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 Part I – 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)
Also Read,
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.
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Chapter 3 - Matrices
Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.1
In this article, we will be going to solve the entire Miscellaneous Exercise 3.1 of Chapter 3 of the NCERT textbook. A matrix is a rectangular array of numbers, symbols, or expressions, arranged in rows and columns. The numbers, symbols, or expressions in the matrix are called elements or entries. M
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.2
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 enclo
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.2 | Set 2
The content of this article has been merged with Chapter 3 Matrices- Exercise 3.2 as per the revised syllabus of NCERT. Chapter 3 of the Class 12 NCERT Mathematics Part I textbook, titled "Matrices," introduces the fundamental concepts and operations related to matrices. This chapter covers various
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 3 Matrices - Exercise 3.3
Chapter 3 of the Class 12 NCERT Mathematics textbook, titled "Matrices," delves into the fundamental concepts of matrices, including their types, operations, and applications. Exercise 3.3 focuses on practical problems involving matrix operations, such as addition, subtraction, and multiplication of
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.4 | Set 1
Using elementary transformations, find the inverse of each of the matrices, if it exists in Exercises 1 to 6.Question 1.[Tex]\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right][/Tex] Solution: [Tex]\begin{aligned} &\text { Since we know that, } A= IA \Rightarrow\left[\begin{array}
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.4 | Set 2
Matrices are a fundamental concept in mathematics providing a compact way to represent and manipulate data and equations. They are used in various fields including algebra, calculus, and applied mathematics. This chapter focuses on the operations and properties of matrices essential for solving the
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Miscellaneous Exercise on Chapter 3
Chapter 3 of the Class 12 NCERT Mathematics textbook, titled "Matrices," covers fundamental concepts related to matrices, including operations such as addition, multiplication, and finding determinants and inverses. The Miscellaneous Exercise in this chapter provides a range of problems that integra
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Chapter 4 - Determinants
Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.1
Chapter 4 of Class 12 NCERT Mathematics Part I introduces the concept of determinants, a crucial topic in linear algebra. Exercise 4.1 focuses on the fundamentals of determinants, including their definition, calculation for 2x2 and 3x3 matrices, and their basic properties. This exercise lays the gro
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.2 | Set 1
Exercise 4.2 specifically focuses on the properties and applications of determinants, building upon the basic definitions introduced earlier in the chapter. This exercise aims to deepen students' understanding of how determinants can be manipulated and applied to solve various mathematical problems.
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants- Exercise 4.2 | Set 2
Exercise 4.2 of Chapter 4 in the NCERT Class 12 Mathematics Part I textbook delves deeper into the properties and applications of determinants. This set builds upon the foundational concepts introduced earlier, challenging students with more complex problems that require a nuanced understanding of d
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.3
Question 1. Find area of the triangle with vertices at the point given in each of the following : (i) (1, 0), (6, 0), (4, 3) (ii) (2, 7), (1, 1), (10, 8) (iii) (–2, –3), (3, 2), (–1, –8) Solution: (i) (1, 0), (6, 0), (4, 3) [Tex]Area\ of\ triangle=\frac{1}{2}\begin{vmatrix}1 & 0 & 1\\6&0
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.4
Exercise 4.4 of Chapter 4 in the NCERT Class 12 Mathematics Part I textbook focuses on advanced applications of determinants and their properties. This exercise builds upon the foundational knowledge established in previous sections, challenging students to apply determinant techniques to more compl
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.5
NCERT Solutions for Class 9 Maths, Chapter 1, Number Systems, Exercise 1.3, meticulously crafted by our subject experts, facilitating effortless learning for students. These solutions serve as a valuable reference for students tackling exercise problems. Exercise 1.3 delves into the realm of real nu
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Exercise 4.6 | Set 1
Examine the consistency of the system of equations in Exercises 1 to 6.Question 1. x + 2y = 2 2x + 3y = 3 Solution: Matrix form of the given equations is AX = B where, A =[Tex]\begin{bmatrix}1 & 2 \\2 & 3 \\\end{bmatrix} [/Tex] , B = [Tex]\begin{bmatrix}2 \\3 \\\end{bmatrix} [/Tex] and, X =
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 4 Determinants - Exercise 4.6 | Set 2
Chapter 4 Determinants - Exercise 4.6 | Set 1Question 11. 2x + y + z = 1 x - 2y - z = 3/2 3y - 5z = 9 Solution: Matrix form of the given equation is AX = B i.e.[Tex]\begin{bmatrix}2 & 1 & 1\\1 & -2 & -1\\0 & 3 & -5\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatri
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 4 Determinants - Miscellaneous Exercises on Chapter 4
Question 1. Prove that the determinant [Tex]\begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\theta & 1 & x \end{vmatrix} [/Tex]is independent of θ. Solution: A = [Tex]\begin{vmatrix} x & sin\theta & cos\theta \\ -sin\theta & -x & 1\\ cos\
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Chapter 5 - Continuity and Differentiability
Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.1
Question 1. Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5. Solution: To prove the continuity of the function f(x) = 5x – 3, first we have to calculate limits and function value at that point. Continuity at x = 0 Left limit = [Tex]\lim_{x \to 0^-} f(x) = \lim_{
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.1 | Set 2
Chapter 5 on Continuity and Differentiability is a crucial part of calculus in Class 12 mathematics. It builds upon the concept of limits and introduces students to the fundamental ideas of continuous functions and differentiation. This chapter lays the groundwork for understanding rates of change,
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.2
Chapter 5 of the Class 12 NCERT Mathematics textbook focuses on the concepts of continuity and differentiability which are fundamental in understanding calculus. This chapter helps students grasp how functions behave concerning their limits and derivatives. Exercise 5.2 is designed to test and reinf
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.3
Chapter 5 of the Class 12 NCERT Mathematics textbook, "Continuity and Differentiability," introduces the concepts of continuity and differentiability of functions, which are fundamental in calculus. Exercise 5.3 focuses on applying these concepts to solve problems related to the continuity and diffe
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.4
Differentiate the following w.r.t xQuestion 1. y = [Tex]\frac{e^x}{\sin x}[/Tex] Solution: [Tex]\frac{dy}{dx}=\frac{d}{dx}(\frac{e^x}{\sin x})[/Tex] [Tex]\frac{dy}{dx}=\frac{\sin x\frac{d}{dx}e^x-e^x\frac{d}{dx}\sin x}{sin^2x} [/Tex] ([Tex]\frac{d}{dx}(\frac{u}{v})=v\frac{\frac{du}{dx}-u\frac{dv}{dx
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5
Differentiate the functions given in question 1 to 10 with respect to x.Question 1. cos x.cos2x.cos3x Solution: Let us considered y = cos x.cos2x.cos3x Now taking log on both sides, we get log y = log(cos x.cos2x.cos3x) log y = log(cos x) + log(cos 2x) + log (cos 3x) Now, on differentiating w.r.t x,
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.5 | Set 2
Exercise 5.5 focuses on the Mean Value Theorem and Rolle's Theorem, which are fundamental concepts in calculus. These theorems provide powerful tools for analyzing functions and their behavior over intervals. This exercise helps students understand how to apply these theorems to solve various mathem
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.6
If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find [Tex]\frac{dy}{dx}[/Tex]Question 1. x = 2at2, y = at4 Solution: Here, x = 2at2, y = at4 [Tex]\frac{dx}{dt} = \frac{d(2at^2)}{dt}[/Tex] = 2a [Tex]\frac{d(t^2)}{dt}[/Tex] = 2a (
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.7
Exercise 5.7 focuses on the application of derivatives in approximation and errors. This exercise explores how derivatives can be used to estimate function values and calculate errors in measurements or approximations. It introduces students to concepts like absolute and relative errors, percentage
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Exercise 5.8
Note: Please note that Exercise 5.8 from Chapter 5, "Continuity and Differentiability" in the NCERT Solutions, has been removed from the revised syllabus. As a result, this exercise will no longer be a part of your study curriculum. Exercise 5.8 focuses on the application of derivatives to find tang
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 5 Continuity And Differentiability - Miscellaneous Exercise on Chapter 5
In Chapter 5 of the Class 12 NCERT Mathematics textbook, titled Continuity and Differentiability, students explore fundamental concepts related to the behavior of functions. The chapter emphasizes understanding continuity, differentiability, and their implications in calculus. It includes various ex
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Chapter 6 - Applications of Derivatives
Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.1
The study of derivatives is a cornerstone in calculus providing the essential tools for understanding and analyzing functions. The Application of Derivatives a key topic in Class 12 Mathematics involves using the derivatives to solve real-world problems. This topic helps in understanding how rates o
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Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.2
The Application of derivatives is a crucial topic in calculus that involves using derivatives to solve practical problems and understand the various aspects of functions beyond their basic behavior. It helps in analyzing the rates of change optimizing the functions and understanding the geometric pr
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Class 12 NCERT Solutions- Mathematics Part I - Application of Derivatives - Exercise 6.2| Set 2
The chapter "Application of Derivatives" in Class 12 Mathematics is a critical part of the NCERT curriculum. It focuses on using derivatives to the solve real-world problems including rate of change, maxima and minima, tangents and normals. Exercise 6.2 specifically deals with the problems related t
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives -Exercise 6.3 | Set 2
In this article, we will see some problems of derivatives a fundamental concept in calculus, and mathematical analysis that measures how a function changes as input changes. Exercise 6.3 focuses on the application of derivatives to find the rate of change of quantities. This exercise builds upon the
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.4
Question 1. Using differentials, find the approximate value of each of the following up to 3 places of decimal. (i)√25.3 (ii)√49.5 (iii) √0.6 (iv) (0.009)1/3 (v) (0.999)1/10 (vi) (15)1/4 (vii) (26)1/3 (viii) (255)1/4 (ix) (82)1/4 (x) (401)1/2 (xi) (0.0037)1/2 (xii) (26.57)1/3 (xiii) (81.5)1/4 (xiv)
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.5 | Set 1
Question 1. Find the maximum and minimum values, if any, of the following function given by(i) f(x) = (2x - 1)2 + 3 Solution: Given that, f(x) = (2x - 1)2 + 3 From the given function we observe that (2x - 1)2 ≥ 0 ∀ x∈ R, So, (2x - 1)2 + 3 ≥ 3 ∀ x∈ R, Now we find the minimum value of function f when
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Class 12 NCERT Solutions - Mathematics Part I - Chapter 6 Application of Derivatives - Exercise 6.5 | Set 2
In Class 12 Mathematics, the chapter on Applications of Derivatives is one of the most important topics. This chapter focuses on how derivatives are applied to solve real-world problems such as finding the rate of change determining the slope of a curve, or calculating the maximum and minimum values
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Miscellaneous Exercise on Chapter 6 | Set 1
Chapter 6 of the Class 12 NCERT Mathematics textbook, titled "Application of Derivatives," focuses on how derivatives are used in various real-life situations and mathematical problems. The chapter covers concepts such as finding the rate of change of quantities, determining maxima and minima, and a
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Class 12 NCERT Solutions- Mathematics Part I - Chapter 6 Application of Derivatives - Miscellaneous Exercise on Chapter 6 | Set 2
Content of this article has been merged with Chapter 6 Application of Derivatives - Miscellaneous Exercise as per the revised syllabus of NCERT. Chapter 6 of the Class 12 NCERT Mathematics textbook, titled "Application of Derivatives," is essential for understanding how derivatives are applied in re
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