Matrices are a fundamental concept in linear algebra, used extensively in mathematics, physics, engineering, computer science, and various other fields. A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. The numbers or elements inside a matrix are enclosed within square brackets and can be used to represent systems of linear equations, transformations, and many real-world data structures. Operations such as addition, subtraction, multiplication, and finding determinants and inverses of matrices help solve complex mathematical problems efficiently.
Class 12 NCERT Solutions- Mathematics- Exercise 3.2
Question 1. Let [Tex]A =\begin{bmatrix}3 & 4 \\3 & 2 \\\end{bmatrix},B=\begin{bmatrix}1 & 3 \\-2 & 5 \\\end{bmatrix}, C=\begin{bmatrix}-2 & 5 \\3 & 4 \\\end{bmatrix} [/Tex]
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
Solution:
(i) [Tex]A+B=\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]+\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right] \\ =\left[\begin{array}{ll} 2+1 & 4+3 \\ 3-2 & 2+5 \end{array}\right] \\ =\left[\begin{array}{ll} 3 & 7 \\ 1 & 7 \end{array}\right][/Tex]
(ii) [Tex]A-B=\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]-\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right] \\ =\left[\begin{array}{cc} 2-1 & 4-3 \\ 3 & -(-2) & 2-5 \end{array}\right] \\ =\left[\begin{array}{cc} 1 & 1 \\ 5 & -3 \end{array}\right] [/Tex]
(iii) [Tex]3 A-C=3\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]-\left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right] \\ =\left[\begin{array}{ll} 3 \times 2 & 3 \times 4 \\ 3 \times 3 & 3 \times 2 \end{array}\right]-\left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right] \\ =\left[\begin{array}{lc} 6 & 12 \\ 9 & 6 \end{array}\right]-\left[\begin{array}{cc} -2 & 5 \\ 3 & 4 \end{array}\right] \\ =\left[\begin{array}{ll} 6+2 & 12-5 \\ 9 -3 & 6-4 \end{array}\right] \\ =\left[\begin{array}{ll} 8 & 7 \\ 6 & 2 \end{array}\right] [/Tex]
(iv) [Tex]A B=\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right]\\ =\left[\begin{array}{ll} 2(1)+4(-2) & 2(3)+4(5) \\ 3(1)+2(-2) & 3(3)+2(5) \end{array}\right]\\ =\left[\begin{array}{ll} 2-8 & 6+20 \\ 3-4 & 9+10 \end{array}\right]\\ =\left[\begin{array}{ll} -6 & 26 \\ -1 & 19 \end{array}\right] [/Tex]
(v) [Tex]BA =\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right]\left[\begin{array}{cc} 2 & 4 \\ 3 & 2 \end{array}\right] \\ =\left[\begin{array}{cc} 1(2)+3(3) & 1(4)+3(2) \\ -2(2)+5(3) & -2(4)+5(2) \end{array}\right] \\ =\left[\begin{array}{rr} 2+9 & 4+6 \\ -4+15 & -8+10 \end{array}\right] \\ =\left[\begin{array}{cc} 11 & 10 \\ 11 & 2 \end{array}\right] [/Tex]
Question 2. Compute the following:
[Tex](i)\begin{bmatrix}a & b \\-b & a \\\end{bmatrix}+\begin{bmatrix}a & b \\b & a \\\end{bmatrix}\\ (i)\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}\\ (i)\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}\\ (i)\begin{bmatrix}cos^{2} & sin^{2} \\sin^{2} & cos^{2} \\\end{bmatrix}+\begin{bmatrix}sin^{2} & cos^{2} \\cos^{2} & sin^{2} \\\end{bmatrix}\\[/Tex]
Solution:
(i) [Tex]{\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]+\left[\begin{array}{cc} a & b \\ b & a \end{array}\right]} \\ =\left[\begin{array}{cc} a+a & b+b \\ -b+b & a+a \end{array}\right] \\ =\left[\begin{array}{cc} 2 a & 2 b \\ 0 & 2 a \end{array}\right] [/Tex]
(ii) [Tex]{\left[\begin{array}{l} a^{2}+b^{2} & b^{2}+c^{2} \\ a^{2}+c^{2} & a^{2}+b^{2} \end{array}\right]+\left[\begin{array}{cc} 2 a b & 2 b c \\ -2 a c & -2 a b \end{array}\right]} \\ =\left[\begin{array}{ll} a^{2}+b^{2}+2 a b & b^{2}+c^{2}+2 b c \\ a^{2}+c^{2}-2 a c & a^{2}+b^{2}-2 a b \end{array}\right] \\ =\left[\begin{array}{c} (a+b)^{2}&(b+c)^{2} \\ (a-c)^{2} & (a-b)^{2} \end{array}\right] [/Tex]
(iii) [Tex]{\left[\begin{array}{ccc} -1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5 \end{array}\right]+\left[\begin{array}{ccc} 12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4 \end{array}\right]} \\ =\left[\begin{array}{ccc} -1+12 & 4+7 & -6+6 \\ 8+8 & 5+0 & 16+5 \\ 2+3 & 8+2 & 5+4 \end{array}\right] \\ =\left[\begin{array}{ccc} 11 & 11 & 0 \\ 16 & 5 & 21 \\ 5 & 10 & 9 \end{array}\right] [/Tex]
(iv) [Tex]{\left[\begin{array}{ll} \cos ^{2} x & \sin ^{2} x \\ \sin ^{2} x & \cos ^{2} x \end{array}\right]+\left[\begin{array}{ll} \sin ^{2} x & \cos ^{2} x \\ \cos ^{2} x & \sin ^{2} x \end{array}\right]} \\ =\left[\begin{array}{cc} \cos ^{2} x+\sin ^{2} x & \cos ^{2} x+\sin ^{2} x \\ \sin ^{2} x+\cos ^{2} x & \cos ^{2} x+\sin ^{2} x \end{array}\right] \\ =\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right][/Tex]
Question 3. Compute the indicated products.
[Tex](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}\\[/Tex]
Solution:
(i) [Tex]{\left[\begin{array}{cc} a & b \\ -b & a \end{array}\right]\left[\begin{array}{cc} a & -b \\ b & a \end{array}\right]} \\ =\left[\begin{array}{cc} a(a)+b(b) & a(-b)+b(a) \\ -b(a)+a(b) & -b(-b)+a(a) \end{array}\right] \\ =\left[\begin{array}{cc} a^{2}+b^{2} & -a b+a b \\ -a b+a b & b^{2}+a^{2} \end{array}\right] \\ =\left[\begin{array}{cc} a^{2}+b^{2} & 0 \\ 0 & b^{2}+a^{2} \end{array}\right] [/Tex]
(ii) [Tex]{\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]\left[\begin{array}{lll} 2 & 3 & 4 \end{array}\right]} \\ =\left[\begin{array}{lll} 1(2) & 1(3) & 1(4) \\ 2(2) & 2(3) & 2(4) \\ 3(2) & 3(3) & 3(4) \end{array}\right] \\ =\left[\begin{array}{lll} 2 & 3 & 4 \\ 4 & 6 & 8 \\ 6 & 9 & 12 \end{array}\right] [/Tex]
(iii) [Tex]{\left[\begin{array}{cc} 1 & -2 \\ 2 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right]} \\ =\left[\begin{array}{llll} 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{array}\right] \\ =\left[\begin{array}{lll} 1-4 & 2-6 & 3-2 \\ 2+6 & 4+9 & 6+3 \end{array}\right] \\ =\left[\begin{array}{ccc} -3 & -4 & 1 \\ 8 & 13 & 9 \end{array}\right] [/Tex]
(iv) [Tex]\left[\begin{array}{ccc} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{array}\right]\left[\begin{array}{rrr} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{array}\right] \\ =\left[\begin{array}{l} 2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\ 3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\ 4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5) \end{array}\right] \\ =\left[\begin{array}{lll} 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{array}\right] \\ =\left[\begin{array}{lll} 14 & 0 & 42 \\ 18 & -1 & 56 \\ 22 & -2 & 70 \end{array}\right] [/Tex]
(v) [Tex]\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 1 \\ -1 & 2 & 1 \end{array}\right] \\ =\left[\begin{array}{cccc} 2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\ 3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\ -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 \\ 3-2 & 0+4 & 3+2 \\ -1-1 & 0+2 & -1+1 \end{array}\right] \\ =\left[\begin{array}{ccc} 1 & 2 & 3 \\ 1 & 4 & 5 \\ -2 & 2 & 0 \end{array}\right] [/Tex]
(vi) [Tex]{\left[\begin{array}{ccc} 3 & -1 & 3 \\ -1 & 0 & 2 \end{array}\right]\left[\begin{array}{cc} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{array}\right]} \\ =\left[\begin{array}{cc} 3(2)-1(1)+3(3) & 3(-3)-1(0)+3(1) \\ -1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1) \end{array}\right] \\ =\left[\begin{array}{cc} 6-1+9 & -9-0+3 \\ -2+0+6 & 3+0+2 \end{array}\right] \\ =\left[\begin{array}{cc} 14 & -6 \\ 4 & 5 \end{array}\right] [/Tex]
Question 4. If [Tex]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} [/Tex], then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Solution:
[Tex]\begin{array}{l} A+B=\left[\begin{array}{ccc} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right]+\left[\begin{array}{ccc} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{array}\right] \\ {\left[\begin{array}{ccc} 1+3 & 2-1 & -3+2 \\ 5+4 & 0+2 & 2+5 \\ 1+2 & -1+0 & 1+3 \end{array}\right]=\left[\begin{array}{cccc} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{array}\right]} & \\ B-C=\left[\begin{array}{ccc} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{array}\right]-\left[\begin{array}{ccc} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right]=\left[\begin{array}{ccc} 3-4 & -1-1 & 2-2 \\ 4-0 & 2-3 & 5-2 \\ 2-1 & 0+2 & 3-3 \end{array}\right]=\left[\begin{array}{ccc} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{array}\right] \end{array}[/Tex]
Now we have to show A + (B – C) = (A + B) – C
[Tex]\Rightarrow\left[\begin{array}{ccc} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{array}\right]+\left[\begin{array}{ccc} -1 & -2 & 0 \\ 4 & -1 & 3 \\ 1 & 2 & 0 \end{array}\right]=\left[\begin{array}{ccc} 4 & 1 & -1 \\ 9 & 2 & 7 \\ 3 & -1 & 4 \end{array}\right]-\left[\begin{array}{ccc} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{array}\right]\\ \Rightarrow\left[\begin{array}{ccc} 1-1 & 2-2 & -3+0 \\ 5+4 & 0-1 & 2+3 \\ 1+1 & -1+2 & 1+0 \end{array}\right]=\left[\begin{array}{ccc} 4-4 & 1-1 & -1-2 \\ 9-0 & 2-3 & 7-2 \\ 3-1 & -1+2 & 4-3 \end{array}\right]\\ \Rightarrow\left[\begin{array}{ccc} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{array}\right]=\left[\begin{array}{ccc} 0 & 0 & -3 \\ 9 & -1 & 5 \\ 2 & 1 & 1 \end{array}\right][/Tex]
L.H.S = R.H.S.
Hence, Proved
Question 5. If[Tex] A=\begin{bmatrix}2/3 & 1 & 5/3\\1/3 & 2/3 & 4/3\\7/3 & 2 & 2/3\end{bmatrix}and \ B=\begin{bmatrix}2/5 & 3/5 & 1\\1/5 & 2/5 & 4/5\\7/5 & 6/5 & 2/5\end{bmatrix} [/Tex], then compute 3A – 5B.
Solution:
[Tex]\begin{array}{l} 3 A -5 B =3\left[\begin{array}{ccc} \frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{array}\right]-5\left[\begin{array}{ccc} \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{array}\right] \\ =\left[\begin{array}{rrr} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{array}\right]-\left[\begin{array}{ccc}\\ 2 -2 & 3-3 & 5 -5 \\ 1 -1 & 2-2 & 4 -4 \\ 7 -7 & 6-6 & 2 -2 \end{array}\right]=\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \end{array}[/Tex]
Question 6. Simplify [Tex]cosθ\begin{bmatrix}cosθ & sinθ \\-sinθ & cosθ \\\end{bmatrix}+sinθ\begin{bmatrix}sinθ& -cosθ\\cosθ & sinθ\\\end{bmatrix}[/Tex]
Solution:
[Tex]\begin{aligned} &\cos \theta\left[\begin{array}{cc} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]+\sin \theta\left[\begin{array}{cc} \sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]\\ &=\left[\begin{array}{cc} \cos ^{2} \theta & \sin \theta \cos \theta \\ -\sin \theta \cos \theta & \cos ^{2} \theta \end{array}\right]+\left[\begin{array}{cc} \sin ^{2} \theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin ^{2} \theta \end{array}\right]\\ &=\left[\begin{array}{cc} \cos ^{2} \theta+\sin ^{2} \theta & \sin \theta \cos \theta-\sin \theta \cos \theta \\ -\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta+\sin ^{2} \theta \end{array}\right]\\ &=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1 \mid\right.\\ \end{aligned}[/Tex]
= 1 = identity matrix
Question 7. Find X and Y if
(i) [Tex]X + Y =\begin{bmatrix}7 & 0 \\2 & 5 \\\end{bmatrix}\:and\:X-Y=\begin{bmatrix}3 & 0 \\0 & 3 \\\end{bmatrix}\\ [/Tex]
(ii) [Tex]2X+3Y=\begin{bmatrix}2 & 3\\4 & 0 \\\end{bmatrix}\:and\:3X+2Y=\begin{bmatrix}2 & -2 \\-1 & 5 \\\end{bmatrix}[/Tex]
Solution:
(i) Given: [Tex]X+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right] \ \ \ -(1)[/Tex]
[Tex]X-Y=\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right] \ \ \ -(2)[/Tex]
Adding (1) and (2), we get
[Tex]2 X=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]+\left[\begin{array}{ll} 3 & 0 \\ 0 & 3 \end{array}\right]\\ =\left[\begin{array}{ll} 7+3 & 0+0 \\ 2+0 & 5+3 \end{array}\right]\\ =\left[\begin{array}{cc} 10 & 0 \\ 2 & 8 \end{array}\right]\\ \Rightarrow X=\frac{1}{2}\left[\begin{array}{ll} 10 & 0 \\ 2 & 8 \end{array}\right]=\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right]\\ X+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]\\ \Rightarrow\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right]+Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]\\ \Rightarrow Y=\left[\begin{array}{ll} 7 & 0 \\ 2 & 5 \end{array}\right]-\left[\begin{array}{ll} 5 & 0 \\ 1 & 4 \end{array}\right]\\ \Rightarrow Y=\left[\begin{array}{ll} 2 & 0 \\ 1 & 1 \end{array}\right][/Tex]
(ii) Given: [Tex]2 X+3 Y=\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \ \ \ -(1)\\ 3 X+2 Y=\left[\begin{array}{cc} 2 & -2 \\ -1 & 5 \end{array}\right] \ \ \ -(2)[/Tex]
Now, multiply equation (1) by 2 and equation (2) by 3 we get
[Tex]4 X+6 Y=\left[\begin{array}{ll} 4 & 6 \\ 8 & 0 \end{array}\right] \ \ \ -(3)\\ 9 X+6 Y=\left[\begin{array}{cc} 6 & -6 \\ -3 & 15 \end{array}\right] \ \ \ -(4)[/Tex]
Subtracting equation (4) from (3), we get,
[Tex](4 X+6 Y)-(9 X+6 Y)=\left[\begin{array}{ll} 4 & 6 \\ 8 & 0 \end{array}\right]-\left[\begin{array}{cc} 6 & -6 \\ -3 & 15 \end{array}\right]\\ \Rightarrow-5 X=\left[\begin{array}{cc} 4-6 & 6-(-6) \\ 8-(-3) & 0-15 \end{array}\right]\\ =\left[\begin{array}{cc} -2 & 12 \\ 11 & -15 \end{array}\right]\\ \Rightarrow X=-\frac{1}{5}\left[\begin{array}{cc} -2 & 12 \\ 11 & -15 \end{array}\right]=\left[\begin{array}{cc} \frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3 \end{array}\right] [/Tex]
[Tex]2 X +3 Y =\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \\ \Rightarrow 2\left[\begin{array}{ll} \frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3 \end{array}\right]+3 Y =\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \\ \Rightarrow\left[\begin{array}{ll} \frac{4}{5} & \frac{-24}{5} \\ \frac{-22}{5} & 6 \end{array}\right]+3 Y =\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right] \\ \Rightarrow 3 Y =\left[\begin{array}{ll} 2 & 3 \\ 4 & 0 \end{array}\right]-\left[\begin{array}{cc} \frac{4}{5} & \frac{-24}{5} \\ \frac{-22}{5} & 6 \end{array}\right] \\ \Rightarrow Y=\frac{1}{3}\left[\begin{array}{cc} \frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6 \end{array}\right] \\ \Rightarrow Y=\left[\begin{array}{cc} \frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2 \end{array}\right] [/Tex]
Question 8. Find X, if [Tex]Y=\begin{bmatrix}3 & 2 \\1 & 4 \\\end{bmatrix} [/Tex]and [Tex]2X + Y=\begin{bmatrix}1 &0\\-3 & 2 \\\end{bmatrix}[/Tex]
Solution:
[Tex]\begin{array}{l} 2 X+Y=\left[\begin{array}{cc} 1 & 0 \\ -3 & 2 \end{array}\right] \\ \Rightarrow 2 X+\left[\begin{array}{cc} 3 & 2 \\ 1 & 4 \end{array}\right]=\left[\begin{array}{cc} 1 & 0 \\ -3 & 2 \end{array}\right] \\ \Rightarrow 2 X=\left[\begin{array}{cc} 1 & 0 \\ -3 & 2 \end{array}\right]-\left[\begin{array}{cc} 3 & 2 \\ 1 & 4 \end{array}\right] \\ \Rightarrow 2 X=\left[\begin{array}{cc} 1-3 & 0-2 \\ -3 & -1 & 2-4 \end{array}\right] \\ \Rightarrow 2 X=\left[\begin{array}{cc} -2 & -2 \\ -4 & -2 \end{array}\right] \\ \Rightarrow X=\frac{1}{2}\left[\begin{array}{cc} -2 & -2 \\ -4 & -2 \end{array}\right] \\ ∴ X=\left[\begin{array}{cc} -1 & -1 \\ -2 & -1 \end{array}\right] \end{array}[/Tex]
Question 9. Find X and Y, if [Tex]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}[/Tex]
Solution:
Given: [Tex]2\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right]\\ \Rightarrow\left[\begin{array}{ll} 2 & 6 \\ 0 & 2 x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2 \end{array}\right]=\left[\begin{array}{ll} 5 & 6 \\ 1 & 8 \end{array}\right]\\ \Rightarrow\left[\begin{array}{cc} 2+y & 6 \\ 1 & 2 x+x \end{array}\right]^{2}=\left[\begin{array}{cc} 5 & 6 \\ 1 & 8 \end{array}\right][/Tex]
Equating corresponding entries, we have
2 + y = 5 and 2x + 2 = 8
y = 5 – 2 and 2(x + 1) = 8
y = 3 and x + 1 = 4
Therefore, y = 3 and x = 3
Question 10. Solve the equation for x, y, z and t, if [Tex]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}[/Tex]
Solution:
Given: [Tex]2\left[\begin{array}{ll} x & z \\ y & t \end{array}\right]+3\left[\begin{array}{cc} 1 & -1 \\ 0 & 2 \end{array}\right]=3\left[\begin{array}{ll} 3 & 5 \\ 4 & 6 \end{array}\right]\\ \Rightarrow\left[\begin{array}{ll} 2 x & 2 z \\ 2 y & 2 t \end{array}\right]+\left[\begin{array}{cc} 3 & -3 \\ 0 & 6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right]\\ \Rightarrow\left[\begin{array}{ll} 2 x+3 & 2 z-3 \\ 2 y+0 & 2 t+6 \end{array}\right]=\left[\begin{array}{cc} 9 & 15 \\ 12 & 18 \end{array}\right][/Tex]
On comparing both sides, we have
2x + 3 = 9 ⇒ 2x = 9 – 3 ⇒ 2x = 6 ⇒ x = 3
2z – 3 = 15 ⇒ 2z = 15 + 3 ⇒ 2z = 18 ⇒ z = 9
2y = 12 ⇒ y = 6
2t + 6 = 18 ⇒ 2t = 18 – 6 ⇒ 2t = 12 ⇒ t = 6
Therefore, x = 3, y = 6, z = 9, t = 6
Question 11. If [Tex]x\left[\begin{array}{l} 2 \\ 3 \end{array}\right]+y\left[\begin{array}{c} -1 \\ 1 \end{array}\right]=\left[\begin{array}{l} 10 \\ 5 \end{array}\right] [/Tex], find the values of x and y.
Solution:
Given: [Tex]x\left[\begin{array}{l} 2 \\ 3 \end{array}\right]+y\left[\begin{array}{c} -1 \\ 1 \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right] \\ \Rightarrow\left[\begin{array}{c} 2 x \\ 3 x \end{array}\right]+\left[\begin{array}{c} -y \\ y \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right] \\ \Rightarrow\left[\begin{array}{c} 2 x-y \\ 3 x+y \end{array}\right]=\left[\begin{array}{c} 10 \\ 5 \end{array}\right] [/Tex]
Equating corresponding entries, we have
2x – y = 10 -(1)
3x + y = 5 -(2)
Adding eq.(1) and (2), we have 5x = 15 ⇒ x = 3
Putting x = 3 in eq.(2)
9 + y = 5 ⇒ y = -4
Therefore, x = 3 and y = -4
Question 12. Given [Tex]3\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]=\left[\begin{array}{cc} x & 0 \\ -1 & 2 w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right] [/Tex], find the values of x, y, z and w.
Solution:
Given: [Tex]3\left[\begin{array}{cc} x & y \\ z & w \end{array}\right]=\left[\begin{array}{cc} x & 0 \\ -1 & 2 w \end{array}\right]+\left[\begin{array}{cc} 4 & x+y \\ z+w & 3 \end{array}\right][/Tex]
[Tex]\Rightarrow\left[\begin{array}{ll} 3 x & 3 y \\ 3 z & 3 w \end{array}\right]=\left[\begin{array}{cc} x+4 & 6+x+y \\ -1+z+w & 2 w+3 \end{array}\right][/Tex]
Equating corresponding entries, we have
3x = x + 4 ⇒ 2x = 4 ⇒ x = 2
and 3y = 6 + x + y
⇒ 2y = 6 + 2
⇒ 2y = 8
⇒ y = 4
and 3z = -1 + z + w ⇒ 2z – w = – 1 -(1)
and 3w = 2w + 3 ⇒ w = 3
Putting w = 3 in eq(i), 2z – 3 = -1
⇒ 2z = 2 ⇒ z = 1
Therefore, x = 2, y = 4, z = 1, w = 3
Question 13. If [Tex]F(x)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right] [/Tex], show that F(x) F(y) = F(x + y).
Solution:
[Tex]\begin{aligned} &\text { } F(x)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right], F(y)=\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &F(x+y)=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &F(x) F(y)=\left[\begin{array}{ccc} \cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} \cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1 \end{array}\right] \end{aligned}[/Tex]
[Tex]=\left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1 \end{array}\right][/Tex]
[Tex]=\left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y+0 & -\cos x \sin y-\sin x \cos y+0 & 0 \\ \sin x \cos y+\cos x \sin y+0 & -\sin x \sin y+\cos x \cos y+0 & 0 \\ 0 & 0 & 0 \end{array}\right][/Tex]
= F(x + y)
= F(x) F(y) = F(x + y)
Question 14. Show that
[Tex](i) \left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right] \neq\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right][/Tex]
[Tex]\text { (ii) }\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \neq\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right][/Tex]
Solution:
(i) L.H.S =[Tex]\left[\begin{array}{cc} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\\ =\left[\begin{array}{ll} 5(2)-1(3) & 5(1)-1(4) \\ 6(2)+7(3) & 6(1)+7(4) \end{array}\right]\\ =\left[\begin{array}{cc} 10-3 & 5-4 \\ 12+21 & 6+28 \end{array}\right]\\ =\left[\begin{array}{cc} 7 & 1 \\ 33 & 34 \end{array}\right] \ \ \ -(1)[/Tex]
R.H.S = [Tex]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{ll} 5 & -1 \\ 6 & 7 \end{array}\right]\\ =\left[\begin{array}{ll} 2(5)+1(6) & 2(-1)+1(7) \\ 3(5)+4(6) & 3(-1)+4(7) \end{array}\right]\\ =\left[\begin{array}{cc} 10+6 & -2+7 \\ 15+24 & -3+28 \end{array}\right]\\ =\left[\begin{array}{ll} 16 & 5 \\ 39 & 25 \end{array}\right] \ \ \ -(2) [/Tex]
Therefore, from (1) and (2), we get
[Tex]\text { }\left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right]\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right] \neq\left[\begin{array}{ll} 2 & 1 \\ 3 & 4 \end{array}\right]\left[\begin{array}{rr} 5 & -1 \\ 6 & 7 \end{array}\right][/Tex]
i.e. L.H.S. ≠ R.H.S
(ii) L.H.S = [Tex]\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] [/Tex]
Multiply both the matrices
[Tex]=\left[\begin{array}{lll} 1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \\ 0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\ 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} 5 & 8 & 14 \\ 0 & -1 & 1 \\ -1 & 0 & 1 \end{array}\right] [/Tex]
R.H.S.= [Tex]\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right][/Tex]
[Tex]\begin{array}{l} =\left[\begin{array}{ccc} -1(1)+1(0)+0(1) & (-1) 2+1(1)+0(1) & (-1) 3+1(0)+0(0) \\ 0(1)+(-1) 0+1(1) & (0) 2+1(-1)+1(1) & (0) 3+0(-1)+1(0) \\ 2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0) \end{array}\right] \\ =\left[\begin{array}{ccc} -1 & -1 & -3 \\ 1 & 0 & 0 \\ 6 & 11 & 6 \end{array}\right] \end{array}[/Tex]
Therefore,
L.H.S. ≠ R.H.S.
i.e.[Tex]\text { }\left[\begin{array}{rrr} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right]\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right] \neq\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{array}\right]\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{array}\right][/Tex]
Question 15. Find A2 – 5A + 6I, if [Tex]A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&0\\\end{bmatrix}[/Tex]
Solution:
[Tex]\begin{aligned} &A^{2}-5 A+6 I=\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]-5\left[\begin{array}{ccc} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{array}\right]+6\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\\ &=\left[\begin{array}{lll} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1-0 & 1-3+0 \end{array}\right]-\left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]+\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \end{aligned}[/Tex]
[Tex]\left.\begin{array}{l} =\left[\begin{array}{ccc} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{array}\right]-\left[\begin{array}{ccc} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{array}\right]+\left[\begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \end{array}\right][/Tex]
[Tex]\begin{array}{l} =\left[\begin{array}{ccc} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1+5+0 & -2+0+6 \end{array}\right] \\ =\left[\begin{array}{ccc} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{array}\right] \end{array}[/Tex]
Question 16. If [Tex]A =\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\\\end{bmatrix} [/Tex], prove that A3 – 6A2 + 7A + 2I = 0
Solution:
[Tex]A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\\\end{bmatrix} A^{2} \\=A * A=\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] \\ =\left[\begin{array}{lll} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{array}\right] \\ =\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right] [/Tex]
[Tex]6 A^{2} =6\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]=\left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right] \\ A^{3} =A^{2} \times A \\ =\left[\begin{array}{lll} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right] \\ =\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right][/Tex]
[Tex]A^{3} – 6 A^{2}+7 A+2 I=\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-\left[\begin{array}{ccc} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{array}\right]+\left[\begin{array}{ccc} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{array}\right]+\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right]\\ =\left[\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] [/Tex]
= 0 (Zero matrix)
= R.H.S.
Hence Proved
Question 17. If [Tex]A=\begin{bmatrix}3&-2\\4&-2\\\end{bmatrix} and \:I=\begin{bmatrix}1&0\\0&1\\\end{bmatrix} [/Tex], find k so that A2 = kA – 2I
Solution:
Given:
[Tex]A=\left[\begin{array}{rr} 3 & -2 \\ 4 & -2 \end{array}\right] \text { and } I=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ A^{2}=k A-2 I \Rightarrow\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]=k\left[\begin{array}{ll} 3 & -2 \\ 4 & -2 \end{array}\right]-2\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\\ \Rightarrow\left[\begin{array}{cc} 9-8 & -6+4 \\ 12-8 & -8+4 \end{array}\right]=\left[\begin{array}{cc} 3 k & -2 k \\ 4 k & -2 k \end{array}\right]-\left[\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right]\\ \Rightarrow\left[\begin{array}{ll} 1 & -2 \\ 4 & -4 \end{array}\right]=\left[\begin{array}{ll} 3 k-2 & -2 k-0 \\ 4 k-0 & -2 k-2 \end{array}\right][/Tex]
Equating corresponding entries, we have
3k – 2 = 1
3k = 3
k = 1
and 4k = 4
k = 1
and -4 = -2k – 2
2k = 2
k = 1
Therefore, k = 1
Question 18. If [Tex]A =\begin{bmatrix}0&-tan\frac{α}{2}\\tan\frac{α}{2}&0\\\end{bmatrix} [/Tex]and I is the identity matrix of order 2, show that I + A = (I – A)[Tex]\begin{bmatrix}cosα&-sinα\\sinα&cosα\\\end{bmatrix}[/Tex]
Solution:
[Tex]\begin{array}{l} \text { L.H.S. } I+A=\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]+\left[\begin{array}{cc} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right] \\ \text { Now, } I-A=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]-\left[\begin{array}{cc} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{array}\right]=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right] \\ \text { R.H.S. }=(I-A)\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right]=\left[\begin{array}{cc} 1 & \tan \frac{\alpha}{2} \\ -\tan \frac{\alpha}{2} & 1 \end{array}\right]\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] \end{array}[/Tex]
[Tex]\begin{aligned} &=\left[\begin{array}{ccc} \cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\ -\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha \\ \end{array}\right]\\ &\text {} \end{aligned}[/Tex]
[Tex]=\left[\begin{array}{ccc} \cos \alpha \cos \frac{\alpha}{2}+\sin \alpha \sin \frac{\alpha}{2}{\cos \frac{\alpha}{2}} & \frac{\alpha \sin \alpha \cos \frac{\alpha}{2}+\cos \alpha \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} & \ \\ \hline \frac{-\cos \alpha \sin \frac{\alpha}{2}+\sin \alpha \cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} & \frac{\sin \alpha \sin \frac{\alpha}{2}+\cos \alpha \cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \end{array}\right][/Tex]
[Tex]\begin{aligned} &=\left[\begin{array}{cc} \frac{\cos \left(\alpha-\frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}} & \frac{-\sin \left(\alpha-\frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}} \\ \frac{\sin \left(\alpha-\frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}} & \frac{\cos \left(\alpha-\frac{\alpha}{2}\right)}{\cos \frac{\alpha}{2}} \end{array}\right]=\left[\begin{array}{ccc} \frac{\cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} & \frac{-\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \\ \frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} & \frac{\cos \frac{\alpha}{2}}{\cos \frac{\alpha}{2}} \end{array}\right]=\left[\begin{array}{cc} 1 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 1 \end{array}\right]\end{aligned}[/Tex]
L.H.S. = R.H.S.
Hence, Proved.
Question 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) Rs.1800
(b) Rs.2000
Solution:
Let invested in the first bond = Rs x
Then, the sum of money invested in the second bond = ₹(30000 – x)
It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.
Thus, in order to obtain an annual total interest of ₹1800, we get:
[Tex]\begin{bmatrix}x&30000-x\end{bmatrix}\begin{bmatrix}5/100\\7/100\end{bmatrix}=1800[/Tex]
⇒ 5x/100 + 7(30000 − x)/100 = 1800
⇒ 5x + 210000 -7x = 180000
⇒ 210000 -2x = 180000
⇒ 2x = 210000 – 180000
⇒ 2x = 30000
⇒ x = 15000
Therefore, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹15000 in the first bond and the remaining ₹15000 in the second bond.
Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:
X = [Tex]\begin{bmatrix}x&30000-x\end{bmatrix}[/Tex]
Hence, the interest obtained after one year can be expressed in matrix representation as:
[Tex]\begin{bmatrix}x&30000-x\end{bmatrix}\begin{bmatrix}5/100\\7/100\end{bmatrix}=2000[/Tex]
⇒ 5x/100 + 7(30000 − x)/100 = 2000
⇒ 5x + 210000 − 7x = 200000
⇒ 210000 − 2x = 200000
⇒ 2x = 210000 – 200000
⇒ 2x = 10000
⇒ x = 5000
Therefore, in order to obtain an annual total interest of ₹2000, the trust fund should invest ₹5000 in the first bond and the remaining ₹(30000 − 5000) = ₹25000 in the second bond.
Question 20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs.80, Rs.60 and Rs.40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution:
Let the number of books as 1 × 3 matrix = B = [Tex]\begin{bmatrix}10 dozen&8dozen&10dozen\\10*12=120&8*12=96&10*12=120\\\end{bmatrix}[/Tex]
Let the selling prices of each book is a 3 × 1 matrix S = [Tex]\begin{bmatrix}80\\60\\40\end{bmatrix}[/Tex]
Therefore, Total amount received by selling all books = BS = [Tex]\begin{bmatrix}120&96&120\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}[/Tex]
[Tex]\begin{bmatrix}120(80)&96(60)&120(40)\end{bmatrix}=\begin{bmatrix}9600&5760&4800\end{bmatrix}=\begin{bmatrix}20160\end{bmatrix}[/Tex]
Therefore, Total amount received by selling all the books = Rs 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.
Question 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
Solution:
Since, Matrices P and Y are of the orders p × k and 3 × k respectively.
Therefore, matrix PY will be defined if k = 3.
Then, PY will be of the order p × k = p × 3.
Matrices W and Y are of the orders n × 3 and 3 × k = 3 × 3 respectively.
As, the number of columns in W is equal to the number of rows in Y, Matrix WY is well-defined and is of the order n × 3.
Matrices PY and WY can be added only when their orders are the same.
Therefore, PY is of the order p × 3 and WY is of the order n × 3.
Thus, we must have p = n.
Therefore, k = 3 and p = n are the restrictions on n, k and p so that PY + WY will be defined.
Therefore, answer is (A)
Question 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
Solution:
Matrix X is of the order 2 × n.
Therefore, matrix 7X is also of the same order.
Matrix Z is of order 2 × p = 2 × n -(∵ p = n)
Then, Matrix 5Z is also of the same order.
Now, both the matrices 7X and 5Z are of the order 2 × n.
Thus, matrix 7X – 5Z is well- defined and is of the order 2 × n.
Therefore, answer is (B)
Conclusion
Matrices are a vital tool for solving real-world problems, from representing data to performing transformations in graphics and solving systems of linear equations. Exercise 3.2 deepens the understanding of fundamental matrix operations, helping students learn how to add, subtract, and multiply matrices. These operations are crucial for many advanced applications, including physics simulations, computer graphics, data analysis, and solving linear equations. By mastering these concepts, students can approach more complex problems involving matrices with confidence.
Class 12 NCERT Solutions- Mathematics Part I – Matrices – FAQs
What are the conditions for adding or subtracting two matrices?
Matrices can be added or subtracted only if they have the same dimensions, meaning they must have the same number of rows and columns. The operations are performed element-wise.
When is matrix multiplication possible?
Matrix multiplication is possible when the number of columns in the first matrix equals the number of rows in the second matrix. The resulting product matrix will have dimensions equal to the number of rows of the first matrix and the number of columns of the second matrix.
What is scalar multiplication in matrices?
Scalar multiplication involves multiplying each element of a matrix by a fixed number or scalar. This operation scales the matrix by that constant factor, maintaining the same dimensions.
What is the significance of matrices in real-life applications?
Matrices are used in various real-life applications, including computer graphics for image transformations, solving systems of linear equations, economic modeling, physics simulations, and data analysis. They help represent and manipulate data in a structured way.
