# Exercise 4.1 Solutions

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Exercise 4.1 Solutions
Evaluate the determinants in Exercises 1 and 2.
1. {\left|\begin{array}{cc} 2 & 4 \\[5pt] -5 & -1 \end{array}\right|}
{\left|\begin{array}{cc} 2 & 4 \\[5pt] -5 & -1 \end{array}\right|}
= (2 × -1) – (4 × -5)
= 2(-1) – 4(-5)
= -2 + 20
= 18
{\left|\begin{array}{cc} 2 & 4 \\[5pt] -5 & -1 \end{array}\right| = } 18
Evaluate the determinants in Exercises 1 and 2.
2. (i) {\left|\begin{array}{cc} \cos θ & -\sin θ \\[5pt] \sin θ & \cos θ \end{array}\right|}
(ii) {\left|\begin{array}{cc} x^2 - x + 1 & x - 1 \\[5pt] x + 1 & x + 1 \end{array}\right|}
(i) To evaluate the determinant of {\left|\begin{array}{cc} \cos θ & -\sin θ \\[5pt] \sin θ & \cos θ \end{array}\right|}
{\left|\begin{array}{cc} \cos θ & -\sin θ \\[5pt] \sin θ & \cos θ \end{array}\right|}
{= (\cos θ) × (\cos θ) - (-\sin θ) × (\sin θ)}
{= \cos^2 θ + \sin^2 θ}
= 1
{\left|\begin{array}{cc} \cos θ & -\sin θ \\[5pt] \sin θ & \cos θ \end{array}\right| =} 1
(ii) To evaluate the determinant of {\left|\begin{array}{cc} x^2 - x + 1 & x - 1 \\[5pt] x + 1 & x + 1 \end{array}\right|}
{\left|\begin{array}{cc} x^2 - x + 1 & x - 1 \\[5pt] x + 1 & x + 1 \end{array}\right|}
{= (x^2 - x + 1)(x + 1) - (x - 1)(x + 1)}
{= (x^3 + x^2 - x^2 - x + x + 1) - (x^2 - 1)}
{= x^3 + 1 - x^2 + 1}
{= x^3 - x^2 + 2}
{\left|\begin{array}{cc} x^2 - x + 1 & x - 1 \\[5pt] x + 1 & x + 1 \end{array}\right| =} {x^3 - x^2 + 2}
3. If {\text{A} = \left|\begin{array}{cc} 1 & 2 \\[5pt] 4 & 2 \end{array}\right|,} then show that |2A| = 4|A|
Given that
A
{= \left[\begin{array}{cc} 1 & 2 \\[5pt] 4 & 2 \end{array}\right]}
⇒ 2A
{= 2 × \left[\begin{array}{cc} 1 & 2 \\[5pt] 4 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 2(1) & 2(2) \\[5pt] 2(4) & 2(2) \end{array}\right]}
{= \left[\begin{array}{cc} 2 & 4 \\[5pt] 8 & 4 \end{array}\right]}
|A|
{= \left|\begin{array}{cc} 1 & 2 \\[5pt] 4 & 2 \end{array}\right|}
= 1(2) – 2(4)
= 2 – 8
= -6 ———-❶
|2A|
{= \left|\begin{array}{cc} 2 & 4 \\[5pt] 8 & 4 \end{array}\right|}
= 2(4) – 8(4)
= 8 – 32
= -24
= 4 × -6
= 4 |A| (from ❶)
∴ It is showed that |2A| = 4|A|
Note: Remember that if A is any square matrix of the order {n,} then {|k\text{A}| = k^n |A|}
4. If {\text{A} = \left[\begin{array}{cc} 1 & 0 & 1 \\[5pt] 0 & 1 & 2 \\[5pt] 0 & 0 & 4 \end{array}\right],} then show that |3A| = 27 |A|
Given that
A
{= \left[\begin{array}{ccc} 1 & 0 & 1 \\[5pt] 0 & 1 & 2 \\[5pt] 0 & 0 & 4 \end{array}\right]}
⇒ 3A
{= 3 × \left[\begin{array}{ccc} 1 & 0 & 1 \\[5pt] 0 & 1 & 2 \\[5pt] 0 & 0 & 4 \end{array}\right]}
{= \left[\begin{array}{ccc} 3(1) & 3(0) & 3(1) \\[5pt] 3(0) & 3(1) & 3(2) \\[5pt] 3(0) & 3(0) & 3(4) \end{array}\right]}
{= \left[\begin{array}{ccc} 3 & 0 & 3 \\[5pt] 0 & 3 & 6 \\[5pt] 0 & 0 & 12 \end{array}\right]}
Expanding along third row,
|A|
{= \left|\begin{array}{cc} 1 & 0 & 0 \\[5pt] 0 & 1 & 2 \\[5pt] 0 & 0 & 4 \end{array}\right|}
{= 0 \left|\begin{array}{cc} 0 & 1 \\[5pt] 1 & 2 \end{array}\right| - 0 \left|\begin{array}{cc} 1 & 1 \\[5pt] 0 & 2 \end{array}\right| + 4 \left|\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right|}
= 0 – 0 + 4[1(1) – 0(0)]
= 4(1 – 0)
= 4(1)
= 4 ———-❶
|3A|
{= \left|\begin{array}{cc} 3 & 0 & 3 \\[5pt] 0 & 3 & 6 \\[5pt] 0 & 0 & 12 \end{array}\right|}
{= 0 \left|\begin{array}{cc} 0 & 3 \\[5pt] 3 & 6 \end{array}\right| - 0 \left|\begin{array}{cc} 3 & 3 \\[5pt] 0 & 6 \end{array}\right| + 12 \left|\begin{array}{cc} 3 & 0 \\[5pt] 0 & 3 \end{array}\right|}
= 0 – 0 + 12[3(3) – 0(0)]
= 12(9)
= 108
= 27 × 4
= 27|A| (from ❶)
∴ It is showed that |3A| = 27 |A|
5. Evaluate the determinants
{\text{(i)} \left|\begin{array}{ccc} 3 & -1 & -2 \\[5pt] 0 & 0 & -1 \\[5pt] 3 & -5 & 0 \end{array}\right|}
{\text{(ii)} \left|\begin{array}{ccc} 3 & -4 & 5 \\[5pt] 1 & 1 & -2 \\[5pt] 2 & 3 & 1 \end{array}\right|}
{(iii) \left|\begin{array}{ccc} 0 & 1 & 2 \\[5pt] -1 & 0 & -3 \\[5pt] -2 & 3 & 0 \end{array}\right|}
{(iv) \left|\begin{array}{ccc} 2 & -1 & -2 \\[5pt] 0 & 2 & -1 \\[5pt] 3 & -5 & 0 \end{array}\right|}
(i) To evaluate the determinant of {\left|\begin{array}{ccc} 3 & -1 & -2 \\[5pt] 0 & 0 & -1 \\[5pt] 3 & -5 & 0 \end{array}\right|}
Expanding along second row (as it has two zeros)
Δ
{= \left|\begin{array}{ccc} 3 & -1 & -2 \\[5pt] 0 & 0 & -1 \\[5pt] 3 & -5 & 0 \end{array}\right|}
{= - 0 \left|\begin{array}{cc} -1 & -2 \\[5pt] -5 & 0 \end{array}\right| + 0 \left|\begin{array}{cc} 3 & -2 \\[5pt] 3 & 0 \end{array}\right| - (-1) \left|\begin{array}{cc} 3 & -1 \\[5pt] 3 & -5 \end{array}\right|}
= – 0 + 0 + 1[3(-5) – (-1)3]
= -15 + 3
= -12
{\left|\begin{array}{ccc} 3 & -1 & -2 \\[5pt] 0 & 0 & -1 \\[5pt] 3 & -5 & 0 \end{array}\right| = -12}
(ii) To evaluate the determinant of {\left|\begin{array}{ccc} 3 & -4 & 5 \\[5pt] 1 & 1 & -2 \\[5pt] 2 & 3 & 1 \end{array}\right|}
Expanding along second row (multiplication with 1 is faster)
Δ
=
{\left|\begin{array}{ccc} 3 & -4 & 5 \\[5pt] 1 & 1 & -2 \\[5pt] 2 & 3 & 1 \end{array}\right|}
=
{- 1 \left|\begin{array}{cc} -5 & 5 \\[5pt] 3 & 1 \end{array}\right| + 1 \left|\begin{array}{cc} 3 & 5 \\[5pt] 2 & 1 \end{array}\right| - (-2) \left|\begin{array}{cc} 3 & -4 \\[5pt] 2 & 3 \end{array}\right|}
=
– 1[(-4)1 – 5(3)] + 1[3(1) – 5(2)] – (-2)[3(3) – (-4)2]
=
(-1)(-4 – 15) + 1(3 – 10) + 2(9 + 8)
=
(-1)(-19) + (1)(-7) + (2)(17)
=
19 – 7 + 34
=
46
{\left|\begin{array}{ccc} 3 & -4 & 5 \\[5pt] 1 & 1 & -2 \\[5pt] 2 & 3 & 1 \end{array}\right| = 46}
(iii) To evaluate the determinant of {\left|\begin{array}{ccc} 0 & 1 & 2 \\[5pt] -1 & 0 & -3 \\[5pt] -2 & 3 & 0 \end{array}\right|}
Expanding along first row
Δ
{= \left|\begin{array}{ccc} 0 & 1 & 2 \\[5pt] -1 & 0 & -3 \\[5pt] -2 & 3 & 0 \end{array}\right|}
{= 0 \left|\begin{array}{cc} 0 & -3 \\[5pt] 3 & 0 \end{array}\right| - 1 \left|\begin{array}{cc} -1 & -3 \\[5pt] -2 & 0 \end{array}\right| + 2 \left|\begin{array}{cc} -1 & 0 \\[5pt] -2 & 3 \end{array}\right|}
= 0 – 1[(-1)0 – (-3)(-2)] + 2[(-1)3 + 0(-2)]
= (-1)(-6) + 2(-3)
= 6 – 6
= 0
{\left|\begin{array}{ccc} 0 & 1 & 2 \\[5pt] -1 & 0 & -3 \\[5pt] -2 & 3 & 0 \end{array}\right| = 0}
(vi) To evaluate the determinant of {\left|\begin{array}{ccc} 2 & -1 & -2 \\[5pt] 0 & 2 & -1 \\[5pt] 3 & -5 & 0 \end{array}\right|}
Expanding along first row
Δ
{= \left|\begin{array}{ccc} 2 & -1 & -2 \\[5pt] 0 & 2 & -1 \\[5pt] 3 & -5 & 0 \end{array}\right|}
{= 2 \left|\begin{array}{cc} 2 & -1 \\[5pt] -5 & 0 \end{array}\right| - 0 \left|\begin{array}{cc} -1 & -2 \\[5pt] -5 & 0 \end{array}\right| + 3 \left|\begin{array}{cc} -1 & -2 \\[5pt] 2 & -1 \end{array}\right|}
= 2[2(0) – (-1)(-5)] – 0 + 3[(-1)(-1) – (-2)2]
= 2(0 -5) + 3(1 + 4)
= 2(-5) + 3(5)
= -10 + 15
= 5
{\left|\begin{array}{ccc} 2 & -1 & -2 \\[5pt] 0 & 2 & -1 \\[5pt] 3 & -5 & 0 \end{array}\right| = 5}
6. If {\text{A} = \left[\begin{array}{ccc} 1 & 1 & -2 \\[5pt] 2 & 1 & -3 \\[5pt] 4 & 4 & -9 \end{array}\right],} find |A|
Given that
A
{= \left[\begin{array}{ccc} 1 & 1 & -2 \\[5pt] 2 & 1 & -3 \\[5pt] 4 & 4 & -9 \end{array}\right]}
∴ |A|
{= \left|\begin{array}{ccc} 1 & 1 & -2 \\[5pt] 2 & 1 & -3 \\[5pt] 4 & 4 & -9 \end{array}\right|}
Expanding along row one,
|A|
=
{1 \left|\begin{array}{cc} 1 & -3 \\[5pt] 4 & -9 \end{array}\right| - 1 \left|\begin{array}{cc} 2 & -3 \\[5pt] 5 & -9 \end{array}\right|+ (-2) \left|\begin{array}{cc} 2 & 1 \\[5pt] 5 & 4 \end{array}\right|}
=
1[1(-9) – (-3)4)] – 1[2(-9) – (-3)5] + (-2)[2(4) – 1(5)]
=
(-9 + 12) – (-18 + 15) – 2(8 – 5)
=
3 – (-3) – 2(3)
=
3 + 3 – 6
=
0
|A| = 0
Note: A matrix whose determinant is zero is called as a singular matrix. So, as per the definition the given matrix A is a singular matrix.
7. Find values of {x,} if
(i) {\left|\begin{array}{cc} 2 & 4 \\[5pt] 5 & 1 \end{array}\right| = \left|\begin{array}{cc} 2x & 4 \\[5pt] 6 & x \end{array}\right|}
(ii) {\left|\begin{array}{cc} 2 & 3 \\[5pt] 4 & 5 \end{array}\right| = \left|\begin{array}{cc} x & 3 \\[5pt] 2x & 5 \end{array}\right|}
(i) To find the value of {x,} if {\left|\begin{array}{cc} 2 & 4 \\[5pt] 5 & 1 \end{array}\right| = \left|\begin{array}{cc} 2x & 4 \\[5pt] 6 & x \end{array}\right|}
Given that
{\left|\begin{array}{cc} 2 & 4 \\[5pt] 5 & 1 \end{array}\right| = \left|\begin{array}{cc} 2x & 4 \\[5pt] 6 & x \end{array}\right|}
⇒ 2(1) – 4(5) = {(2x × x) - (4 × 6)}
⇒ 2 – 20 = {2x^2 - 24}
{2x^2 - 24} = 2 – 20
{2x^2} = 24 + 2 – 20
{2x^2} = 6
{x^2 = \dfrac62}
{x^2 = 3}
{x = \pm\sqrt{3}}
{x = \pm\sqrt{3}}
(ii) To find the value of {x,} if {\left|\begin{array}{cc} 2 & 3 \\[5pt] 4 & 5 \end{array}\right| = \left|\begin{array}{cc} x & 3 \\[5pt] 2x & 5 \end{array}\right|}
Given that
{\left|\begin{array}{cc} 2 & 3 \\[5pt] 4 & 5 \end{array}\right| = \left|\begin{array}{cc} x & 3 \\[5pt] 2x & 5 \end{array}\right|}
⇒ 2(5) – 3(4) = {x(5) - 3(2x)}
⇒ 10 – 12 = {5x - 6x}
⇒ -2 = {-x}
{x = 2}
{x = 2}
8. If {\left|\begin{array}{cc} x & 2 \\[5pt] 18 & x \end{array}\right| = \left|\begin{array}{cc} 6 & 2 \\[5pt] 18 & 6 \end{array}\right|,} then x is equal to
(A) 6
(B) ±6 ✔
(C) -6
(D) 0
As we’re dealing with determinants here, we can not say that As the given matrices are equal, therefore, their corresponding elements must be equal.
So, we can not compare the corresponding elements and say that {x = 6}
So, option A is not correct
So, let’s expand the determinants.
Given that
{\left|\begin{array}{cc} x & 2 \\[5pt] 18 & x \end{array}\right| = \left|\begin{array}{cc} 6 & 2 \\[5pt] 18 & 6 \end{array}\right|,}
{(x × x) - (2 × 18) = (6 × 6) - (2 × 18)}
{x^2 - 36 = 36 - 36}
{x^2 - 36 = 0}
{x^2 = 36}
{x = \sqrt{36}}
{x = \pm6}
{x = \pm6}
option B is the correct answer.