 2(c - b) \end{vmatrix}}
Applying the row transformations {\text{R}_2 → \text{R}_2 - \text{R}_1,} we have,
Δ
{= \begin{vmatrix} x + 2 & x + 3 & x + 2a \\[5pt] x + 3 - (x + 2) & x + 4 - (x + 3) & x + 2b - (x + 2a) \\[5pt] 1 & 1 & 2(c - b) \end{vmatrix}}
{= \begin{vmatrix} x + 2 & x + 3 & x + 2a \\[5pt] 1 & 1 & 2(b - a) \\[5pt] 1 & 1 & 2(c - b) \end{vmatrix}}
{= \begin{vmatrix} x + 2 & x + 3 & x + 2a \\[5pt] 1 & 1 & 2d \\[5pt] 1 & 1 & 2d \end{vmatrix}} (from ❶)
As {\text{R}_2} and {\text{R}_3} are identical, from the properties of determinants, we have
Δ = 0
So, option A is the correct answer.
Choose the correct answer in Exercise 17 to 19.
18. If {x,} {y} and z are nonzero real numbers, then the inverse of matrix {\text{A} = \begin{bmatrix} x & 0 & 0 \\[5pt] 0 & y & 0 \\[5pt] 0 & 0 & z \end{bmatrix}} is
(A) {\begin{bmatrix} x^{-1} & 0 & 0 \\[5pt] 0 & y^{-1} & 0 \\[5pt] 0 & 0 & z^{-1} \end{bmatrix}}
(B) {xyz\begin{bmatrix} x^{-1} & 0 & 0 \\[5pt] 0 & y^{-1} & 0 \\[5pt] 0 & 0 & z^{-1} \end{bmatrix}}
(C) {\dfrac{1}{xyz}\begin{bmatrix} x & 0 & 0 \\[5pt] 0 & y & 0 \\[5pt] 0 & 0 & z \end{bmatrix}}
(D) {\dfrac{1}{xyz}\begin{bmatrix} 1 & 0 & 0 \\[5pt] 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \end{bmatrix}}
Given that
{\text{A} = \begin{bmatrix} x & 0 & 0 \\[5pt] 0 & y & 0 \\[5pt] 0 & 0 & z \end{bmatrix}}
Expanding |A| along {\text{R}_1,} we have,
|A|
{= x \begin{vmatrix} y & 0 \\[5pt] 0 & z \end{vmatrix} - 0 \begin{vmatrix} 0 & 0 \\[5pt] 0 & z \end{vmatrix} + 0 \begin{vmatrix} 0 & y \\[5pt] 0 & 0 \end{vmatrix}}
{= x \left[yz - 0(0)\right] - 0 + 0}
{= xyz} ———-❶
≠ 0
As |A| ≠ 0, A is nonsingular and is invertible.
In otherwords, {\text{A}^{-1}} exists.
As {\text{A}^{-1} = \dfrac{1}{\text{|A|}}(adj \text{ A)},} lets’s first find {adj \text{ A},} by first finding the Minors and then the Cofactors of the elements of A.
Element
{b_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{11} = x}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{x}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{y} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{0} & \textcolor{green}{z} \end{vmatrix}}
{= \begin{vmatrix} y & 0 \\[5pt] 0 & z \end{vmatrix}}
{= yz - 0(0)}
{= yz - 0}
{= yz}
{a_{12} = 0}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{x}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{y}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{0}} & \textcolor{green}{z} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 0 & z \end{vmatrix}}
{= 0(z) - 0(0)}
= 0 – 0
= 0
{a_{13} = 0}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{x}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{y} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{0} & \textcolor{red}{\cancel{z}} \end{vmatrix}}
{= \begin{vmatrix} 0 & y \\[5pt] 0 & 0 \end{vmatrix}}
{= 0(0) - y(0)}
= 0 – 0
= 0
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{x}} & \textcolor{green}{0} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{y}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{0} & \textcolor{green}{z} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 0 & z \end{vmatrix}}
{= 0(z) - 0(0)}
= 0 – 0
= 0
{a_{22} = y}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{x} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{y}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{0}} & \textcolor{green}{z} \end{vmatrix}}
{= \begin{vmatrix} x & 0 \\[5pt] 0 & z \end{vmatrix}}
{= xz - 0(0)}
{= xz - 0}
{= xz}
{a_{23} = 0}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{x} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{y}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{0} & \textcolor{red}{\cancel{z}} \end{vmatrix}}
{= \begin{vmatrix} x & 0 \\[5pt] 0 & 0 \end{vmatrix}}
{= x(0) - 0(0)}
= 0 – 0
= 0
{a_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{x}} & \textcolor{green}{0} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{y} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{z}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] y & 0 \end{vmatrix}}
{= 0(0) - 0(y)}
= 0 – 0
= 0
{a_{32} = 0}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{x} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{y}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{z}} \end{vmatrix}}
{= \begin{vmatrix} x & 0 \\[5pt] 0 & 0 \end{vmatrix}}
{= x(0) - 0(0)}
= 0 – 0
= 0
{a_{33} = z}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{x} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{y} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{z}} \end{vmatrix}}
{= \begin{vmatrix} x & 0 \\[5pt] 0 & y \end{vmatrix}}
{= xy - 0(0)}
{= xy - 0}
{= xy}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = x}
{\text{M}_{11} = yz}
{\text{A}_{11} = (-1)^{1 + 1} × \text{M}_{11} = 1 × yz = yz}
{a_{12} = 0}
{\text{M}_{12} = 0}
{\text{A}_{12} = (-1)^{1 + 2} × \text{M}_{12} = -1 × 0 = 0}
{a_{13} = 0}
{\text{M}_{13} = 0}
{\text{A}_{13} = (-1)^{1 + 3} × \text{M}_{13} = 1 × 0 = 0}
{a_{21} = 0}
{\text{M}_{21} = 0}
{\text{A}_{21} = (-1)^{2 + 1} × \text{M}_{21} = -1 × 0 = 0}
{a_{22} = y}
{\text{M}_{22} = xz}
{\text{A}_{22} = (-1)^{2 + 2} × \text{M}_{22} = 1 × zx = zx}
{a_{23} = 0}
{\text{M}_{23} = 0}
{\text{A}_{23} = (-1)^{2 + 3} × \text{M}_{23} = -1 × 0 = 0}
{a_{31} = 0}
{\text{M}_{31} = 0}
{\text{A}_{31} = (-1)^{3 + 1} × \text{M}_{31} = 1 × 0 = 0}
{a_{32} = 0}
{\text{M}_{32} = 0}
{\text{A}_{32} = (-1)^{3 + 2} × \text{M}_{32} = -1 × 0 = 0}
{a_{33} = y}
{\text{M}_{33} = xy}
{\text{A}_{33} = (-1)^{3 + 3} × \text{M}_{33} = 1 × xy = xy}
So,
{= \begin{bmatrix} \text{A}_{11} & \text{A}_{21} & \text{A}_{31} \\[5pt] \text{A}_{12} & \text{A}_{22} & \text{A}_{32} \\[5pt] \text{A}_{13} & \text{A}_{23} & \text{A}_{33} \end{bmatrix}}
{= \begin{bmatrix} yz & 0 & 0 \\[5pt] 0 & zx & 0 \\[5pt] 0 & 0 & xy \end{bmatrix}} ———-❷
{\text{A}^{-1}}
{= \dfrac{1}{xyz}\begin{bmatrix} yz & 0 & 0 \\[5pt] 0 & xz & 0 \\[5pt] 0 & 0 & xy \end{bmatrix}} (from ❶ and ❷)
{= \begin{bmatrix} \dfrac{1}{xyz} × yz & \dfrac{1}{xyz} × 0 & \dfrac{1}{xyz} × 0 \\[10pt] \dfrac{1}{xyz} × 0 & \dfrac{1}{xyz} × xz & \dfrac{1}{xyz} × 0 \\[10pt] \dfrac{1}{xyz} × 0 & \dfrac{1}{xyz} × 0 & \dfrac{1}{xyz} × xy \end{bmatrix}}
{= \begin{bmatrix} \dfrac{1}{x} & 0 & 0 \\[5pt] 0 & \dfrac{1}{y} & 0 \\[5pt] 0 & 0 & \dfrac{1}{z} \end{bmatrix}}
{= \begin{bmatrix} x^{-1} & 0 & 0 \\[5pt] 0 & y^{-1} & 0 \\[5pt] 0 & 0 & z^{-1} \end{bmatrix}}
option A is the correct answer.
Choose the correct answer in Exercise 17 to 19.
19. Let {\text{A} = \begin{bmatrix} 1 & \sin \text{θ} & 1 \\[5pt] -\sin \text{θ} & 1 & \sin \text{θ} \\[5pt] -1 & -\sin \text{θ} & 1 \end{bmatrix},} where 0 ≤ θ ≤ 2π
(A) Det (A) = 0
(B) Det (A) ∈ (2, ∞)
(C) Det (A) ∈ (2, 4)
(D) Det (A) ∈ [2, 4] ✔
Given that
{\text{A} = \begin{bmatrix} 1 & \sin \text{θ} & 1 \\[5pt] -\sin \text{θ} & 1 & \sin \text{θ} \\[5pt] -1 & -\sin \text{θ} & 1 \end{bmatrix}}
Expanding along {\text{R}_1,} we have,
Det A
{= 1 \begin{vmatrix} 1 & \sin \text{θ} \\[5pt] -\sin \text{θ} & 1 \end{vmatrix} - \sin \text{θ} \begin{vmatrix} -\sin \text{θ} & \sin \text{θ} \\[5pt] -1 & 1 \end{vmatrix} + 1 \begin{vmatrix} -\sin \text{θ} & 1 \\[5pt] -1 & -\sin \text{θ} \end{vmatrix}}
{= 1 \left[1(1) - \sin \text{θ}(-\sin \text{θ})\right]\ - \sin \text{θ} \left[(-\sin \text{θ})1 - (\sin \text{θ})(-1)\right] + 1 \left[(-\sin \text{θ})(-\sin \text{θ}) - 1(-1)\right]}
{= 1(1 + \sin^2 \text{θ}) - \sin \text{θ} (-\sin \text{θ} + \sin \text{θ}) + 1(\sin^2 \text{θ} + 1)}
{= 1 + \sin^2 \text{θ} - (\sin \text{θ})0 + \sin^2 \text{θ} + 1}
{= 2 + 2\sin^2 \text{θ}}
As it is given in the problem that {0 ≤ θ ≤ 2π,} so {0 ≤ \sin^2 \text{θ} ≤ 1}
⇒ The range of Det A is {(0 × 2) ≤ 2\sin^2 \text{θ} ≤ (2 × 1)}
⇒ The range of Det A is {0 ≤ 2\sin^2 \text{θ} ≤ 2}
⇒ The range of Det A is {(2 + 0) ≤ 2 + 2\sin^2 \text{θ} ≤ 2 + 2}
⇒ The range of Det A is {2 ≤ 2 + 2\sin^2 \text{θ} ≤ 4}
So, option D is the correct answer.