# Exercise 3.2 Solutions

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Exercise 3.2 Solutions
1. Let {\text{A} = \left[\begin{array}{cc} 2 & 4 \\[5pt] 3 & 2 \end{array}\right],} {\text{B} = \left[\begin{array}{cc} 1 & 3 \\[5pt] -2 & 5 \end{array}\right],} {\text{C} = \left[\begin{array}{cc} -2 & 5 \\[5pt] 3 & 4 \end{array}\right].} Find each of the following:
(i) A + B (ii) A – B (iii) 3A – C (iv) AB (v) BA
(i) To find A + B
A + B
{= \left[\begin{array}{cc} 2 & 4 \\[5pt] 3 & 2 \end{array}\right] + \left[\begin{array}{cc} 1 & 3 \\[5pt] -2 & 5 \end{array}\right]}
{= \left[\begin{array}{cc} 2 + 1 & 4 + 3 \\[5pt] 3 - 2 & 2 + 5 \end{array}\right]}
{= \left[\begin{array}{cc} 3 & 7 \\[5pt] 1 & 7 \end{array}\right]}
(ii) To find A – B
A – B
{= \left[\begin{array}{cc} 2 & 4 \\[5pt] 3 & 2 \end{array}\right] - \left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right]}
{= \left[\begin{array}{cc} 2 - 1 & 4 - 3 \\[5pt] 3 - (-2) & 2 - 5 \end{array}\right]}
{= \left[\begin{array}{cc} 1 & 1 \\[5pt] 5 & -3 \end{array}\right]}
(iii) To find 3A – C
3A – C
{= 3 × \left[\begin{array}{cc} 2 & 4 \\[5pt] 3 & 2 \end{array}\right] - \left[\begin{array}{cc} -2 & 5 \\[5pt] 3 & 4 \end{array}\right]}
{= \left[\begin{array}{cc} 3 × 2 & 3 × 4 \\[5pt] 3 × 3 & 3 × 2 \end{array}\right] - \left[\begin{array}{cc} -2 & 5 \\[5pt] 3 & 4 \end{array}\right]}
{= \left[\begin{array}{cc} 6 & 12 \\[5pt] 9 & 6 \end{array}\right] - \left[\begin{array}{cc} -2 & 5 \\[5pt] 3 & 4 \end{array}\right]}
{= \left[\begin{array}{cc} 6 - (-2) & 12 - 5 \\[5pt] 9 - 3 & 6 - 4 \end{array}\right]}
{= \left[\begin{array}{cc} 8 & 7 \\[5pt] 6 & 2 \end{array}\right]}
(iv) To find AB
The order of the matrix A is 2 × 2 and the order of the matrix B is 2 × 2. Here
No. of columns of A = No. of rows of B
⇒ AB is defined.
Now,
AB
{= \left[\begin{array}{cc} 2 & 4 \\[5pt] 3 & 2 \end{array}\right] \left[\begin{array}{cc} 1 & 3 \\[5pt] -2 & 5 \end{array}\right]}
{= \left[\begin{array}{cc} (2 × 1) + (4 × (-2)) & (2 × 3) + (4 × 5) \\[5pt] (3 × 1) + (2 × (-2)) & (3 × 3) + (2 × 5) \end{array}\right]}
{= \left[\begin{array}{cc} 2 - 8 & 6 + 20 \\[5pt] 3 - 4 & 9 + 10 \end{array}\right]}
{= \left[\begin{array}{cc} -6 & 26 \\[5pt] -1 & 19 \end{array}\right]}
(v) To find BA
The order of the matrix A is 2 × 2 and the order of the matrix B is 2 × 2. Here
No. of columns of B = No. of rows of A
⇒ BA is defined.
Now,
BA
{= \left[\begin{array}{cc} 1 & 3 \\[5pt] -2 & 5 \end{array}\right] \left[\begin{array}{cc} 2 & 4 \\[5pt] 3 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} (1 × 2) + (3 × 3) & (1 × 4) + (3 × 2) \\[5pt] (-2 × 2) + (5 × 3) & (-2 × 4) + (5 × 2) \end{array}\right]}
{= \left[\begin{array}{cc} 2 + 9 & 4 + 6 \\[5pt] -4 + 15 & -8 + 10 \end{array}\right]}
{= \left[\begin{array}{cc} 11 & 10 \\[5pt] 11 & 2 \end{array}\right]}

2. Compute the following:
(i) {\left[\begin{array}{cc} a & b \\[5pt] -b & a \end{array}\right] + \left[\begin{array}{cc} a & b \\[5pt] b & a \end{array}\right]}
(ii) {\left[\begin{array}{cc} a^2 + b^2 & b^2 + c^2 \\[5pt] a^2 + c^2 & a^2 + b^2 \end{array}\right] + \left[\begin{array}{cc} 2ab & 2bc \\[5pt] -2ac & -2ab \end{array}\right]}
(iii) {\left[\begin{array}{ccc} -1 & 4 & -6 \\[5pt] 8 & 5 & 16 \\[5pt] 2 & 8 & 5 \end{array}\right] + \left[\begin{array}{ccc} 12 & 7 & 6 \\[5pt] 8 & 0 & 5 \\[5pt] 3 & 2 & 4 \end{array}\right]}
(iv) {\left[\begin{array}{cc} \cos^2 x & \sin^2 x \\[5pt] \sin^2 x & \cos^2 x \end{array}\right] + \left[\begin{array}{cc} \sin^2 x & \cos^2 x \\[5pt] \cos^2 x & \sin^2 x \end{array}\right]}
(i) To compute {\left[\begin{array}{cc} a & b \\[5pt] -b & a \end{array}\right] + \left[\begin{array}{cc} a & b \\[5pt] b & a \end{array}\right]}
{\left[\begin{array}{cc} a & b \\[5pt] -b & a \end{array}\right] + \left[\begin{array}{cc} a & b \\[5pt] b & a \end{array}\right]}
{= \left[\begin{array}{cc} a + a & b + b \\[5pt] -b + b & a + a \end{array}\right]}
{= \left[\begin{array}{cc} 2a & 2b \\[5pt] 0 & 2a \end{array}\right]}
(ii) To compute {\left[\begin{array}{cc} a^2 + b^2 & b^2 + c^2 \\[5pt] a^2 + c^2 & a^2 + b^2 \end{array}\right] + \left[\begin{array}{cc} 2ab & 2bc \\[5pt] -2ac & -2ab \end{array}\right]}
{= \left[\begin{array}{cc} a^2 + b^2 & b^2 + c^2 \\[5pt] a^2 + c^2 & a^2 + b^2 \end{array}\right] + \left[\begin{array}{cc} 2ab & 2bc \\[5pt] -2ac & -2ab \end{array}\right]}
{= \left[\begin{array}{cc} a^2 + b^2 + 2ab & b^2 + c^2 + 2bc \\[5pt] a^2 + c^2 - 2ac & a^2 + b^2 - 2ab \end{array}\right]}
{= \left[\begin{array}{cc} (a + b)^2 & (b + c)^2 \\[5pt] (a - c)^2 & (a - b)^2 \end{array}\right]}
(iii) To compute {\left[\begin{array}{ccc} -1 & 4 & -6 \\[5pt] 8 & 5 & 16 \\[5pt] 2 & 8 & 5 \end{array}\right] + \left[\begin{array}{ccc} 12 & 7 & 6 \\[5pt] 8 & 0 & 5 \\[5pt] 3 & 2 & 4 \end{array}\right]}
{\left[\begin{array}{ccc} -1 & 4 & -6 \\[5pt] 8 & 5 & 16 \\[5pt] 2 & 8 & 5 \end{array}\right] + \left[\begin{array}{ccc} 12 & 7 & 6 \\[5pt] 8 & 0 & 5 \\[5pt] 3 & 2 & 4 \end{array}\right]}
{= \left[\begin{array}{ccc} -1 + 12 & 4 + 7 & -6 + 6 \\[5pt] 8 + 8 & 5 + 0 & 16 + 5 \\[5pt] 2 + 3 & 8 + 2 & 5 + 4 \end{array}\right]}
{= \left[\begin{array}{ccc} 11 & 11 & 0 \\[5pt] 16 & 5 & 21 \\[5pt] 5 & 10 & 9 \end{array}\right]}
(iv) To compute {\left[\begin{array}{cc} \cos^2 x & \sin^2 x \\[5pt] \sin^2 x & \cos^2 x \end{array}\right] + \left[\begin{array}{cc} \sin^2 x & \cos^2 x \\[5pt] \cos^2 x & \sin^2 x \end{array}\right]}
{\left[\begin{array}{cc} \cos^2 x & \sin^2 x \\[5pt] \sin^2 x & \cos^2 x \end{array}\right] + \left[\begin{array}{cc} \sin^2 x & \cos^2 x \\[5pt] \cos^2 x & \sin^2 x \end{array}\right]}
{= \left[\begin{array}{cc} \cos^2 x + \sin^2 x & \sin^2 x + \cos^2 x \\[5pt] \sin^2 x + \cos^2 x & \cos^2 x + \sin^2 x \end{array}\right]}
{= \left[\begin{array}{cc} 1 & 1 \\[5pt] 1 & 1 \end{array}\right]} {\left(∵ \cos^2 x + \sin^2 x = 1\right)}

3. Compute the indicated products.
(i) {\left[\begin{array}{cc} a & b \\[5pt] -b & a \end{array}\right] \left[\begin{array}{cc} a & -b \\[5pt] b & a \end{array}\right]}
(ii) {\left[\begin{array}{c} 1 \\[5pt] 2 \\[5pt] 3 \end{array}\right] \left[\begin{array}{ccc} 2 & 3 & 4 \end{array}\right]}
(iii) {\left[\begin{array}{cc} 1 & -2 \\[5pt] 2 & 3 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 2 & 3 & 1 \end{array}\right]}
(iv) {\left[\begin{array}{ccc} 2 & 3 & 4 \\[5pt] 3 & 4 & 5 \\[5pt] 4 & 5 & 6 \end{array}\right] \left[\begin{array}{ccc} 1 & -3 & 5 \\[5pt] 0 & 2 & 4 \\[5pt] 3 & 0 & 5 \end{array}\right]}
(v) {\left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 2 \\[5pt] -1 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 1 \\[5pt] -1 & 2 & 1 \end{array}\right]}
(vi) {\left[\begin{array}{ccc} 3 & -1 & 3 \\[5pt] -1 & 0 & 2 \end{array}\right] \left[\begin{array}{cc} 2 & -3 \\[5pt] 1 & 0 \\[5pt] 3 & 1 \end{array}\right]}
(i) To compute {\left[\begin{array}{cc} a & b \\[5pt] -b & a \end{array}\right] \left[\begin{array}{cc} a & -b \\[5pt] b & a \end{array}\right]}
{\left[\begin{array}{cc} a & b \\[5pt] -b & a \end{array}\right] \left[\begin{array}{cc} a & -b \\[5pt] b & a \end{array}\right]}
{= \left[\begin{array}{cc} (a × a) + (b × b) & (a × (-b)) + (b × a) \\[5pt] (-b × a) + (a × b) & ((-b) × (-b)) + (a × a) \end{array}\right]}
{= \left[\begin{array}{cc} a^2 + b^2 & -ab + ab \\[5pt] -ab + ab & b^2 + a^2 \end{array}\right]}
{= \left[\begin{array}{cc} a^2 + b^2 & 0 \\[5pt] 0 & a^2 + b^2 \end{array}\right]}
(ii) To compute {\left[\begin{array}{c} 1 \\[5pt] 2 \\[5pt] 3 \end{array}\right] \left[\begin{array}{ccc} 2 & 3 & 4 \end{array}\right]}
{= \left[\begin{array}{c} 1 \\[5pt] 2 \\[5pt] 3 \end{array}\right] \left[\begin{array}{ccc} 2 & 3 & 4 \end{array}\right]}
{= \left[\begin{array}{ccc} (1 × 2) & (1 × 3) & (1 × 4) \\[5pt] (2 × 2) & (2 × 3) & (2 × 4) \\[5pt] (3 × 2) & (3 × 3) & (3 × 4) \end{array}\right]}
{\left[\begin{array}{cc} 2 & 3 & 4 \\[5pt] 4 & 6 & 8 \\[5pt] 6 & 9 & 12 \end{array}\right]}
(iii) To compute {\left[\begin{array}{cc} 1 & -2 \\[5pt] 2 & 3 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 2 & 3 & 1 \end{array}\right]}
{\left[\begin{array}{cc} 1 & -2 \\[5pt] 2 & 3 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 2 & 3 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} (1 × 1) + (-2 × 2) & (1 × 2) + (-2 × 3) & (1 × 3) + (-2 × 1) \\[5pt] (2 × 1) + (3 × 2) & (2 × 2) + (3 × 3) & (2 × 3) + (3 × 1) \end{array}\right]}
{= \left[\begin{array}{ccc} 1 - 4 & 2 - 6 & 3 - 2 \\[5pt] 2 + 6 & 4 + 9 & 6 + 3 \end{array}\right]}
{= \left[\begin{array}{ccc} -3 & -4 & 1 \\[5pt] 8 & 13 & 9 \end{array}\right]}
(iv) To compute {\left[\begin{array}{ccc} 2 & 3 & 4 \\[5pt] 3 & 4 & 5 \\[5pt] 4 & 5 & 6 \end{array}\right] \left[\begin{array}{ccc} 1 & -3 & 5 \\[5pt] 0 & 2 & 4 \\[5pt] 3 & 0 & 5 \end{array}\right]}
{\left[\begin{array}{ccc} 2 & 3 & 4 \\[5pt] 3 & 4 & 5 \\[5pt] 4 & 5 & 6 \end{array}\right] \left[\begin{array}{ccc} 1 & -3 & 5 \\[5pt] 0 & 2 & 4 \\[5pt] 3 & 0 & 5 \end{array}\right]}
{= \left[\begin{array}{ccc} (2 × 1) + (3 × 0) + (4 × 3) & (2 × (-3)) + (3 × 2) + (4 × 0) & (2 × 5) + (3 × 4) + (4 × 5) \\[5pt] (3 × 1) + (4 × 0) + (5 × 3) & (3 × (-3)) + (4 × 2) + (5 × 0) & (3 × 5) + (4 × 4) + (5 × 5) \\[5pt] (4 × 1) + (5 × 0) + (6 × 3) & (4 × (-3)) + (5 × 2) + (6 × 0) & (4 × 5) + (5 × 4) + (6 × 5) \end{array}\right]}
{= \left[\begin{array}{ccc} 2 + 0 + 12 & -6 + 6 + 0 & 10 + 12 + 20 \\[5pt] 3 + 0 + 15 & -9 + 8 + 0 & 15 + 16 + 25 \\[5pt] 4 + 0 + 18 & -12 + 10 + 0 & 20 + 20 + 30 \end{array}\right]}
{= \left[\begin{array}{ccc} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{array}\right]}
(v) To compute {\left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 2 \\[5pt] -1 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 1 \\[5pt] -1 & 2 & 1 \end{array}\right]}
{\left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 2 \\[5pt] -1 & 1 \end{array}\right] \left[\begin{array}{ccc} 1 & 0 & 1 \\[5pt] -1 & 2 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} (2 × 1) + (1 × (-1)) & (2 × 0) + (1 × 2) & (2 × 1) + (1 × 1) \\[5pt] (3 × 1) + (2 × (-1)) & (3 × 0) + (2 × 2) & (3 × 1) + (2 × 1) \\[5pt] ((-1)) × 1) + (1 × (-1)) & ((-1) × 0) + (1 × 2) & ((-1) × 1) + (1 × 1) \end{array}\right]}
{= \left[\begin{array}{ccc} 2 - 1 & 0 + 2 & 2 + 1 \\[5pt] 3 - 2 & 0 + 4 & 3 + 2 \\[5pt] -1 - 1 & 0 + 2 & -1 + 1 \end{array}\right]}
= {\left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 1 & 4 & 5 \\[5pt] -2 & 2 & 0 \end{array}\right]}

4. If {\text{A} = \left[\begin{array}{ccc} 1 & 2 & -3 \\[5pt] 5 & 0 & 2 \\[5pt] 1 & -1 & 1 \end{array}\right],} {\text{B} = \left[\begin{array}{ccc} 3 & -1 & 2 \\[5pt] 4 & 2 & 5 \\[5pt] 2 & 0 & 3 \end{array}\right]} and {\text{C} = \left[\begin{array}{ccc} 4 & 1 & 2 \\[5pt] 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right],} then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Given that {\text{A} = \left[\begin{array}{ccc} 1 & 2 & -3 \\[5pt] 5 & 0 & 2 \\[5pt] 1 & -1 & 1 \end{array}\right],}
{\text{B} = \left[\begin{array}{ccc} 3 & -1 & 2 \\[5pt] 4 & 2 & 5 \\[5pt] 2 & 0 & 3 \end{array}\right]} and
{\text{C} = \left[\begin{array}{ccc} 4 & 1 & 2 \\[5pt] 0 & 3 & 2 \\[5pt] 1 & -2 & 3 \end{array}\right]}
To compute A + B
A + B
=
{\left[\begin{array}{ccc} 1 & 2 & -3 \\[5pt] 5 & 0 & 2 \\[5pt] 1 & -1 & 1 \end{array}\right] + \left[\begin{array}{ccc} 3 & -1 & 2 \\[5pt] 4 & 2 & 5 \\[5pt] 2 & 0 & 3 \end{array}\right]}
=
{\left[\begin{array}{ccc} 1 + 3 & 2 + (-1) & -3 + 2 \\[5pt] 5 + 4 & 0 + 2 & 2 + 5 \\[5pt] 1 + 2 & -1 + 0 & 1 + 3 \end{array}\right]}
=
{\left[\begin{array}{ccc} 4 & 1 & -1 \\[5pt] 9 & 2 & 7 \\[5pt] 3 & -1 & 4 \end{array}\right]}
To compute B – C
B – C
=
{\left[\begin{array}{ccc} 3 & -1 & 2 \\[5pt] 4 & 2 & 5 \\[5pt] 2 & 0 & 3 \end{array}\right] - \left[\begin{array}{ccc} 4 & 1 & 2 \\[5pt] 0 & 3 & 2 \\[5pt] 1 & -2 & 3 \end{array}\right]}
=
{\left[\begin{array}{ccc} 3 - 4 & -1 - 1 & 2 - 2 \\[5pt] 4 - 0 & 2 - 3 & 5 - 2 \\[5pt] 2 - 1 & 0 - (-2) & 3 - 3 \end{array}\right]}
=
{\left[\begin{array}{ccc} -1 & -2 & 0 \\[5pt] 4 & -1 & 3 \\[5pt] 1 & 2 & 0 \end{array}\right]}
To prove that A + (B – C) = (A + B) – C
L.H.S.
= A + (B – C)
{= \left[\begin{array}{ccc} 1 & 2 & -3 \\[5pt] 5 & 0 & 2 \\[5pt] 1 & -1 & 1 \end{array}\right] + \left[\begin{array}{ccc} -1 & -2 & 0 \\[5pt] 4 & -1 & 3 \\[5pt] 1 & 2 & 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 1 + (-1) & 2 + (-2) & -3 + 0 \\[5pt] 5 + 4 & 0 + (-1) & 2 + 3 \\[5pt] 1 + 1 & -1 + 2 & 1 + 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 0 & 0 & -3 \\[5pt] 9 & -1 & 5 \\[5pt] 2 & 1 & 1 \end{array}\right]}
R.H.S.
= (A + B) – C
{= \left[\begin{array}{ccc} 4 & 1 & -1 \\[5pt] 9 & 2 & 7 \\[5pt] 3 & -1 & 4 \end{array}\right] - \left[\begin{array}{ccc} 4 & 1 & 2 \\[5pt] 0 & 3 & 2 \\[5pt] 1 & -2 & 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 4 - 4 & 1 - 1 & -1 - 2 \\[5pt] 9 - 0 & 2 - 3 & 7 - 2 \\[5pt] 3 - 1 & -1 -(-2) & 4 - 3 \end{array}\right]}
{= \left[\begin{array}{ccc} 0 & 0 & -3 \\[5pt] 9 & -1 & 5 \\[5pt] 2 & 1 & 1 \end{array}\right]}

L.H.S. = R.H.S.
5. If {\text{A} = \left[\begin{array}{ccc} \dfrac{2}{3} & 1 & \dfrac{5}{3} \\[10pt] \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3} \\[10pt] \dfrac{7}{3} & 2 & \dfrac{2}{3} \end{array}\right]} {\text{B} = \left[\begin{array}{ccc} \dfrac{2}{5} & \dfrac{3}{5} & 1 \\[10pt] \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5} \\[10pt] \dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5} \end{array}\right],} then compute 3A – 5B.
3A – 5B
{= 3 × \left[\begin{array}{ccc} \dfrac{2}{3} & 1 & \dfrac{5}{3} \\[10pt] \dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3} \\[10pt] \dfrac{7}{3} & 2 & \dfrac{2}{3} \end{array}\right] - 5 × \left[\begin{array}{ccc} \dfrac{2}{5} & \dfrac{3}{5} & 1 \\[10pt] \dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5} \\[10pt] \dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5} \end{array}\right]}
{= \left[\begin{array}{ccc} 3 × \dfrac{2}{3} & 3 × 1 & 3 × \dfrac{5}{3} \\[10pt] 3 × \dfrac{1}{3} & 3 × \dfrac{2}{3} & 3 × \dfrac{4}{3} \\[10pt] 3 × \dfrac{7}{3} & 3 × 2 & 3 × \dfrac{2}{3} \end{array}\right] - \left[\begin{array}{ccc} 5 × \dfrac{2}{5} & 5 × \dfrac{3}{5} & 5 × 1 \\[10pt] 5 × \dfrac{1}{5} & 5 × \dfrac{2}{5} & 5 × \dfrac{4}{5} \\[10pt] 5 × \dfrac{7}{5} & 5 × \dfrac{6}{5} & 5 × \dfrac{2}{5} \end{array}\right]}
{= \left[\begin{array}{ccc} 2 & 3 & 5 \\[5pt] 1 & 2 & 4 \\[5pt] 7 & 6 & 2 \end{array}\right] - \left[\begin{array}{ccc} 2 & 3 & 5 \\[5pt] 1 & 2 & 4 \\[5pt] 7 & 6 & 2 \end{array}\right]}
{= \left[\begin{array}{ccc} 2 - 2 & 3 - 3 & 5 - 5 \\[5pt] 1 - 1 & 2 - 2 & 4 - 4 \\[5pt] 7 - 7 & 6 - 6 & 2 - 2\end{array}\right]}
{= \left[\begin{array}{ccc} 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \end{array}\right]}

6. Simplify {\cos θ \left[\begin{array}{cc} \cos θ & \sin θ \\[5pt] -\sin θ & \cos θ \end{array}\right] + \sin θ \left[\begin{array}{cc} \sin θ & -\cos θ \\[5pt] \cos θ & \sin θ \end{array}\right]}
{\cos θ \left[\begin{array}{cc} \cos θ & \sin θ \\[5pt] -\sin θ & \cos θ \end{array}\right] + \sin θ \left[\begin{array}{cc} \sin θ & -\cos θ \\[5pt] \cos θ & \sin θ \end{array}\right]}
{= \left[\begin{array}{cc} \left(\cos θ\right) × \left(\cos θ\right) & \left(\cos θ\right) × \left(\sin θ\right) \\[5pt] \left(\cos θ\right) × \left(-\sin θ\right) & \left(\cos θ\right) × \left(\cos θ\right) \end{array}\right] + \left[\begin{array}{cc} \left(\sin θ\right) × \left(\sin θ\right) & \left(\sin θ\right) × \left(-\cos θ\right) \\[5pt] \left(\sin θ\right) × \left(\cos θ\right) & \left(\sin θ\right) × \left(\sin θ\right) \end{array}\right]}
{= \left[\begin{array}{cc} \cos^2 θ & \cos θ \sin θ \\[5pt] -\cos θ \sin θ & \cos^2 θ \end{array}\right] + \left[\begin{array}{cc} \sin^2 θ & -\sin θ \cos θ \\[5pt] \sin θ \cos θ & \sin^2 θ \end{array}\right]}
{= \left[\begin{array}{cc} \cos^2 θ + \sin^2 θ & \cos θ \sin θ - \sin θ \cos θ \\[5pt] -\cos θ \sin θ + \sin θ \cos θ & \cos^2 θ + \sin^2 θ \end{array}\right]}
= {\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}

7. Find X and Y, if
(i)
X + Y = {\left[\begin{array}{cc} 7 & 0 \\[5pt] 2 & 5 \end{array}\right]} and X – Y = {\left[\begin{array}{cc} 3 & 0 \\[5pt] 0 & 3 \end{array}\right]}
(ii)
2X + 3Y = {\left[\begin{array}{cc} 2 & 3 \\[5pt] 4 & 0 \end{array}\right]} and 3X + 2Y = {\left[\begin{array}{cc} 2 & -2 \\[5pt] -1 & 5 \end{array}\right]}
(i) To find X and Y when X + Y = {\left[\begin{array}{cc} 7 & 0 \\[5pt] 2 & 5 \end{array}\right]} and X – Y = {\left[\begin{array}{cc} 3 & 0 \\[5pt] 0 & 3 \end{array}\right]}
Give that
{\text{X + Y =} \left[\begin{array}{cc} 7 & 0 \\[5pt] 2 & 5 \end{array}\right]}
———-❶
{\text{X - Y =} \left[\begin{array}{cc} 3 & 0 \\[5pt] 0 & 3 \end{array}\right]}
———-❷
Adding equations ❶ and ❷, we get
X + Y + X – Y
{= \left[\begin{array}{cc} 7 & 0 \\[5pt] 2 & 5 \end{array}\right] + \left[\begin{array}{cc} 3 & 0 \\[5pt] 0 & 3 \end{array}\right]}
⇒ 2X
{= \left[\begin{array}{cc} 7 + 3 & 0 + 0 \\[5pt] 2 + 0 & 5 + 3 \end{array}\right]}
{= \left[\begin{array}{cc} 10 & 0 \\[5pt] 2 & 8 \end{array}\right]}
⇒ X
{= \dfrac{1}{2} × \left[\begin{array}{cc} 10 & 0 \\[5pt] 2 & 8 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{1}{2} × 10 & \dfrac{1}{2} × 0 \\[10pt] \dfrac{1}{2} × 2 & \dfrac{1}{2} × 8 \end{array}\right]}
{= \left[\begin{array}{cc} 5 & 0 \\[5pt] 1 & 4 \end{array}\right]}
Similarly, subtracting equation ❷ from equation ❶, we get
(X + Y) – (X – Y)
{= \left[\begin{array}{cc} 7 & 0 \\[5pt] 2 & 5 \end{array}\right] - \left[\begin{array}{cc} 3 & 0 \\[5pt] 0 & 3 \end{array}\right]}
⇒ X + Y – X + Y
{= \left[\begin{array}{cc} 7 - 3 & 0 - 0 \\[5pt] 2 - 0 & 5 - 3 \end{array}\right]}
⇒ 2Y
{= \left[\begin{array}{cc} 4 & 0 \\[5pt] 2 & 2 \end{array}\right]}
⇒ Y
{= \dfrac{1}{2} × \left[\begin{array}{cc} 4 & 0 \\[5pt] 2 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{1}{2} × 4 & \dfrac{1}{2} × 0 \\[10pt] \dfrac{1}{2} × 2 & \dfrac{1}{2} × 2 \end{array}\right]}
{= \left[\begin{array}{cc} 2 & 0 \\[5pt] 1 & 1 \end{array}\right]}
{\text{X} = \left[\begin{array}{cc} 5 & 0 \\[5pt] 1 & 4 \end{array}\right]} and {\text{Y} = \left[\begin{array}{cc} 2 & 0 \\[5pt] 1 & 1 \end{array}\right]}
(ii) To find X and Y when 2X + 3Y = {\left[\begin{array}{cc} 2 & 3 \\[5pt] 4 & 0 \end{array}\right]} and 3X + 2Y = {\left[\begin{array}{cc} 2 & -2 \\[5pt] -1 & 5 \end{array}\right]}
Given that
{\text{2X + 3Y =} \left[\begin{array}{cc} 2 & 3 \\[5pt] 4 & 0 \end{array}\right]}
———-❶
{\text{3X + 2Y =} \left[\begin{array}{cc} 2 & -2 \\[5pt] -1 & 5 \end{array}\right]}
———-❷
{\text{❶ × 2 ➜ 4X + 6Y =} \left[\begin{array}{cc} 4 & 6 \\[5pt] 8 & 0 \end{array}\right]}
———-❸
{\text{❷ × 3 ➜ 9X + 6Y =} \left[\begin{array}{cc} 6 & -6 \\[5pt] -3 & 15 \end{array}\right]}
———-❹
Subtracting equation ❸ from equation ❹, we get,
(9X + 6Y) – (4X + 6Y)
{= \left[\begin{array}{cc} 6 & -6 \\[5pt] -3 & 15 \end{array}\right] - \left[\begin{array}{cc} 4 & 6 \\[5pt] 8 & 0 \end{array}\right]}
⇒ 9X + 6Y – 4X – 6Y
{= \left[\begin{array}{cc} 6 - 4 & -6 - 6 \\[5pt] -3 - 8 & 15 - 0 \end{array}\right]}
⇒ 5X
{= \left[\begin{array}{cc} 2 & -12 \\[5pt] -11 & 15 \end{array}\right]}
⇒ X
{= \dfrac{1}{5} × \left[\begin{array}{cc} 2 & -12 \\[5pt] -11 & 15 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{1}{5} × 2 & \dfrac{1}{5} × -12 \\[10pt] \dfrac{1}{5} × -11 & \dfrac{1}{5} × 15 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{2}{5} & -\dfrac{12}{5} \\[10pt] -\dfrac{11}{5} & 3 \end{array}\right]}
Now,
{\text{❶ × 3 ➜ 6X + 9Y =} \left[\begin{array}{cc} 6 & 9 \\[5pt] 12 & 0 \end{array}\right]}
———-❺
{\text{❷ × 2 ➜ 6X + 4Y =} \left[\begin{array}{cc} 4 & -4 \\[5pt] -2 & 10 \end{array}\right]}
———-❻
Subtracting equation ❻ from equation ❺, we get
(6X + 9Y) – (6X + 4Y)
{= \left[\begin{array}{cc} 6 & 9 \\[5pt] 12 & 0 \end{array}\right] - \left[\begin{array}{cc} 4 & -4 \\[5pt] -2 & 10 \end{array}\right]}
⇒ 6X + 9Y – 6X – 4Y
{= \left[\begin{array}{cc} 6 - 4 & 9 - (-4) \\[5pt] 12 - (-2) & 0 - 10 \end{array}\right]}
⇒ 5Y
{= \left[\begin{array}{cc} 2 & 13 \\[5pt] 14 & -10 \end{array}\right]}
⇒ Y
{= \dfrac{1}{5} × \left[\begin{array}{cc} 2 & 13 \\[5pt] 14 & -10 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{1}{5} × 2 & \dfrac{1}{5} × 13 \\[10pt] \dfrac{1}{5} × 14 & \dfrac{1}{5} × -10 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{2}{5} & \dfrac{13}{5} \\[10pt] \dfrac{14}{5} & -2 \end{array}\right]}
{\text{X} = \left[\begin{array}{cc} \dfrac{2}{5} & -\dfrac{12}{5} \\[10pt] -\dfrac{11}{5} & 3 \end{array}\right]} and {\text{Y} = \left[\begin{array}{cc} \dfrac{2}{5} & \dfrac{13}{5} \\[10pt] \dfrac{14}{5} & -2 \end{array}\right]}

8. Find X, if {\text{Y} = \left[\begin{array}{cc} 3 & 2 \\[5pt] 1 & 4 \end{array}\right]} and {\text{2X + Y} = \left[\begin{array}{cc} 1 & 0 \\[5pt] -3 & 2 \end{array}\right]}
Given that
Y
{= \left[\begin{array}{cc} 3 & 2 \\[5pt] 1 & 4 \end{array}\right]}
———-❶
2X + Y
{= \left[\begin{array}{cc} 1 & 0 \\[5pt] -3 & 2 \end{array}\right]}
———-❷
Substituting ❶ in ❷, we have
{2\text{X} + \left[\begin{array}{cc} 3 & 2 \\[5pt] 1 & 4 \end{array}\right] = \left[\begin{array}{cc} 1 & 0 \\[5pt] -3 & 2 \end{array}\right]}
⇒ 2X
{= \left[\begin{array}{cc} 1 & 0 \\[5pt] -3 & 2 \end{array}\right] - \left[\begin{array}{cc} 3 & 2 \\[5pt] 1 & 4 \end{array}\right]}
⇒ X
{= \dfrac{1}{2} × \left[\begin{array}{cc} 1 - 3 & 0 - 2 \\[5pt] -3 - 1 & 2 - 4 \end{array}\right]}
{= \dfrac{1}{2} × \left[\begin{array}{cc} -2 & -2 \\[5pt] -4 & -2 \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{1}{2} × (-2) & \dfrac{1}{2} × (-2) \\[10pt] \dfrac{1}{2} × (-4) & \dfrac{1}{2} × (-2) \end{array}\right]}
{= \left[\begin{array}{cc} -1 & -1 \\[5pt] -2 & -1 \end{array}\right]}

9. Find x and y if {2\left[\begin{array}{cc} 1 & 3 \\[5pt] 0 & x \end{array}\right] + \left[\begin{array}{cc} y & 0 \\[5pt] 1 & 2 \end{array}\right] = \left[\begin{array}{cc} 5 & 6 \\[5pt] 1 & 8 \end{array}\right]}
Given that
{2\left[\begin{array}{cc} 1 & 3 \\[5pt] 0 & x \end{array}\right] + \left[\begin{array}{cc} y & 0 \\[5pt] 1 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 5 & 6 \\[5pt] 1 & 8 \end{array}\right]}
{⇒ \left[\begin{array}{cc} 2 × 1 & 2 × 3 \\[5pt] 2 × 0 & 2x \end{array}\right] + \left[\begin{array}{cc} y & 0 \\[5pt] 1 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 5 & 6 \\[5pt] 1 & 8 \end{array}\right]}
{⇒ \left[\begin{array}{cc} 2 & 6 \\[5pt] 0 & 2x \end{array}\right] + \left[\begin{array}{cc} y & 0 \\[5pt] 1 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 5 & 6 \\[5pt] 1 & 8 \end{array}\right]}
{⇒ \left[\begin{array}{cc} 2 + y & 6 + 0 \\[5pt] 0 + 1 & 2x + 2 \end{array}\right]}
{= \left[\begin{array}{cc} 5 & 6 \\[5pt] 1 & 8 \end{array}\right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
{2 + y}
= 5
y
= 5 – 2
= 3
Also
{2x + 2}
= 8
{⇒ 2x}
= 8 – 2
= 6
{⇒ x}
{= \dfrac{6}{2}}
= 3
So, {x = 3} and {y = 3}

10. Solve the equation for x, y, z and t, if {2\left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right] + 3\left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 2 \end{array}\right] = 3\left[\begin{array}{cc} 3 & 5 \\[5pt] 4 & 6 \end{array}\right]}
Given that
{2\left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right] + 3\left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 2 \end{array}\right] = 3\left[\begin{array}{cc} 3 & 5 \\[5pt] 4 & 6 \end{array}\right]}
{⇒ 2\left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right]}
{= 3\left[\begin{array}{cc} 3 & 5 \\[5pt] 4 & 6 \end{array}\right] - 3\left[\begin{array}{cc} 1 & -1 \\[5pt] 0 & 2 \end{array}\right]}
{⇒ 2\left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right]}
{= \left[\begin{array}{cc} 3 × 3 & 3 × 5 \\[5pt] 3 × 4 & 3 × 6 \end{array}\right] - \left[\begin{array}{cc} 3 × 1 & 3 × (-1) \\[5pt] 3 × 0 & 3 × 2 \end{array}\right]}
{⇒ 2\left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right]}
{= \left[\begin{array}{cc} 9 & 15 \\[5pt] 12 & 18 \end{array}\right] - \left[\begin{array}{cc} 3 & -3 \\[5pt] 0 & 6 \end{array}\right]}
{⇒ 2\left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right]}
{= \left[\begin{array}{cc} 9 - 3 & 15 - (-3) \\[5pt] 12 - 0 & 18 - 6 \end{array}\right]}
{⇒ \left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right]}
{= \dfrac{1}{2} × \left[\begin{array}{cc} 6 & 18 \\[5pt] 12 & 12 \end{array}\right]}
{⇒ \left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{1}{2} × 6 & \dfrac{1}{2} × 18 \\[10pt] \dfrac{1}{2} × 12 & \dfrac{1}{2} × 12 \end{array}\right]}
{⇒ \left[\begin{array}{cc} x & z \\[5pt] y & t \end{array}\right]}
{= \left[\begin{array}{cc} 3 & 9 \\[5pt] 6 & 6 \end{array}\right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
{x = 3,} {y = 6,} {z = 9} and {t = 6}

11. If {x\left[\begin{array}{c} 2 \\[5pt] 3 \end{array}\right] + y\left[\begin{array}{c} -1 \\[5pt] 1 \end{array}\right] = \left[\begin{array}{c} 10 \\[5pt] 5 \end{array}\right],} find the values of x and y.
Given that
{x\left[\begin{array}{c} 2 \\[5pt] 3 \end{array}\right] + y\left[\begin{array}{c} -1 \\[5pt] 1 \end{array}\right]}
{= \left[\begin{array}{c} 10 \\[5pt] 5 \end{array}\right]}
{⇒ \left[\begin{array}{c} 2x - y \\[5pt] 3x + y \end{array}\right]}
{= \left[\begin{array}{c} 10 \\[5pt] 5 \end{array}\right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
{2x - y =}
10 ———–❶
{3x + y =}
5 ———–❷
Adding equations ❶ and ❷, we get
5x
= 15
⇒ x
{= \dfrac{15}{5}}
= 3
From equation ❷, we have
{y}
{= 5 - 3x}
= 5 – (3 × 3) {(∵ x = 3)}
= 5 – 9
= -4
So, we have {x = 3} and {y = -4}
12. Given {3\left[\begin{array}{cc} x & y \\[5pt] z & w \end{array}\right] = \left[\begin{array}{cc} x & 6 \\[5pt] -1 & 2w \end{array}\right] + \left[\begin{array}{cc} 4 & x + y \\[5pt] z + w & 3 \end{array}\right],} find the values of x, y, z and w.
Given that
{3\left[\begin{array}{cc} x & y \\[5pt] z & w \end{array}\right]}
{= \left[\begin{array}{cc} x & 6 \\[5pt] -1 & 2w \end{array}\right] + \left[\begin{array}{cc} 4 & x + y \\[5pt] z + w & 3 \end{array}\right]}
{\left[\begin{array}{cc} 3x & 3y \\[5pt] 3z & 3w \end{array}\right]}
{= \left[\begin{array}{cc} x + 4 & 6 + x + y \\[5pt] -1 + z + w & 2w + 3 \end{array}\right]}
As the given matrices are equal, therefore, their corresponding elements must be equal. Comparing the corresponding elements, we get
3x
{= x + 4}
———–❶
3y
{= 6 + x + y}
———–❷
z
{= -1 + z + w}
———–❸
w
{= 2w + 3}
———–❹
From equation ❶, we get
3x
{= x + 4}
{⇒ 3x - x}
= 4
{⇒ 2x}
= 4
{⇒ x}
{= \dfrac{4}{2}}
= 2
Substituting {x = 2} into equation ❷, we have
{3y}
{= 6 + x + y}
{⇒ 3y - y}
{= 6 + x}
{⇒ 2y}
{= 6 + x}
= 6 + 2 {(∵ x = 2)}
= 8
{⇒ y}
{= \dfrac{8}{2}}
= 4
From equation ❹, we get
{3w}
{= 2w + 3}
{⇒ 3w - 2w}
= 3
{⇒ w}
= 3
Substituting {w = 3,} into equation ❸, we get
{⇒ 3z}
{= -1 + z + w}
{⇒ 3z - z}
= -1 + 3 {(∵ w = 3)}
{⇒2z}
= 2
{⇒ z}
{= \dfrac{2}{2}}
= 1
So, we have {x = 2,} {y = 4,} {z = 1} and {w = 3}
13. If {\text{F}(x) = \left[\begin{array}{ccc} \cos x & -\sin x & 0 \\[5pt] \sin x & \cos x & 0 \\[5pt] 0 & 0 & 1\end{array}\right],} show that {\text{F}(x) \text{F}(y) = \text{F}(x + y).}
We have
{\text{F}(x)}
{= \left[\begin{array}{ccc} \cos x & -\sin x & 0 \\[5pt] \sin x & \cos x & 0 \\[5pt] 0 & 0 & 1\end{array}\right]}
{\text{F}(y)}
{= \left[\begin{array}{ccc} \cos y & -\sin y & 0 \\[5pt] \sin y & \cos y & 0 \\[5pt] 0 & 0 & 1\end{array}\right]}
Now,
L.H.S.
{= \text{F}(x) \text{F}(y)}
{= \left[\begin{array}{ccc} \cos x & -\sin x & 0 \\[5pt] \sin x & \cos x & 0 \\[5pt] 0 & 0 & 1\end{array}\right]\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\[5pt] \sin y & \cos y & 0 \\[5pt] 0 & 0 & 1\end{array}\right]}
{= \left[\begin{array}{ccc} \cos x \cos y - \sin x \sin y + 0 & -\cos x \sin y - \sin x \cos y + 0 & 0 + 0 + 0 \\[5pt] \sin x \cos y + \cos x \sin y + 0 & -\sin x \sin y + \cos x \cos y + 0 & 0 + 0 + 0 \\[5pt] 0 + 0 + 0 & 0 + 0 + 0 & 0 + 0 + 1 \end{array}\right]}
{= \left[\begin{array}{ccc} \cos x \cos y - \sin x \sin y & -(\sin x \cos y + \cos x \sin y) & 0 \\[5pt] \sin x \cos y + \cos x \sin y & -(\cos x \cos y - \sin x \sin y) & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} \cos (x + y) & -\sin (x + y) & 0 \\[5pt] \sin (x + y) & -\cos (x + y) & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \text{F}(x + y)}
= R.H.S.
14. Show that
(i)
{\left[\begin{array}{cc} 5 & -1 \\[5pt] 6 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 4 \end{array}\right]}{\left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 4 \end{array}\right] \left[\begin{array}{cc} 5 & -1 \\[5pt] 6 & 7 \end{array}\right]}
(ii)
{\left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 1 & 0 \end{array}\right] \left[\begin{array}{ccc} -1 & 1 & 0 \\[5pt] 0 & -1 & 1 \\[5pt] 2 & 3 & 4 \end{array}\right]}{\left[\begin{array}{ccc} -1 & 1 & 0 \\[5pt] 0 & -1 & 1 \\[5pt] 2 & 3 & 4 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 1 & 0 \end{array}\right]}
(i) To show that {\left[\begin{array}{cc} 5 & -1 \\[5pt] 6 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 4 \end{array}\right]}{\left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 4 \end{array}\right] \left[\begin{array}{cc} 5 & -1 \\[5pt] 6 & 7 \end{array}\right]}
L.H.S.
{= \left[\begin{array}{cc} 5 & -1 \\[5pt] 6 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 4 \end{array}\right]}
{= \left[\begin{array}{cc} (5 × 2) + ((-1) × 3) & (5 × 1) + (-1 × 4) \\[5pt] (6 × 2) + (7 × 3) & (6 × 1) + (7 × 4) \end{array}\right]}
{= \left[\begin{array}{cc} 10 - 3 & 5 - 4 \\[5pt] 12 + 21 & 6 + 28 \end{array}\right]}
{= \left[\begin{array}{cc} 7 & 1 \\[5pt] 33 & 34 \end{array}\right]}
R.H.S.
{= \left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 4 \end{array}\right] \left[\begin{array}{cc} 5 & -1 \\[5pt] 6 & 7 \end{array}\right]}
{= \left[\begin{array}{cc} (2 × 5) + (1 × 6) & (2 × (-1)) + (1 × 7) \\[5pt] (3 × 5) + (4 × 6) & (3 × (-1)) + (4 × 7) \end{array}\right]}
{= \left[\begin{array}{cc} 10 + 6 & -2 + 7 \\[5pt] 15 + 24 & -3 + 28 \end{array}\right]}
{= \left[\begin{array}{cc} 16 & 5 \\[5pt] 39 & 25 \end{array}\right]}
As L.H.S ≠ R.H.S., it is proved that {\left[\begin{array}{cc} 5 & -1 \\[5pt] 6 & 7 \end{array}\right] \left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 4 \end{array}\right]}{\left[\begin{array}{cc} 2 & 1 \\[5pt] 3 & 4 \end{array}\right] \left[\begin{array}{cc} 5 & -1 \\[5pt] 6 & 7 \end{array}\right]}
(ii) To show that {\left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 1 & 0 \end{array}\right] \left[\begin{array}{ccc} -1 & 1 & 0 \\[5pt] 0 & -1 & 1 \\[5pt] 2 & 3 & 4 \end{array}\right]}{\left[\begin{array}{ccc} -1 & 1 & 0 \\[5pt] 0 & -1 & 1 \\[5pt] 2 & 3 & 4 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 1 & 0 \end{array}\right]}
L.H.S.
{= {\left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 1 & 0 \end{array}\right] \left[\begin{array}{ccc} -1 & 1 & 0 \\[5pt] 0 & -1 & 1 \\[5pt] 2 & 3 & 4 \end{array}\right]}}
{= \left[\begin{array}{ccc} (1 × (-1)) + (2 × 0) + (3 × 2) & (1 × 1) + (2 × (-1)) + (3 × 3) & (1 × 0) + (2 × 1) + (3 × 4) \\[5pt] (0 × (-1)) + (1 × 0) + (0 × 2) & (0 × 1) + (1 × (-1)) + (0 × 3) & (0 × 0) + (1 × 1) + (0 × 4) \\[5pt] (1 × -1) + (1 × 0) + (0 × 2) & (1 × 1) + (1 × (-1)) + (0 × 3) & (1 × 0) + (1 × 1) + (0 × 4) \end{array}\right]}
{= \left[\begin{array}{ccc} -1 + 0 + 6 & 1 - 2 + 9 & 0 + 2 + 12 \\[5pt] 0 + 0 + 0 & 0 - 1 + 0 & 0 + 1 + 0 \\[5pt] -1 + 0 + 0 & 1 - 1 + 0 & 0 + 1 + 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 5 & 8 & 14 \\[5pt] 0 & -1 & 1 \\[5pt] -1 & -1 & 1 \end{array}\right]}
R.H.S.
{= \left[\begin{array}{ccc} -1 & 1 & 0 \\[5pt] 0 & -1 & 1 \\[5pt] 2 & 3 & 4 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 1 & 0 \end{array}\right]}
{= \left[\begin{array}{cc} (-1 × 1) + (1 × 0) + (0 × 1) & (-1 × 2) + (1 × 1) + (0 × 1) & (-1 × 3) + (1 × 0) + (0 × 0) \\[5pt] (0 × 1) + (-1 × 0) + (1 × 1) & (0 × 2) + (-1 × 1) + (1 × 1) & (0 × 3) + (-1 × 0) + (1 × 0) \\[5pt] (2 × 1) + (3 × 0) + (4 × 1) & (2 × 2) + (3 × 1) + (4 × 1) & (2 × 3) + (3 × 0) + (4 × 0) \end{array}\right]}
{= \left[\begin{array}{cc} -1 + 0 + 0 & -2 + 1 + 0 & -3 + 0 + 0 \\[5pt] 0 + 0 + 1 & 0 - 1 + 1 & 0 + 0 + 0 \\[5pt] 2 + 0 + 4 & 4 + 3 + 4 & 6 + 0 + 0 \end{array}\right]}
{= \left[\begin{array}{cc} -1 & -1 & -3 \\[5pt] 1 & 0 & 0 \\[5pt] 6 & 11 & 6 \end{array}\right]}
As L.H.S ≠ R.H.S., it is proved that {\left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 1 & 0 \end{array}\right] \left[\begin{array}{ccc} -1 & 1 & 0 \\[5pt] 0 & -1 & 1 \\[5pt] 2 & 3 & 4 \end{array}\right]}{\left[\begin{array}{ccc} -1 & 1 & 0 \\[5pt] 0 & -1 & 1 \\[5pt] 2 & 3 & 4 \end{array}\right] \left[\begin{array}{ccc} 1 & 2 & 3 \\[5pt] 0 & 1 & 0 \\[5pt] 1 & 1 & 0 \end{array}\right]}
15. Find {\text{A}^2 - 5\text{A} + 6\text{I},} if {\text{A} = \left[\begin{array}{ccc} 2 & 0 & 1 \\[5pt] 2 & 1 & 3 \\[5pt] 1 & -1 & 0 \end{array}\right]}
Let’s first find out {\text{A}^2,} 5A and 6I individually.
{\text{A}^2}
= A.A
{= \left[\begin{array}{ccc} 2 & 0 & 1 \\[5pt] 2 & 1 & 3 \\[5pt] 1 & -1 & 0 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 1 \\[5pt] 2 & 1 & 3 \\[5pt] 1 & -1 & 0 \end{array}\right]}
{= \left[\begin{array}{ccc} (2 × 2) + (0 × 2) + (1 × 1) & (2 × 0) + (0 × 1) + (1 × (-1)) & (2 × 1) + (0 × 3) + (1 × 0) \\[5pt] (2 × 2) + (1 × 2) + (3 × 1) & (2 × 0) + (1 × 1) + (3 × (-1)) & (2 × 1) + (1 × 3) + (3 × 0) \\[5pt] (1 × 2) + ((-1) × 2) + (0 × 1) & (1 × 0) + ((-1) × 1) + (0 × (-1)) & (1 × 1) + ((-1) × 3) + (0 × 0) \end{array}\right]}
{= \left[\begin{array}{ccc} 4 + 0 + 1 & 0 + 0 - 1 & 2 + 0 + 0 \\[5pt] 4 + 2 + 3 & 0 + 1 - 3 & 2 + 3 + 0 \\[5pt] 2 - 2 + 0 & 0 - 1 + 0 & 1 - 3 + 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 5 & -1 & 2 \\[5pt] 9 & -2 & 5 \\[5pt] 0 & -1 & -2 \end{array}\right]}
5A
{= 5\left[\begin{array}{ccc} 2 & 0 & 1 \\[5pt] 2 & 1 & 3 \\[5pt] 1 & -1 & 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 5 × 2 & 5 × 0 & 5 × 1 \\[5pt] 5 × 2 & 5 × 1 & 5 × 3 \\[5pt] 5 × 1 & 5 × (-1) & 5 × 0 \end{array}\right]}
{= \left[\begin{array}{ccc} 10 & 0 & 5 \\[5pt] 10 & 5 & 15 \\[5pt] 5 & -5 & 0 \end{array}\right]}
6I
{= 6\left[\begin{array}{ccc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{ccc} 6 & 0 & 0 \\[5pt] 0 & 6 & 0 \\[5pt] 0 & 0 & 6 \end{array}\right]}
{\text{A}^2 - 5\text{A} + 6\text{I}}
{= \left[\begin{array}{ccc} 5 & -1 & 2 \\[5pt] 9 & -2 & 5 \\[5pt] 0 & -1 & -2 \end{array}\right] - \left[\begin{array}{ccc} 10 & 0 & 5 \\[5pt] 10 & 5 & 15 \\[5pt] 5 & -5 & 0 \end{array}\right] + \left[\begin{array}{ccc} 6 & 0 & 0 \\[5pt] 0 & 6 & 0 \\[5pt] 0 & 0 & 6 \end{array}\right]}
{= \left[\begin{array}{ccc} 5 - 10 + 6 & -1 - 0 + 0 & 2 - 5 + 0 \\[5pt] 9 - 10 + 0 & -2 - 5 + 6 & 5 - 15 + 0 \\[5pt] 0 - 5 + 0 & -1 + 5 + 0 & -2 - 0 + 6 \end{array}\right]}
= {\left[\begin{array}{ccc} 1 & -1 & -3 \\[5pt] -1 & -1 & -10 \\[5pt] -5 & 4 & 4 \end{array}\right]}
16. If {\text{A} = \left[\begin{array}{cc} 1 & 0 & 2 \\[5pt] 0 & 2 & 1 \\[5pt] 2 & 0 & 3 \end{array}\right],} prove that {\text{A}^3 - 6\text{A}^2 + 7\text{A} + 2\text{I} = \text{O}}
{\text{A}^2}
= A.A
{= \left[\begin{array}{cc} 1 & 0 & 2 \\[5pt] 0 & 2 & 1 \\[5pt] 2 & 0 & 3 \end{array}\right] \left[\begin{array}{cc} 1 & 0 & 2 \\[5pt] 0 & 2 & 1 \\[5pt] 2 & 0 & 3 \end{array}\right]}
{= \left[\begin{array}{cc} (1 × 1) + (0 × 0) + (2 × 2) & (1 × 0) + (0 × 2) + (2 × 0) & (1 × 2) + (0 × 1) + (2 × 3) \\[5pt] (0 × 1) + (2 × 0) + (1 × 2) & (0 × 0) + (2 × 2) + (1 × 0) & (0 × 2) + (2 × 1) + (1 × 3) \\[5pt] (2 × 1) + (0 × 0) + (3 × 2) & (2 × 0) + (0 × 2) + (3 × 0) & (2 × 2) + (0 × 1) + (3 × 3) \end{array}\right]}
{= \left[\begin{array}{cc} 1 + 0 + 4 & 0 + 0 + 0 & 2 + 0 + 6 \\[5pt] 0 + 0 + 2 & 0 + 4 + 0 & 0 + 2 + 3 \\[5pt] 2 + 0 + 6 & 0 + 0 + 0 & 4 + 0 + 9 \end{array}\right]}
{= \left[\begin{array}{cc} 5 & 0 & 8 \\[5pt] 2 & 4 & 5 \\[5pt] 8 & 0 & 13 \end{array}\right]}
{\text{A}^3}
{= \text{A}^2.\text{A}}
{= \left[\begin{array}{cc} 5 & 0 & 8 \\[5pt] 2 & 4 & 5 \\[5pt] 8 & 0 & 13 \end{array}\right]\left[\begin{array}{cc} 1 & 0 & 2 \\[5pt] 0 & 2 & 1 \\[5pt] 2 & 0 & 3 \end{array}\right]}
{= \left[\begin{array}{cc} (5 × 1) + (0 × 0) + (8 × 2) & (5 × 0) + (0 × 2) + (8 × 0) & (5 × 2) + (0 × 1) + (8 × 3) \\[5pt] (2 × 1) + (4 × 0) + (5 × 2) & (2 × 0) + (4 × 2) + (5 × 0) & (2 × 2) + (4 × 1) + (5 × 3) \\[5pt] (8 × 1) + (0 × 0) + (13 × 2) & (8 × 0) + (0 × 2) + (13 × 0) & (8 × 2) + (0 × 1) + (13 × 3) \end{array}\right]}
{= \left[\begin{array}{cc} 5 + 0 + 16 & 0 + 0 + 0 & 10 + 0 + 24 \\[5pt] 2 + 0 + 10 & 0 + 8 + 0 & 4 + 4 + 15 \\[5pt] 8 + 0 + 26 & 0 + 0 + 0 & 16 + 0 + 39 \end{array}\right]}
{= \left[\begin{array}{cc} 21 & 0 & 34 \\[5pt] 12 & 8 & 23 \\[5pt] 34 & 0 & 55 \end{array}\right]}
{6\text{A}^2}
{= 6\left[\begin{array}{cc} 5 & 0 & 8 \\[5pt] 2 & 4 & 5 \\[5pt] 8 & 0 & 13 \end{array}\right]}
{= \left[\begin{array}{cc} 6 × 5 & 6 × 0 & 6 × 8 \\[5pt] 6 × 2 & 6 × 4 & 6 × 5 \\[5pt] 6 × 8 & 6 × 0 & 6 × 13 \end{array}\right]}
{= \left[\begin{array}{cc} 30 & 0 & 48 \\[5pt] 12 & 24 & 30 \\[5pt] 48 & 0 & 78 \end{array}\right]}
7A
{= 7\left[\begin{array}{cc} 1 & 0 & 2 \\[5pt] 0 & 2 & 1 \\[5pt] 2 & 0 & 3 \end{array}\right]}
{= \left[\begin{array}{cc} 7 × 1 & 7 × 0 & 7 × 2 \\[5pt] 7 × 0 & 7 × 2 & 7 × 1 \\[5pt] 7 × 2 & 7 × 0 & 7 × 3 \end{array}\right]}
{= \left[\begin{array}{cc} 7 & 0 & 14 \\[5pt] 0 & 14 & 7 \\[5pt] 14 & 0 & 21 \end{array}\right]}
2I
{= 2\left[\begin{array}{cc} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{array}\right]}
{= \left[\begin{array}{cc} 2 & 0 & 0 \\[5pt] 0 & 2 & 0 \\[5pt] 0 & 0 & 2 \end{array}\right]}
Now,
L.H.S.
{= \text{A}^3 - 6\text{A}^2 + 7\text{A} + 2\text{I}}
{= \left[\begin{array}{cc} 21 & 0 & 34 \\[5pt] 12 & 8 & 23 \\[5pt] 34 & 0 & 55 \end{array}\right] - \left[\begin{array}{cc} 30 & 0 & 48 \\[5pt] 12 & 24 & 30 \\[5pt] 48 & 0 & 78 \end{array}\right] + \left[\begin{array}{cc} 7 & 0 & 14 \\[5pt] 0 & 14 & 7 \\[5pt] 14 & 0 & 21 \end{array}\right] + \left[\begin{array}{cc} 2 & 0 & 0 \\[5pt] 0 & 2 & 0 \\[5pt] 0 & 0 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 21 - 30 + 7 + 2 & 0 - 0 + 0 + 0 & 34 - 48 + 14 + 0 \\[5pt] 12 - 12 + 0 + 0 & 8 - 24 + 14 + 2 & 23 - 30 + 7 + 0 \\[5pt] 34 - 48 + 14 + 0 & 0 + 0 + 0 + 0 & 55 - 78 + 21 + 2 \end{array}\right]}
{= \left[\begin{array}{cc} 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \\[5pt] 0 & 0 & 0 \end{array}\right]}
= O
= R.H.S.
Hence it is proved that {\text{A}^3 - 6\text{A}^2 + 7\text{A} + 2\text{I} = \text{O}}
17. If {\text{A} = \left[\begin{array}{cc} 3 & -2 \\[5pt] 4 & -2 \end{array}\right]} and {\text{I} = \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right],} find k so that {\text{A}^2 = k\text{A} - 2\text{I}}
We have to find the value of k such that
{\text{A}^2 = k\text{A} - 2\text{I}}
{⇒ \text{A}^2 + 2\text{I} = k\text{A}}
{⇒ k\text{A}}
{= \text{A}^2 + 2\text{I}}
{= \left[\begin{array}{cc} 3 & -2 \\[5pt] 4 & -2 \end{array}\right] \left[\begin{array}{cc} 3 & -2 \\[5pt] 4 & -2 \end{array}\right] - 2\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right]}
{= \left[\begin{array}{cc} (3 × 3) + (-2 × 4) & (3 × (-2)) + ((-2) × (-2)) \\[5pt] (4 × 3) + (-2 × 4) & (4 × (-2)) + ((-2) × (-2)) \end{array}\right] - \left[\begin{array}{cc} 2 & 0 \\[5pt] 0 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 9 - 8 & -6 + 4 \\[5pt] 12 - 8 & -8 + 4 \end{array}\right] - \left[\begin{array}{cc} 2 & 0 \\[5pt] 0 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 1 & -2 \\[5pt] 4 & -4 \end{array}\right] - \left[\begin{array}{cc} 2 & 0 \\[5pt] 0 & 2 \end{array}\right]}
{= \left[\begin{array}{cc} 1 + 2 & -2 + 0 \\[5pt] 4 + 0 & -4 + 2 \end{array}\right]}
{= \left[\begin{array}{cc} 3 & -2 \\[5pt] 4 & -2 \end{array}\right]}
= A
So, we have {k\text{A} = \text{A}}
{k = 1}
18. If {\text{A} = \left[\begin{array}{cc} 0 & -\tan \dfrac{α}{2} \\[10pt] \tan \dfrac{α}{2} & 0 \end{array}\right]} and I is the identity matrix of order 2, show that {\text{I + A = (I - A)} \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right]}
Let {t = \tan \dfrac{α}{2}}
This implies that
sin α
{= \dfrac{2\tan \dfrac{α}{2}}{1 + \tan^2 \dfrac{α}{2}}}
{= \dfrac{2t}{1 + t^2}}
cos α
{= \dfrac{1 - \tan^2 \dfrac{α}{2}}{1 + \tan^2 \dfrac{α}{2}}}
{= \dfrac{1 - t^2}{1 + t^2}}
Substituting, we have
L.H.S.
= I + A
{= \left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right] + \left[\begin{array}{cc} 0 & -t \\[5pt] t & 0 \end{array}\right]}
{= \left[\begin{array}{cc} 1 + 0 & 0 + (-t) \\[5pt] 0 + t & 1 + 0 \end{array}\right]}
{= \left[\begin{array}{cc} 1 & -t \\[5pt] t & 1 \end{array}\right]}
R.H.S.
{= \text{(I - A)} \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right]}
{= \left(\left[\begin{array}{cc} 1 & 0 \\[5pt] 0 & 1 \end{array}\right] - \left[\begin{array}{cc} 0 & -t \\[5pt] t & 0 \end{array}\right]\right) \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right]}
{= \left[\begin{array}{cc} (1 - 0) & (0 - (-t)) \\[5pt] 0 - t & 1 - 0 \end{array}\right] \left[\begin{array}{cc} \dfrac{1 - t^2}{1 + t^2} & -\dfrac{2t}{1 + t^2} \\[10pt] \dfrac{2t}{1 + t^2} & \dfrac{1 - t^2}{1 + t^2} \end{array}\right]}
{= \left[\begin{array}{cc} 1 & t \\[5pt] -t & 1 \end{array}\right] × \dfrac{1}{1 + t^2} × \left[\begin{array}{cc} 1 - t^2 & -2t \\[5pt] 2t & 1 - t^2 \end{array}\right]} {\left(∵ [ka_{ij}] = k[a_{ij}]\right)}
{= \dfrac{1}{1 + t^2} × \left[\begin{array}{cc} 1 & t \\[5pt] -t & 1 \end{array}\right] × \left[\begin{array}{cc} 1 - t^2 & -2t \\[5pt] 2t & 1 - t^2 \end{array}\right]}
{= \dfrac{1}{1 + t^2} × \left[\begin{array}{cc} \left(1 × \left(1 - t^2\right)\right) + (t × 2t) & (1 × -2t) + \left(t × \left(1 - t^2\right)\right) \\[10pt] \left(-t × \left(1 - t^2\right)\right) + (1 × 2t) & (-t × -2t) + \left(1 × \left(1 - t^2\right)\right) \end{array}\right]}
{= \dfrac{1}{1 + t^2} × \left[\begin{array}{cc} 1 - t^2 + 2t^2 & -2t + t - t^3 \\[10pt] -t + t^3 + 2t & 2t^2 + 1 - t^2 \end{array}\right]}
{= \dfrac{1}{1 + t^2} × \left[\begin{array}{cc} 1 + t^2 & -t - t^3 \\[10pt] t + t^3 & 1 + t^2 \end{array}\right]}
{= \dfrac{1}{1 + t^2} × \left[\begin{array}{cc} 1 + t^2 & -t\left(1 + t^2\right) \\[10pt] t\left(1 + t^2\right) & 1 + t^2 \end{array}\right]}
{= \dfrac{1}{1 + t^2} × \left(1 + t^2\right) × \left[\begin{array}{cc} 1 & -t \\[5pt] t & 1 \end{array}\right]}
{= \left[\begin{array}{cc} 1 & -t \\[5pt] t & 1 \end{array}\right]}
= R.H.S.
As L.H.S. = R.H.S., {\text{I + A = (I - A)} \left[\begin{array}{cc} \cos α & -\sin α \\[5pt] \sin α & \cos α \end{array}\right]}
19. A trust fund has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:
(a) ₹ 1800
(b) ₹ 2000
Given that the total investment is ₹ 30,000
Let investment in type I bond be x
⇒ Investment in type II bond will ₹~30,000 - x
These principals can be represented in the matrix form as
{\text{P} = \left[\begin{array}{cc} x & ₹~30,000 - x \end{array}\right]}
Rate of Interest:
Type I bond
= 5%
{= \dfrac{5}{100}}
Type II bond
= 7%
{= \dfrac{7}{100}}
This interest rates can be represented in the matrix form as
{\text{R} = \left[\begin{array}{c} \dfrac{5}{100} \\[10pt] \dfrac{7}{100} \end{array}\right]}
We know that interest on the principle p and an interest rate of r% for one year is given by
{\dfrac{pnr}{100} = \dfrac{p × 1 × r}{100} = \dfrac{pr}{100}}
The total interest per year can be obtained by matrix multiplication of the principal matrix P and interest rate matrix R
Thus
Total Interest
= PR
{= \left[\begin{array}{cc} x & ₹~30,000 - x \end{array}\right] \left[\begin{array}{c} \dfrac{5}{100} \\[10pt] \dfrac{7}{100} \end{array}\right]}
{= \left[\begin{array}{cc} x × \dfrac{5}{100} + (₹~30,000 - x) × \dfrac{7}{100} \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{5x + (₹~30,000 - x) × 7}{100} \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{5x + ₹~2,10,000 - 7x}{100} \end{array}\right]}
{= \left[\begin{array}{cc} \dfrac{₹~2,10,000 - 2x}{100} \end{array}\right]}
(i) To determine how to divide ₹ 30,000 to obtain an annual interest of ₹ 1,800
{\dfrac{₹~2,10,000 - 2x}{100}}
= ₹ 1,800
{⇒ ₹~2,10,000 - 2x}
= ₹ 1,800 × 100
{⇒ ₹~2,10,000 - 2x}
= ₹ 1,80,000
{⇒ 2x}
= ₹ 2,10,000 – ₹ 1,80,000
{⇒ 2x}
= ₹ 30,000
{⇒ x}
= ₹ 15,000
{⇒ ₹~30,000 - x}
= ₹ 30,000 – ₹ 15,000
= ₹ 15,000
∴ To obtain an annual interest of ₹ 1,800, the trust fund of ₹ 30,000 should be devided as:
Investment in Type I bond
₹ 15,000
Investment in Type II bond
₹ 15,000
(ii) To determine how to divide ₹ 30,000 to obtain an annual interest of ₹ 2,000
{\dfrac{₹~2,10,000 - 2x}{100}}
= ₹ 2,000
{⇒ ₹~2,10,000 - 2x}
= ₹ 2,000 × 100
{⇒ ₹~2,10,000 - 2x}
= ₹ 2,00,000
{⇒ 2x}
= ₹ 2,10,000 – ₹ 2,00,000
{⇒ 2x}
= ₹ 10,000
{⇒ x}
= ₹ 5,000
{⇒ ₹~30,000 - x}
= ₹ 30,000 – ₹ 5,000
= ₹ 25,000
∴ To obtain an annual interest of ₹ 2,000, the trust fund of ₹ 30,000 should be devided as:
Investment in Type I bond
₹ 5,000
Investment in Type II bond
₹ 25,000
20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 respectively. Find the total amount the bookshop will receive from selling all the books using matrix alzebra.
The inventory of books available in the school bookshop is as follows:
Subject
No. of Books
Available
Chemistry
10 dozen
= 10 × 12
= 120
Physics
8 dozen
= 8 × 12
= 96
Economics
10 dozen
= 10 × 12
= 120
The available quantities can be represented in the matrix form as follows:
{\text{A} = \left[\begin{array}{ccc} 120 & 96 & 120 \end{array}\right]}
The cost of each of these books can be represented in a column matrix as follows:
{\text{C} = \left[\begin{array}{c} 80 \\ 60 \\ 40 \end{array}\right]}
The total amount the bookshop will receive from selling all the books can be obtained by doing the matrix multiplication of the availability matrix A and the cost matrix C. So,
Total Amount
= AC
{= \left[\begin{array}{ccc} 120 & 96 & 120 \end{array}\right] \left[\begin{array}{c} 80 \\ 60 \\ 40 \end{array}\right]}
{= \left[\begin{array}{ccc} (120 × 80) + (96 × 60) + (120 × 40) \end{array}\right]}
{= \left[\begin{array}{ccc} 9,600 + 5,760 + 4,800 \end{array}\right]}
{= \left[\begin{array}{ccc} 20,160 \end{array}\right]}
The total amount the bookshop will receive from selling all the books is ₹ 20,160.
Assume X, Y, Z, W and P are matrices of order {2 × n,} {3 × k,} {2 × p,} {n × 3} and {p × k} respectively. Choose the correct answer in Exercises 21 and 22.
21. The restriction on {n,} k and p so that PY + WY will be defined are:
(A) {k = 3,} {p = n}
(B) k is arbitrary, {p = 2}
(C) p is arbitrary, {k = 3}
(D) {k = 2,} {p = 3}
The following is the order of the various matrices involved.
Matrix
Order
X
{2 × n}
Y
{3 × k}
Z
{2 × p}
W
{n × 3}
P
{p × k}
Multiplication of matrices P and Y is possible when
No. columns of P = No. of rows of Y
{⇒ k = 3}
Similarly, multiplication of matrices W and Y is possible when
No. columns of W = No. of rows of Y
{⇒ 3 = p}
The order of PY will be {p × k}
The order of WY will be {n × k}
Both PY and WY can be added only when their orders are equal. So,
Order of PY = Order of WY
{p × k = n × k}
{p = n}
So, PY + WY will be defined when {k = 3} and {p = n} = 3
So, option A is the correct answer.
22. If {n = p,} then the order of the matrix 7X – 5Z is:
(A) p × 2
(B) 2 × n ✔
(C) n × 3
(D) p × n
It is given that
The order of the X is {2 × n}
The order of the Z is {2 × p}
As it is also given that {n = p,} the matrix 7X – 5Z is defined and will have the same order as that of X or Z.
⇒ The order of 7X – 5Z will be either {2 × n} or {2 × p}
So, option B is the correct answer.