NCERT Solutions Class-12-Chapter-3-Matrices
Excercise-3.2
Note: Matrix addition is performed element-wise. Matrix multiplication \( AB \) is defined only if the number of columns in \( A \) equals the number of rows in \( B \). The element \( (AB)_{ij} \) is the sum of products of corresponding elements from row \( i \) of \( A \) and column \( j \) of \( B \).
Q1
Let \( A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \), \( B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \), \( C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} \). Find each of the following:
(i) \( A + B \)
(ii) \( A - B \)
(iii) \( 3A - C \)
(iv) \( AB \)
(v) \( BA \)
(i) \( A + B \)
(ii) \( A - B \)
(iii) \( 3A - C \)
(iv) \( AB \)
(v) \( BA \)
Solution:
(i) \( A + B \)
\[ A + B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2+1 & 4+3 \\ 3-2 & 2+5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix} \]
(ii) \( A - B \)
\[ A - B = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix} \]
(iii) \( 3A - C \)
\[ 3A - C = 3 \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix} - \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 6+2 & 12-5 \\ 9-3 & 6-4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix} \]
(iv) \( AB \)
\[ \begin{aligned} AB &= \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \\ &= \begin{bmatrix} 2(1) + 4(-2) & 2(3) + 4(5) \\ 3(1) + 2(-2) & 3(3) + 2(5) \end{bmatrix} \\ &= \begin{bmatrix} 2-8 & 6+20 \\ 3-4 & 9+10 \end{bmatrix} = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix} \end{aligned} \]
(v) \( BA \)
\[ \begin{aligned} BA &= \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} \\ &= \begin{bmatrix} 1(2) + 3(3) & 1(4) + 3(2) \\ -2(2) + 5(3) & -2(4) + 5(2) \end{bmatrix} \\ &= \begin{bmatrix} 2+9 & 4+6 \\ -4+15 & -8+10 \end{bmatrix} = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix} \end{aligned} \]
Q2
Compute the following:
(i) \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} \)
(ii) \( \begin{bmatrix} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix} \)
(iii) \( \begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix} \)
(iv) \( \begin{bmatrix} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{bmatrix} + \begin{bmatrix} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{bmatrix} \)
(i) \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} + \begin{bmatrix} a & b \\ b & a \end{bmatrix} \)
(ii) \( \begin{bmatrix} a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2 \end{bmatrix} + \begin{bmatrix} 2ab & 2bc \\ -2ac & -2ab \end{bmatrix} \)
(iii) \( \begin{bmatrix} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{bmatrix} \)
(iv) \( \begin{bmatrix} \cos^2 x & \sin^2 x \\ \sin^2 x & \cos^2 x \end{bmatrix} + \begin{bmatrix} \sin^2 x & \cos^2 x \\ \cos^2 x & \sin^2 x \end{bmatrix} \)
Solution:
(i)
\[ \begin{bmatrix} a+a & b+b \\ -b+b & a+a \end{bmatrix} = \begin{bmatrix} 2a & 2b \\ 0 & 2a \end{bmatrix} \]
(ii)
\[ \begin{bmatrix} a^2+b^2+2ab & b^2+c^2+2bc \\ a^2+c^2-2ac & a^2+b^2-2ab \end{bmatrix} = \begin{bmatrix} (a+b)^2 & (b+c)^2 \\ (a-c)^2 & (a-b)^2 \end{bmatrix} \]
(iii)
\[ \begin{bmatrix} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{bmatrix} = \begin{bmatrix} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{bmatrix} \]
(iv)
\[ \begin{bmatrix} \cos^2 x + \sin^2 x & \sin^2 x + \cos^2 x \\ \sin^2 x + \cos^2 x & \cos^2 x + \sin^2 x \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \]
Q3
Compute the indicated products:
(i) \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix} \)
(v) \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \)
(vi) \( \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix} \)
(i) \( \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix} \)
(v) \( \begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix} \)
(vi) \( \begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix} \)
Solution:
(i)
\[ \begin{bmatrix} a^2+b^2 & -ab+ab \\ -ab+ab & b^2+a^2 \end{bmatrix} = \begin{bmatrix} a^2+b^2 & 0 \\ 0 & a^2+b^2 \end{bmatrix} \]
(ii)
\[ \begin{bmatrix} 1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4) \end{bmatrix} = \begin{bmatrix} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{bmatrix} \]
(iii)
\[ \begin{bmatrix} 1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\ 2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1) \end{bmatrix} = \begin{bmatrix} -3 & -4 & 1 \\ 8 & 13 & 9 \end{bmatrix} \]
(iv)
\[ \begin{bmatrix} 2+0+12 & -6+6+0 & 10+12+20 \\ 3+0+15 & -9+8+0 & 15+16+25 \\ 4+0+18 & -12+10+0 & 20+20+30 \end{bmatrix} = \begin{bmatrix} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{bmatrix} \]
(v)
\[ \begin{bmatrix} 2-1 & 0+2 & 2+1 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{bmatrix} \]
(vi)
\[ \begin{bmatrix} 6-1+9 & -9+0+3 \\ -2+0+6 & 3+0+2 \end{bmatrix} = \begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix} \]
Q4
If \( A = \begin{bmatrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{bmatrix} \), \( B = \begin{bmatrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{bmatrix} \) and \( C = \begin{bmatrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{bmatrix} \), then compute \( (A+B) \) and \( (B-C) \). Also, verify that \( A + (B - C) = (A + B) - C \).
Solution:
Compute \( A+B \):
\[ A+B = \begin{bmatrix} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{bmatrix} = \begin{bmatrix} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{bmatrix} \]
Compute \( B-C \):
\[ B-C = \begin{bmatrix} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0-(-2) & 3-3 \end{bmatrix} = \begin{bmatrix} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{bmatrix} \]
Verify \( A + (B - C) = (A + B) - C \):
LHS: \( A + (B-C) = \begin{bmatrix} 1-1 & 2-2 & -3+0 \\ 5+4 & 0-1 & 2+3 \\ 1+1 & -1+2 & 1+0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix} \)
RHS: \( (A+B) - C = \begin{bmatrix} 4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1-(-2) & 4-3 \end{bmatrix} = \begin{bmatrix} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{bmatrix} \)
Since LHS = RHS, verified.
Q5
If \( A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix} \) and \( B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix} \), then compute \( 3A - 5B \).
Solution:
\[ 3A = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix}, \quad 5B = \begin{bmatrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{bmatrix} \]
\[ 3A - 5B = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O \]
Q6
Simplify \( \cos \theta \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} + \sin \theta \begin{bmatrix} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{bmatrix} \)
Solution:
\[ \begin{aligned} &= \begin{bmatrix} \cos^2 \theta & \cos \theta \sin \theta \\ -\sin \theta \cos \theta & \cos^2 \theta \end{bmatrix} + \begin{bmatrix} \sin^2 \theta & -\sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix} \\ &= \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & 0 \\ 0 & \cos^2 \theta + \sin^2 \theta \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I \end{aligned} \]
Q7
Find X and Y, if
(i) \( X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \) and \( X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
(ii) \( 2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \) and \( 3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} \)
(i) \( X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} \) and \( X - Y = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
(ii) \( 2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} \) and \( 3X + 2Y = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix} \)
Solution:
(i) Adding the equations: \( 2X = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} \Rightarrow X = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix} \).
Subtracting: \( 2Y = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix} \Rightarrow Y = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \).
(ii) Multiply first by 3 and second by 2:
\( 6X + 9Y = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix} \)
\( 6X + 4Y = \begin{bmatrix} 4 & -4 \\ -2 & 10 \end{bmatrix} \)
Subtract: \( 5Y = \begin{bmatrix} 2 & 13 \\ 14 & -10 \end{bmatrix} \Rightarrow Y = \begin{bmatrix} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix} \).
Find X using \( 2X = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} - 3Y \). Result: \( X = \begin{bmatrix} \frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3 \end{bmatrix} \).
Q8
Find X, if \( Y = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} \) and \( 2X + Y = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} \)
Solution:
\[ 2X = \begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} - \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix} \]
\( X = \begin{bmatrix} -1 & -1 \\ -2 & -1 \end{bmatrix} \).
Q9
Find x and y, if \( 2 \begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
Solution:
\[ \begin{bmatrix} 2 & 6 \\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \]
\( \begin{bmatrix} 2+y & 6 \\ 1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix} \)
\( 2+y=5 \Rightarrow y=3 \). \( 2x+2=8 \Rightarrow 2x=6 \Rightarrow x=3 \).
Q10
Solve the equation for x, y, z and t, if \( 2 \begin{bmatrix} x & z \\ y & t \end{bmatrix} + 3 \begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix} = 3 \begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix} \)
Solution:
\[ \begin{bmatrix} 2x & 2z \\ 2y & 2t \end{bmatrix} + \begin{bmatrix} 3 & -3 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 9 & 15 \\ 12 & 18 \end{bmatrix} \]
Equations:
- \( 2x+3=9 \Rightarrow 2x=6 \Rightarrow x=3 \)
- \( 2z-3=15 \Rightarrow 2z=18 \Rightarrow z=9 \)
- \( 2y+0=12 \Rightarrow y=6 \)
- \( 2t+6=18 \Rightarrow 2t=12 \Rightarrow t=6 \)
Q11
If \( x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix} \), find the values of x and y.
Solution:
\( 2x - y = 10 \quad \dots(1) \)
\( 3x + y = 5 \quad \dots(2) \)
Adding equations: \( 5x = 15 \Rightarrow x = 3 \).
Substituting x in (1): \( 6 - y = 10 \Rightarrow y = -4 \).
Q12
Given \( 3 \begin{bmatrix} x & y \\ z & w \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2w \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+w & 3 \end{bmatrix} \), find the values of x, y, z and w.
Solution:
\( \begin{bmatrix} 3x & 3y \\ 3z & 3w \end{bmatrix} = \begin{bmatrix} x+4 & 6+x+y \\ -1+z+w & 2w+3 \end{bmatrix} \)
- \( 3x = x+4 \Rightarrow 2x = 4 \Rightarrow x = 2 \)
- \( 3y = 6+x+y \Rightarrow 2y = 6+2 \Rightarrow 2y=8 \Rightarrow y=4 \)
- \( 3w = 2w+3 \Rightarrow w=3 \)
- \( 3z = -1+z+w \Rightarrow 2z = -1+3 \Rightarrow 2z=2 \Rightarrow z=1 \)
Q13
If \( F(x) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \), show that \( F(x) F(y) = F(x+y) \).
Solution:
\[ F(x)F(y) = \begin{bmatrix} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
Multiplying rows by columns:
Row 1, Col 1: \( \cos x \cos y - \sin x \sin y = \cos(x+y) \)
Row 1, Col 2: \( -\cos x \sin y - \sin x \cos y = -(\sin x \cos y + \cos x \sin y) = -\sin(x+y) \)
Rest follows similarly to give \( F(x+y) \).
Q14
Show that:
(i) \( \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \)
(i) \( \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{bmatrix} \)
Solution:
This demonstrates that matrix multiplication is generally not commutative (\( AB \neq BA \)).
Compute LHS and RHS separately for each part and show the resulting matrices have different elements.
Q15
Find \( A^2 - 5A + 6I \), if \( A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \).
Solution:
First compute \( A^2 = A \cdot A \).
\( A^2 = \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} \).
Then compute \( A^2 - 5A + 6I \):
\[ \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} \]
Result: \( \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix} \).
Q16
If \( A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix} \), prove that \( A^3 - 6A^2 + 7A + 2I = 0 \).
Solution:
Calculate \( A^2 \) and \( A^3 \), then substitute into the equation to show it results in a zero matrix.
Q17
If \( A = \begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix} \) and \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \), find \( k \) so that \( A^2 = kA - 2I \).
Solution:
\( A^2 = \begin{bmatrix} 1 & -2 \\ 4 & -4 \end{bmatrix} \).
\( kA - 2I = \begin{bmatrix} 3k-2 & -2k \\ 4k & -2k-2 \end{bmatrix} \).
Equating elements: \( 4k = 4 \Rightarrow k=1 \). Also check \( -2k = -2 \Rightarrow k=1 \).
Q18
If \( A = \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix} \) and \( I \) is the identity matrix of order 2, show that \( I + A = (I - A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix} \).
Solution:
Let \( t = \tan \frac{\alpha}{2} \). Then \( A = \begin{bmatrix} 0 & -t \\ t & 0 \end{bmatrix} \).
LHS \( = I+A = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \).
RHS involves multiplying \( (I-A) \) with rotation matrix. Substitute \( \cos \alpha = \frac{1-t^2}{1+t^2} \) and \( \sin \alpha = \frac{2t}{1+t^2} \) to simplify.
Q19
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
(a) ₹1800
(b) ₹2000
Solution:
Let investment in first bond be \( x \). Then second is \( 30000-x \).
Matrix representation: \( \begin{bmatrix} x & 30000-x \end{bmatrix} \begin{bmatrix} 5/100 \\ 7/100 \end{bmatrix} = [ \text{Total Interest} ] \).
(a) \( \frac{5x}{100} + \frac{7(30000-x)}{100} = 1800 \Rightarrow -2x + 210000 = 180000 \Rightarrow 2x = 30000 \Rightarrow x = 15000 \).
Invest ₹15000 in each.
(b) \( \frac{5x + 210000 - 7x}{100} = 2000 \Rightarrow -2x = -10000 \Rightarrow x = 5000 \).
Invest ₹5000 in first and ₹25000 in second.
Q20
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 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution:
Quantities: \( 120, 96, 120 \). Prices: \( 80, 60, 40 \).
Matrix: \( \begin{bmatrix} 120 & 96 & 120 \end{bmatrix} \begin{bmatrix} 80 \\ 60 \\ 40 \end{bmatrix} \).
Total = \( 120(80) + 96(60) + 120(40) = 9600 + 5760 + 4800 = 20160 \).
Total amount is ₹20,160.
Q21
The restriction on \( n, k \) and \( p \) so that \( PY + WY \) will be defined are:- \( k = 3, p = n \)
- \( k \) is arbitrary, \( p = 2 \)
- \( p \) is arbitrary, \( k = 3 \)
- \( k = 2, p = 3 \)
- \( k = 3, p = n \)
- \( k \) is arbitrary, \( p = 2 \)
- \( p \) is arbitrary, \( k = 3 \)
- \( k = 2, p = 3 \)
Answer: (A)
Solution:
Order of P: \( p \times k \). Order of Y: \( 3 \times k \).
For PY to be defined, cols of P = rows of Y, so \( k = 3 \). Result PY is \( p \times k \).
Order of W: \( n \times 3 \). Order of Y: \( 3 \times k \). WY is \( n \times k \).
For addition PY + WY, orders must match: \( p \times k = n \times k \Rightarrow p = n \).
Q22
If \( n = p \), then the order of the matrix \( 7X - 5Z \) is:- \( p \times 2 \)
- \( 2 \times n \)
- \( n \times 3 \)
- \( p \times n \)
- \( p \times 2 \)
- \( 2 \times n \)
- \( n \times 3 \)
- \( p \times n \)
Answer: (B)
Solution:
Order of X is \( 2 \times n \). Order of Z is \( 2 \times p \).
Since \( n=p \), both are \( 2 \times n \).