Ncert class 12 maths exercise 3.2 solutions | class 12 maths exercise 3.2 solutions | exercise 3.2 class 12 maths | ncert solutions matrices class 12 maths chapter 3 | class 12 math matrix ncert solution
Looking for NCERT Class 12 Maths Exercise 3.2 solutions? You’re in the right place! This section offers complete and step-by-step answers to all questions from Exercise 3.2 Class 12 Maths, based on Chapter 3 – Matrices from the NCERT textbook. These Class 12 Maths Exercise 3.2 solutions focus on important matrix operations like addition, subtraction, and scalar multiplication. Designed according to the latest CBSE guidelines, the NCERT Solutions Matrices Class 12 Maths Chapter 3 make it easier for students to build a solid foundation in matrix algebra. Whether you’re preparing for board exams or brushing up your concepts, these Class 12 Math Matrix NCERT Solutions provide the clarity and practice you need to succeed.
ncert class 12 maths exercise 3.2 solutions || exercise 3.2 class 12 maths || class 12 math matrix ncert solution || class 12 maths exercise 3.2 solutions || ncert solutions matrices class 12 maths chapter 3
Exercise 3.2
1 A. Let \( \mathrm{A}=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], \mathrm{B}=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], \mathrm{C}=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right] \)
Find each of the following:
i) \( A+B \)
ii) \( A-B \)
iii) \( 3 \mathrm{~A}-\mathrm{C} \)
iv) AB
v) BA
Find each of the following:
i) \( A+B \)
ii) \( A-B \)
iii) \( 3 \mathrm{~A}-\mathrm{C} \)
iv) AB
v) BA
Answer
(i) \(A+B=\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]+\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right] \)
\(=\left[\begin{array}{ll}
2+1 & 4+3 \\
3-2 & 2+5
\end{array}\right] \\
=\left[\begin{array}{ll}
3 & 7 \\
1 & 7
\end{array}\right]\)
(ii) \(\mathrm{A}-\mathrm{B}=\left[\begin{array}{ll} 2 & 4 \\
3 & 2 \end{array}\right]+\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\)
\(=\left[\begin{array}{cc}
2-1 & 4-3 \\
3-(-2) & 2-5
\end{array}\right] \)
\(=\left[\begin{array}{cc}
1 & 1 \\
5 & -3
\end{array}\right]\)
(iii) \(3 \mathrm{~A}-\mathrm{C} =3\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]-\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right] \)
\(=\left[\begin{array}{cc}
6 & 12 \\
9 & 6
\end{array}\right]-\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right]\)
\(=\left[\begin{array}{cc}
6-(-2) & 12-5 \\
9-3 & 6-4
\end{array}\right] \)
\(=\left[\begin{array}{ll}
8 & 7 \\
6 & 2
\end{array}\right]\)
(iv) \(AB=\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right] \)
\(=\left[\begin{array}{ll}
2(1)+4(-2) & 2(3)+4(5) \\
3(1)+2(-2) & 3(3)+5(2)
\end{array}\right] \)
\(=\left[\begin{array}{ll}
2-8 & 6+20 \\
3-4 & 9+10
\end{array}\right] \)
\(=\left[\begin{array}{ll}
-6 & 26 \\
-1 & 19
\end{array}\right]\)
(v) \(BA =\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right] \)
\(=\left[\begin{array}{cc}
1(2)+3(3) & 1(4)+3(2) \\
-2(2)+5(3) & -2(4)+5(2)
\end{array}\right] \)
\(=\left[\begin{array}{cc}
2+9 & 4+6 \\
-4+15 & -8+10
\end{array}\right] \)
\(=\left[\begin{array}{cc}
11 & 10 \\
11 & 2
\end{array}\right]\)
1 B. Let \( A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right] \)
Find each of the following:
\(
A-B
\)
Find each of the following:
\(
A-B
\)
Answer
\(\begin{array}{l}
A-B=\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]-\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right] \\
=\left[\begin{array}{cc}
2-1 & 4-3 \\
3-(-2) & 2-5
\end{array}\right] \\
=\left[\begin{array}{cc}
1 & 1 \\
5 & -3
\end{array}\right]
\end{array}
\)
1 C. Let \( A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right] \)
Find each of the following:
\(
3 \mathrm{~A}-\mathrm{C}
\)
Find each of the following:
\(
3 \mathrm{~A}-\mathrm{C}
\)
Answer
\(
\begin{array}{l}
3 \mathrm{~A}-\mathrm{C}=3\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]-\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right] \\
=\left[\begin{array}{cc}
6 & 12 \\
9 & 6
\end{array}\right]-\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right] \\
=\left[\begin{array}{cc}
6-(-2) & 12-5 \\
9-3 & 6-4
\end{array}\right] \\
=\left[\begin{array}{ll}
8 & 7 \\
6 & 2
\end{array}\right]
\end{array}
\)
1 D. Let \( A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right] \)
Find each of the following:
AB
Find each of the following:
AB
Answer
\(\begin{array}{l}
\mathrm{AB}=\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right] \\
=\left[\begin{array}{ll}
2(1)+4(-2) & 2(3)+4(5) \\
3(1)+2(-2) & 3(3)+2(5)
\end{array}\right] \\
=\left[\begin{array}{ll}
2-8 & 6+20 \\
3-4 & 9+10
\end{array}\right] \\
=\left[\begin{array}{ll}
-6 & 26 \\
-1 & 19
\end{array}\right]
\end{array}
\)
1 E. Let \( A=\left[\begin{array}{ll}2 & 4 \\ 3 & 2\end{array}\right], B=\left[\begin{array}{cc}1 & 3 \\ -2 & 5\end{array}\right], C=\left[\begin{array}{cc}-2 & 5 \\ 3 & 4\end{array}\right] \)
Find each of the following:
BA
Find each of the following:
BA
Answer
\(
\begin{array}{l}
\mathrm{BA}=\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right] \\
=\left[\begin{array}{cc}
1(2)+3(3) & 1(4)+3(2) \\
-2(2)+5(3) & -2(4)+5(2)
\end{array}\right] \\
=\left[\begin{array}{cc}
2+9 & 4+6 \\
-4+15 & -8+10
\end{array}\right] \\
=\left[\begin{array}{cc}
11 & 10 \\
11 & 2
\end{array}\right]
\end{array}
\)
ncert class 12 maths exercise 3.2 solutions || exercise 3.2 class 12 maths || class 12 math matrix ncert solution || class 12 maths exercise 3.2 solutions || ncert solutions matrices class 12 maths chapter 3
2 A. Compute the following:
\(
\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]
\)
\(
\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]
\)
Answer
\(\begin{array}{l}
{\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]} \\
=\left[\begin{array}{cc}
a+a & b+b \\
-b+b & a+a
\end{array}\right] \\
=\left[\begin{array}{cc}
2 a & 2 b \\
0 & 2 a
\end{array}\right]
\end{array}\)
2 B. Compute the following:
\(
\left[\begin{array}{ll}
a^{2}+b^{2} & b^{2}+c^{2} \\
a^{2}+c^{2} & a^{2}+b^{2}
\end{array}\right]+\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]
\)
\(
\left[\begin{array}{ll}
a^{2}+b^{2} & b^{2}+c^{2} \\
a^{2}+c^{2} & a^{2}+b^{2}
\end{array}\right]+\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{ll}
a^{2}+b^{2} & b^{2}+c^{2} \\
a^{2}+c^{2} & a^{2}+b^{2}
\end{array}\right]+\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]} \\
=\left[\begin{array}{ll}
a^{2}+b^{2}+2 a b & b^{2}+c^{2}+2 b c \\
a^{2}+c^{2}-2 a c & a^{2}+b^{2}-2 a b
\end{array}\right] \\
=\left[\begin{array}{ll}
(a+b)^{2} & (b+c)^{2} \\
(a-c)^{2} & (a-b)^{2}
\end{array}\right]
\end{array}
\)
2 C. Compute the following:
\(
\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]+\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]
\)
\(
\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]+\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]+\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]} \\
=\left[\begin{array}{ccc}
-1+12 & 4+7 & -6+6 \\
8+8 & 5+0 & 16+5 \\
2+3 & 8+2 & 5+4
\end{array}\right]
\end{array}
\)
\(
=\left[\begin{array}{ccc}
11 & 11 & 0 \\
16 & 5 & 21 \\
5 & 10 & 9
\end{array}\right]
\)
2 D. Compute the following:
\(
\left[\begin{array}{ll}
\cos ^{2} x & \sin ^{2} x \\
\sin ^{2} x & \cos ^{2} x
\end{array}\right]+\left[\begin{array}{ll}
\sin ^{2} x & \cos ^{2} x \\
\cos ^{2} x & \text { sinh } x
\end{array}\right]
\)
\(
\left[\begin{array}{ll}
\cos ^{2} x & \sin ^{2} x \\
\sin ^{2} x & \cos ^{2} x
\end{array}\right]+\left[\begin{array}{ll}
\sin ^{2} x & \cos ^{2} x \\
\cos ^{2} x & \text { sinh } x
\end{array}\right]
\)
Answer
\(\begin{array}{l}
{\left[\begin{array}{cc}
\cos ^{2} x & \sin ^{2} x \\
\sin ^{2} x & \cos ^{2}x
\end{array}\right]+\left[\begin{array}{cc}
\sin ^{2} x & \cos ^{2} x \\
\cos ^{2} x & \sin^{2} x
\end{array}\right]} \\
=\left[\begin{array}{cc}
\cos ^{2} x+\sin ^{2} x & \cos ^{2} x+\sin x \\
\sin ^{2} x+\cos ^{2} x & \cos ^{2} x+\sin ^{2} x
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]
\end{array}
\)
3 A. Compute the indicated products.
\(
\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]
\)
\(
\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]} \\
=\left[\begin{array}{cc}
a(a)+b(b) & a(-b)+b(a) \\
-b(a)+a(b) & -b(-b)+a(a)
\end{array}\right] \\
=\left[\begin{array}{cc}
a^{2}+b^{2} & -a b+a b \\
-a b+a b & a^{2}+b^{2}
\end{array}\right] \\
=\left[\begin{array}{cc}
a^{2}+b^{2} & 0 \\
0 & a^{2}+b^{2}
\end{array}\right]
\end{array}
\)
3 B.
Compute the indicated products.
\(
\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\left[\begin{array}{lll}
2 & 3 & 4
\end{array}\right]
\)
Compute the indicated products.
\(
\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\left[\begin{array}{lll}
2 & 3 & 4
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\left[\begin{array}{lll}
2 & 3 & 4
\end{array}\right]} \\
=\left[\begin{array}{lll}
1(2) & 1(3) & 1(4) \\
2(2) & 2(3) & 2(4) \\
3(2) & 3(3) & 3(4)
\end{array}\right] \\
=\left[\begin{array}{ccc}
2 & 3 & 4 \\
4 & 6 & 8 \\
6 & 9 & 12
\end{array}\right]
\end{array}
\)
3 C. Compute the indicated products.
\(
\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]
\)
\(
\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 2 & 1
\end{array}\right]} \\
=\left[\begin{array}{lll}
1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\
2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1)
\end{array}\right] \\
=\left[\begin{array}{ccc}
1-4 & 2-6 & 3-2 \\
2+6 & 4+9 & 6+3
\end{array}\right] \\
=\left[\begin{array}{ccc}
-3 & -4 & 1 \\
8 & 13 & 9
\end{array}\right]
\end{array}
\)
3 D.
Compute the indicated products.
\(
\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]
\)
Compute the indicated products.
\(
\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]} \\
=\left[\begin{array}{lll}
2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\
3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\
4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5)
\end{array}\right] \\
=\left[\begin{array}{ccc}
2+0+12 & -6+6+0 & 10+12+20 \\
3+0+15 & -9+8+0 & 15+16+25 \\
4+0+18 & -12+10+0 & 20+20+30
\end{array}\right] \\
=\left[\begin{array}{ccc}
14 & 0 & 42 \\
18 & -1 & 56 \\
22 & -2 & 70
\end{array}\right]
\end{array}
\)
3 E. Compute the indicated products.
\(
\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
-1 & 2 & 1
\end{array}\right]
\)
\(
\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
-1 & 2 & 1
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
-1 & 2 & 1
\end{array}\right]} \\
=\left[\begin{array}{ccc}
2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\
3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\
-1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1)
\end{array}\right]
\end{array}
\)
\(
\begin{array}{l}
=\left[\begin{array}{ccc}
2-1 & 0+2 & 2+1 \\
3-2 & 0+4 & 3+2 \\
-1-1 & 0+2 & -1+1
\end{array}\right] \\
=\left[\begin{array}{ccc}
1 & 2 & 3 \\
1 & 4 & 5 \\
-2 & 2 & 0
\end{array}\right]
\end{array}
\)
3 F . Compute the indicated products.
\(
\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]
\)
\(
\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]} \\
=\left[\begin{array}{cc}
3(2)-1(1)+3(3) & 3(-3)-1(0)+3(1) \\
-1(2)+0(1)+2(3) & 1(-3)+0(0)+2(1)
\end{array}\right] \\
=\left[\begin{array}{cc}
6-1+9 & -9-0+3 \\
-2+0+6 & -3+0+2
\end{array}\right] \\
=\left[\begin{array}{cc}
14 & -6 \\
4 & -1
\end{array}\right]
\end{array}
\)
4. If \( A=\left[\begin{array}{ccc}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right], B=\left[\begin{array}{ccc}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right] \), and \( C=\left[\begin{array}{ccc}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right] \) then compute \( (\mathrm{A}+\mathrm{B}) \) and \( (\mathrm{B}-\mathrm{C}) \). Also, verify that \( \mathrm{A}+(\mathrm{B}-\mathrm{C})=(\mathrm{A}+\mathrm{B})- \) C.
Answer
\(
\begin{array}{l}
\text { Now, } A+B=\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]+\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 5 & 5 \\
2 & 0 & 3
\end{array}\right] \\
=\left[\begin{array}{ccc}
1+3 & 2-1 & -3+2 \\
5+4 & 0+2 & 2+5 \\
1+2 & -1+0 & 1+3
\end{array}\right] \\
=\left[\begin{array}{ccc}
4 & 1 & -1 \\
9 & 2 & 7 \\
3 & -1 & 4
\end{array}\right] \\
B-C=\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]-\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right] \\
=\left[\begin{array}{ccc}
3-4 & -1-1 & 2-2 \\
4-0 & 2-3 & 5-2 \\
2-1 & 0-(-2) & 3-3
\end{array}\right] \\
=\left[\begin{array}{ccc}
-1 & -2 & 0 \\
4 & -1 & 3 \\
1 & 2 & 0
\end{array}\right] \\
A+(B-C)=\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]+\left[\begin{array}{ccc}
-1 & -2 & 0 \\
4 & -1 & 3 \\
1 & 2 & 0
\end{array}\right] \\
=\left[\begin{array}{ccc}
1+(-1) & 2+(-2) & -3+0 \\
5+4 & 0+(-1) & 2+3 \\
1+1 & -1+2 & 1+0
\end{array}\right] \\
=\left[\begin{array}{ccc}
0 & 0 & -3 \\
9 & -1 & 5 \\
2 & 1 & 1
\end{array}\right] \\
(A+B)-C=\left[\begin{array}{ccc}
4 & 1 & -1 \\
9 & 2 & 7 \\
3 & -1 & 4
\end{array}\right]-\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right]
\end{array}
\)
\(
\begin{array}{l}
=\left[\begin{array}{ccc}
4-4 & 1-1 & -1-2 \\
9-0 & 2-3 & 7-2 \\
3-1 & -1-(-2) & 4-3
\end{array}\right] \\
=\left[\begin{array}{ccc}
0 & 0 & -3 \\
9 & -1 & 5 \\
2 & 1 & 1
\end{array}\right]
\end{array}
\)
Therefore, \( \mathrm{A}+(\mathrm{B}-\mathrm{C})=(\mathrm{A}+\mathrm{B})-\mathrm{C} \)
ncert class 12 maths exercise 3.2 solutions || exercise 3.2 class 12 maths || class 12 math matrix ncert solution || class 12 maths exercise 3.2 solutions || ncert solutions matrices class 12 maths chapter 3
5. If \( A=\left[\begin{array}{ccc}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right] \), and \( B=\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{3} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{2}\end{array}\right] \) then compute \( 3 A-5 B \).
Answer
\( 3 \mathrm{~A}-5 \mathrm{~B}=3\left[\begin{array}{lll}\frac{2}{3} & 1 & \frac{5}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3}\end{array}\right]-5\left[\begin{array}{ccc}\frac{2}{5} & \frac{3}{5} & 1 \\ \frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5}\end{array}\right] \)
\( =\left[\begin{array}{lll}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right]-\left[\begin{array}{lll}2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2\end{array}\right] \)
\( =\left[\begin{array}{lll}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{array}\right] \)
6. Simplify \( \cos \theta\left[\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right]+\sin \theta\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta\end{array}\right] \)
Answer
\(
\begin{array}{l}
\operatorname{Cos} \theta\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right] \\
=\left[\begin{array}{cc}
\cos ^{2} \theta & \cos \theta \sin \theta \\
-\sin \theta \cos \theta & \cos ^{2} \theta
\end{array}\right]+\left[\begin{array}{cc}
\sin ^{2} \theta & -\sin \theta \cos \theta \\
\sin \theta \cos \theta & \sin^2 \theta
\end{array}\right] \\
=\left[\begin{array}{cc}
\cos ^{2} \theta+\sin ^2 \theta & \cos \theta \sin \theta-\sin \theta \cos \theta \\
-\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^{2} \theta+\sin ^{2} \theta
\end{array}\right] \\
=\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right]
\end{array}
\)
7 A. Find \( X \) and \( Y \), if
\(
X+Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right] \text { and } X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]
\)
\(
X+Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right] \text { and } X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]
\)
Answer
Now, \( X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right] \ldots (1)\)
\(
X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \ldots(2)
\)
Adding (1) and (2), we get,
\(
\begin{array}{l}
2 X=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right] \\
=\left[\begin{array}{cc}
7+3 & 0+0 \\
2+0 & 5+3
\end{array}\right] \\
=\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right] \\
=X=\frac{1}{2}\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]
\end{array}
\)
Now, \( X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right] \)
\(
=\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]+Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]
\)
\(
\begin{array}{l}
=Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]-\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right] \\
=Y=\left[\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right]
\end{array}
\)
7 B. Find \( X \) and \( Y \), if \( 2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right] \) and \( 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc}2 & -2 \\ -1 & 5\end{array}\right] \)
Answer
\( 2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 2\end{array}\right] \ldots(1) \)
\( 3 \mathrm{X}+2 \mathrm{Y}=\left[\begin{array}{cc}2 & -2 \\ -1 & 5\end{array}\right] \ldots(2)\)
Now, multiply equation (1) by 2 and equation (2) by 3 , we get,
\( 4 \mathrm{X}+6 \mathrm{Y}=\left[\begin{array}{ll}4 & 6 \\ 8 & 0\end{array}\right] \ldots(3)\)
\( 9 X+6 Y=\left[\begin{array}{cc}6 & -6 \\ -3 & 15\end{array}\right] \ldots(4)\)
Subtracting equation (4) from (3), we get,
\(
\begin{array}{l}
(4 \mathrm{X}+6 \mathrm{Y})-(9 \mathrm{X}+6 \mathrm{Y})=\left[\begin{array}{cc}
4 & 6 \\
8 & 0
\end{array}\right]-\left[\begin{array}{cc}
6 & -6 \\
-3 & 15
\end{array}\right] \\
=5 \mathrm{X}=\left[\begin{array}{cc}
4-6 & 6-(-6) \\
8-(-3) & 0-15
\end{array}\right] \\
=\left[\begin{array}{cc}
-2 & 12 \\
11 & -15
\end{array}\right] \\
=X=-\frac{1}{5}\left[\begin{array}{cc}
-2 & 12 \\
11 & -15
\end{array}\right]=\left[\begin{array}{cc}
\frac{2}{5} & \frac{-12}{5} \\
\frac{-11}{5} & 3
\end{array}\right]
\end{array}
\)
Now, \( 2 \mathrm{X}+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right] \)
\( =2\left[\begin{array}{cc}\frac{2}{5} & \frac{-12}{5} \\ \frac{-11}{5} & 3\end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right] \)
\( =\left[\begin{array}{cc}\frac{4}{5} & \frac{-24}{5} \\ \frac{-22}{5} & 6\end{array}\right]+3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right] \)
\( =3 \mathrm{Y}=\left[\begin{array}{ll}2 & 3 \\ 4 & 0\end{array}\right]-\left[\begin{array}{cc}\frac{4}{5} & \frac{-24}{5} \\ \frac{-22}{5} & 6\end{array}\right] \)
\( =3 \mathrm{Y}=\left[\begin{array}{ll}2-\frac{4}{5} & 3+\frac{24}{5} \\ 4+\frac{22}{5} & 0-6\end{array}\right]=\left[\begin{array}{cc}6 & \frac{39}{5} \\ \frac{42}{5} & -6\end{array}\right] \)
\( =\mathrm{Y}=\frac{1}{3}\left[\begin{array}{cc}\frac{6}{5} & \frac{39}{5} \\ \frac{42}{5} & -6\end{array}\right] \)
\( =\mathrm{Y}=\left[\begin{array}{cc}\frac{2}{5} & \frac{13}{5} \\ \frac{14}{5} & -2\end{array}\right] \)
8. Find \( X \), if \( Y=\left[\begin{array}{ll}3 & 2 \\ 1 & 4\end{array}\right] \) and \( 2 X+Y=\left[\begin{array}{cc}1 & 0 \\ -3 & 2\end{array}\right] \)
Answer
\(
\begin{array}{l}
2 X+Y=\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right] \\
=2 X+\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]
\end{array}
\)
\(
\begin{array}{l}
=2 X=\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right] \\
=2 X=\left[\begin{array}{cc}
1-3 & 0-2 \\
-3-1 & 2-4
\end{array}\right] \\
=2 X=\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right] \\
=X=\frac{1}{2}\left[\begin{array}{cc}
-2 & -2 \\
-4 & -2
\end{array}\right] \\
=X=\left[\begin{array}{ll}
-1 & -1 \\
-2 & -1
\end{array}\right]
\end{array}
\)
9. Find \( x \) and \( y \), if \( 2\left[\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right] \)
Answer
\(
\begin{array}{l}
2\left[\begin{array}{cc}
1 & 3 \\
0 & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right] \\
=\left[\begin{array}{cc}
2 & 6 \\
0 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right] \\
=\left[\begin{array}{cc}
2+y & 6 \\
1 & 2 x+2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]
\end{array}
\)
Now, on comparing elements of these two matrices, we get,
\(
\begin{array}{l}
2+y=5 \\
\Rightarrow y=3
\end{array}
\)
And \( 2 x+2=8 \)
\(
\Rightarrow x=3
\)
Therefore, \( x=3 \) and \( y=3 \).
10. Solve the equation for \( \mathrm{x}, \mathrm{y}, \mathrm{z} \) and t , if \( 2\left[\begin{array}{ll}x & z \\ y & t\end{array}\right]+3\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]=3\left[\begin{array}{ll}3 & 5 \\ 4 & 6\end{array}\right] \)
Answer
\(
\begin{array}{l}
2\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]+3\left[\begin{array}{cc}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right] \\
=\left[\begin{array}{cc}
2 x & 2 z \\
2 y & 2 t
\end{array}\right]+\left[\begin{array}{cc}
3 & -3 \\
0 & 6
\end{array}\right]=\left[\begin{array}{cc}
9 & 15 \\
12 & 18
\end{array}\right] \\
=\left[\begin{array}{cc}
2 x+3 & 2 z-3 \\
2 y & 2 t+6
\end{array}\right]=\left[\begin{array}{cc}
9 & 15 \\
12 & 18
\end{array}\right]
\end{array}
\)
On comparing the elements of these two matrices, we get,
\(
2 x+3=9\)
\(\Rightarrow 2 x=6\)
\(\Rightarrow x=3\)
\(2 y=12\)
\(\Rightarrow y=6\)
\(2 z-3=15\)
\(\Rightarrow 2 z=18\)
\(\Rightarrow z=9\)
\(2 \mathrm{t}+6=18\)
\(\Rightarrow 2 \mathrm{t}=12\)
\(\Rightarrow \mathrm{t}=6
\)
Therefore, \( x=3, y=6, z=9 \) and \( t=6 \).
11. If \( x\left[\begin{array}{l}2 \\ 3\end{array}\right]+y\left[\begin{array}{c}-1 \\ 1\end{array}\right]=\left[\begin{array}{c}10 \\ 5\end{array}\right] \), find the values of \( x \) and \( y \).
Answer
\(
\begin{array}{l}
x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{c}
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right] \\
=\left[\begin{array}{l}
2 x \\
3 x
\end{array}\right]+\left[\begin{array}{c}
-y \\
y
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right] \\
=\left[\begin{array}{l}
2 x-y \\
3 x+y
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]
\end{array}
\)
On comparing the corresponding elements of these two matrices, we get, \( 2 x-y=10 \) and \( 3 x+y=5 \)
Now, adding above two equations, we get
\(5 x=15\)
\(\Rightarrow x=3
\)
Now, \( 3 x+y=5 \)
\(\Rightarrow y=5-3 x\)
\(\Rightarrow y=5-9=-4
\)
Therefore, \( x=3 \) and \( y=-4 \).
12. Given \( 3\left[\begin{array}{cc}x & y \\ z & w\end{array}\right]=\left[\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right] \), find the values of \( x \), \( \mathrm{y}, \mathrm{z} \) and w .
Answer
\(
\begin{array}{l}
3\left[\begin{array}{cc}
x & y \\
z & w
\end{array}\right]=\left[\begin{array}{cc}
x & 6 \\
-1 & 2 w
\end{array}\right]+\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right] \\
=\left[\begin{array}{cc}
3 x & 3 y \\
3 z & 3 w
\end{array}\right]=\left[\begin{array}{cc}
x+4 & 6+w+y \\
-1+z+w & 2 w+3
\end{array}\right]
\end{array}
\)
On comparing the corresponding elements of these two matrices, we get,
\(
3 x=x+4\)
\(\Rightarrow 2 x+4\)
\(\Rightarrow x=2\)
\(3 y=6+x+y\)
\(\Rightarrow 2 y=6+x=6+2=8\)
\(\Rightarrow y=4\)
\(3 w=2 w+3\)
\(\Rightarrow w=3\)
\(3 z=-1+z+w\)
\(\Rightarrow 2 z=-1+w=-1+3=2\)
\(\Rightarrow z=1
\)
Therefore, \( x=2, y=4, z=1 \) and \( w=3 \).
13. If \( \mathrm{F}(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0
\\ 0 & 0 & 1\end{array}\right] \) show that \( \mathrm{F}(x) \mathrm{F}(y)=\mathrm{F}(x+y) \).
\\ 0 & 0 & 1\end{array}\right] \) show that \( \mathrm{F}(x) \mathrm{F}(y)=\mathrm{F}(x+y) \).
Answer
\(
\begin{array}{l}
\mathrm{F}(x)=\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right], \mathrm{F}(y)=\left[\begin{array}{ccc}
\cos y & -\sin y & 0 \\
\sin x & \cos y & 0 \\
0 & 0 & 1
\end{array}\right] \\
\mathrm{F}(x+y)=\left[\begin{array}{ccc}
\cos (x+y) & -\sin (x+y) & 0 \\
\sin (x+y) & \cos (x+y) & 0 \\
1 & 1 & 1
\end{array}\right] \\
\mathrm{F}(x) \mathrm{F}(y)=\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\operatorname{sn} x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
\cos y & -\sin y & 0 \\
\sin y & \cos y & 0 \\
0 & 0 & 1
\end{array}\right]
\end{array}
\)
\( \begin{array}{l}=\left[\begin{array}{ccc}\cos x \cos y-\sin x \sin y+0 & -\cos x \sin y-\sin x \cos y+0 & 0 \\ \sin x \cos y+\cos x \sin y+0 & -\sin x \sin y+\cos x \cos y \ 0 & 0 \\ 0 & 0 & 1\end{array}\right] \\ =\left[\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1\end{array}\right] \\ =\mathrm{F}(x+y)\end{array} \)
Therefore, \( F(x) \mathrm{F}(y)=\mathrm{F}(x+y) \)
14 A. Show that \( \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right] \neq\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right] \)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right]} \\
=\left[\begin{array}{cc}
5(2)-1(3) & 5(-1)-1(4) \\
6(2)+7(3) & 6(1)+7(4)
\end{array}\right] \\
=\left[\begin{array}{cc}
10-3 & 5-4 \\
12+21 & 6+28
\end{array}\right] \\
=\left[\begin{array}{cc}
7 & 1 \\
33 & 34
\end{array}\right] \\
{\left[\begin{array}{cc}
2 & 1 \\
3 & 4
\end{array}\right]\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]} \\
=\left[\begin{array}{cc}
2(5)+1(6) & 2(-1)+1(7) \\
3(5)+4(6) & 3(-1)+4(7)
\end{array}\right] \\
=\left[\begin{array}{cc}
10+6 & -2+7 \\
15+24 & -3+28
\end{array}\right] \\
=\left[\begin{array}{cc}
16 & 5 \\
39 & 25
\end{array}\right]
\end{array}
\)
Therefore, \( \left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right] \neq\left[\begin{array}{cc}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{cc}5 & -1 \\ 6 & 7\end{array}\right] \)
14 B. Show that
\(
\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right] \neq\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]
\)
\(
\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right] \neq\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]
\)
Answer
\(
\begin{array}{l}
{\left[\begin{array}{ccc}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]} \\
{\left[\begin{array}{ccc}
1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \\
0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\
1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4)
\end{array}\right]} \\
=\left[\begin{array}{ccc}
5 & 8 & 14 \\
0 & -1 & 1 \\
-1 & 0 & 1
\end{array}\right] \\
{\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]} \\
{\left[\begin{array}{ccc}
-1(1)+1(0)+0(1) & -1(2)+1(1)+0(1) & -1(3)+1(0)+0(0) \\
0(1)+(-1)(0)+1(1) & 0(2)+(-1)(1)+1(1) & 0(3)+(-1)(0)+1(0) \\
2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0)
\end{array}\right]} \\
=\left[\begin{array}{ccc}
-1 & -1 & -3 \\
1 & 0 & 0 \\
6 & 11 & 6
\end{array}\right]
\end{array}
\)
Therefore, \( \left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right] \neq\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right] \)
ncert class 12 maths exercise 3.2 solutions || exercise 3.2 class 12 maths || class 12 math matrix ncert solution || class 12 maths exercise 3.2 solutions || ncert solutions matrices class 12 maths chapter 3
15. Find \( A^{2}-5 A+6 I \), if \( A=\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right] \)
Answer
\(
\begin{array}{l}
\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A}=\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right] \\
=\left[\begin{array}{ccc}
4+0+1 & 0+0-1 & 2+0+0 \\
4+2+3 & 0+1-3 & 2+3+0 \\
2-2+0 & 0-1+0 & 1-3+0
\end{array}\right] \\
=\left[\begin{array}{ccc}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]
\end{array}
\)
Now, \( \mathrm{A}^{2}-5 \mathrm{~A}+6 \mathrm{I}=\left[\begin{array}{ccc}5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2\end{array}\right]-5\left[\begin{array}{ccc}2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0\end{array}\right]+6\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \)
\(
\begin{array}{l}
=\left[\begin{array}{ccc}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]-\left[\begin{array}{ccc}
10 & 0 & 5 \\
10 & 5 & 15 \\
5 & -5 & 0
\end{array}\right]+\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right] \\
=\left[\begin{array}{ccc}
5-10 & -1-0 & 2-5 \\
9-10 & -2-5 & 5-15 \\
0-5 & -1+5 & -2-0
\end{array}\right]+\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right] \\
=\left[\begin{array}{ccc}
-5 & -1 & -3 \\
-1 & -7 & -10 \\
-5 & 4 & -2
\end{array}\right]+\left[\begin{array}{ccc}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right] \\
=\left[\begin{array}{ccc}
5+6 & -1+0 & -3+0 \\
-1+0 & -7+6 & -10+0 \\
-5+0 & 4+0 & -2+6
\end{array}\right]
\end{array}
\)
\(
=\left[\begin{array}{ccc}
1 & -1 & -3 \\
-1 & -1 & -10 \\
-5 & 4 & 4
\end{array}\right]
\)
16. If \( A=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right] \), prove that \( A^{3}-6 A^{2}+7 A+2 I=0 \)
Answer
\(
\begin{array}{l}
\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A}=\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right] \\
=\left[\begin{array}{lll}
1+0+4 & 0+0+0 & 2+0+6 \\
0+0+2 & 0+4+0 & 0+2+3 \\
2+0+6 & 0+0+0 & 4+0+9
\end{array}\right] \\
=\left[\begin{array}{ccc}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right]
\end{array}
\)
Now, \( A^{3}=A^{2} \cdot A=\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right] \)
\(
\begin{array}{l}
=\left[\begin{array}{ccc}
5+0+16 & 0+0+0 & 10+0+24 \\
2+0+10 & 0+8+0 & 4+4+15 \\
8+0+26 & 0+0+0 & 16+0+39
\end{array}\right] \\
=\left[\begin{array}{ccc}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]
\end{array}
\)
Now, \( \mathrm{A}^{3}-6 \mathrm{~A}^{2}+7 \mathrm{~A}+2 \mathrm{I}=\left[\begin{array}{ccc}21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55\end{array}\right]-6\left[\begin{array}{ccc}5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13\end{array}\right]+ \)
\(
7\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]+2\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]
\)
\(
\begin{array}{l}
=\left[\begin{array}{ccc}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]-\left[\begin{array}{ccc}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right]+\left[\begin{array}{ccc}
7 & 0 & 14 \\
0 & 14 & 7 \\
4 & 0 & 21
\end{array}\right]+\left[\begin{array}{ccc}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right] \\
=\left[\begin{array}{ccc}
21+7+2 & 0+0+0 & 34+14+0 \\
12+0+0 & 8+14+2 & 23+7+0 \\
34+14+0 & 0+0+0 & 55+21+2
\end{array}\right]-\left[\begin{array}{ccc}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right] \\
=\left[\begin{array}{ccc}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right]-\left[\begin{array}{ccc}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right] \\
=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=0
\end{array}
\)
Therefore, \( \mathrm{A}^{3}-6 \mathrm{~A}^{2}+7 \mathrm{~A}+2 \mathrm{I}=0 \)
17. If \( A=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right] \) and \( I=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \), find \( k \) so that \( A^{2}=k A-2 I \)
Answer
\(
\begin{array}{l}
\mathrm{A}^{2}=\mathrm{A} \cdot \mathrm{A}=\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right] \\
=\left[\begin{array}{ll}
3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\
4(3)+(-2)(4) & 4(-2)+(-2)(-2)
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]
\end{array}
\)
Now, \( \mathrm{A}^{2}=\mathrm{kA}-2 \mathrm{I} \)
\(
\begin{array}{l}
=\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=k\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=\left[\begin{array}{ll}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]-2\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=\left[\begin{array}{cc}
3 k-2 & -2 k \\
4 k & -2 k-2
\end{array}\right]
\end{array}
\)
Comparing the corresponding elements, we get,
\(
3 k-2=1\)
\(\Rightarrow 3 k=3\)
\(\Rightarrow k=1
\)
Therefore, the value of k is 1 .
18. If \( A=\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right] \) and I is the identity matrix of order 2, show that
\(
\mathrm{I}+\mathrm{A}=(\mathrm{I}-\mathrm{A})\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]
\)
\(
\mathrm{I}+\mathrm{A}=(\mathrm{I}-\mathrm{A})\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]
\)
Answer
We know that \( \mathrm{I}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \)
Now, LHS \( =\mathrm{I}+\mathrm{A}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right] \)
\(
=\left[\begin{array}{cc}
1 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 1
\end{array}\right]
\)
And on RHS \( =(\mathrm{I}-\mathrm{A})\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right] \)
\(
\begin{array}{l}
=\left(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\right)\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
=\left[\begin{array}{cc}
1 & \tan \frac{\alpha}{2} \\
-\tan \frac{\alpha}{2} & 1
\end{array}\right]\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]
\end{array}
\)
\(
=\left[\begin{array}{cc}
\cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\
-\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha
\end{array}\right]
\)
As we know, \( \cos 2 \theta=1-2 \sin 2 \theta \)
\( \cos 2 \theta=2 \cos 2 \theta-1 \) and \( \sin 2 \theta=2 \sin \theta \cos \theta \)
\( \left[\begin{array}{cc}1-2 \sin^{2} \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\left(2 \cos ^{2} \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2} \\ -\left(2 \cos ^{2}\right) \tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \tan \frac{\alpha}{2}+1-2 \sin \frac{\alpha}{2}\end{array}\right] \)
As \( \tan \theta=\sin \theta / \cos \theta \)
\(
\begin{array}{l}
{\left[\begin{array}{cc}
1-2 \sin \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} & -2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}-\tan \frac{\alpha}{2} \\
-2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}+\tan \frac{\alpha}{2}+2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} & 2 \sin \frac{\alpha}{2}+1-2 \sin \frac{\alpha}{2}
\end{array}\right]} \\
=\left[\begin{array}{cc}
1 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 1
\end{array}\right] \\
=\mathrm{LHS}
\end{array}
\)
Hence Proved.
19 A. A trust fund has Rs. 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 Rs. 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: Rs 1800
Answer
(a) Let Rs \(x\) be invested in the first bond.
Then, the sum of money invested in the second bond will be Rs. ( 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 Rs. 1800, we get:
\(
\begin{array}{l}
{[x(3000-x)]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]=1800} \\ \end{array}
\)
\(=\frac{5 x}{100}+\frac{7(3000-x)}{100}=1800\)
\(\Rightarrow 5 x+210000-7 x=180000\)
\(\Rightarrow 210000-2 x=180000\)
\(\Rightarrow 2 x=210000-180000\)
\(\Rightarrow 2 x=30000\)
\(\Rightarrow x=15000\)
Therefore, in order to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs. 15000 in the first bond and the remaining Rs. 15000 in the second bond.
19 B. A trust fund has Rs. 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 Rs. 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: Rs. 2000
Answer
(a) Let Rs \( x \) be invested in the first bond.
Then, the sum of money invested in the second bond will be Rs ( \( 30000- \) \( x) \).
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 Rs 1800 , we get:
\(
\begin{array}{l}
{[x(30000-x)]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]=2000} \\ \end{array}
\)
\(=\frac{5 x}{100}+\frac{7(30000-x)}{100}=2000\)
\(\Rightarrow 5 x+210000-7 x=200000\)
\(\Rightarrow 210000-2 x=200000\)
\(\Rightarrow 2 x=210000-200000\)
\(\Rightarrow 2 x=10000\)
\(\Rightarrow x=5000\)
Therefore, in order to obtain an annual total interest of Rs 2000, the trust fund should invest Rs 5000 in the first bond and the remaining Rs 25000 in the second bond.
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.
Assume X, Y, Z, W and Pare matrices of order \( 2 \times \mathrm{n}, 3 \times \mathrm{k}, 2 \times \mathrm{p}, \mathrm{n} \times 3 \) and \( \mathrm{p} \times \mathrm{k} \), respectively.
Assume X, Y, Z, W and Pare matrices of order \( 2 \times \mathrm{n}, 3 \times \mathrm{k}, 2 \times \mathrm{p}, \mathrm{n} \times 3 \) and \( \mathrm{p} \times \mathrm{k} \), respectively.
Answer
It is given that the bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.
Number of chemistry book \( =10 \times 12=120 \) Number of physics book \( =8 \) \( \times 12=96 \) Number of economics book \( =10 \times 12=120 \)
Their selling prices are Rs 80 , Rs 60 and Rs 40 each respectively.
Let A be the matrix of books per subject. 4
\( A=\left[\begin{array}{lll}120 & 96 & 120\end{array}\right] \) Let \( B \) denotes selling price of the books. \( B=\left[\begin{array}{l}80 \\ 60 \\ 40\end{array}\right] \)
The total amount of money \( = \) number of books \( \times \) selling price
\( \left[\begin{array}{lll}120 & 96 & 120\end{array}\right]\left[\begin{array}{l}80 \\ 60 \\ 40\end{array}\right] \)
\(=[120 \times 80+96 \times 60+120 \times 40]\)
\(=(9600+5760+4800)\)
\(=20160\)
Therefore, the bookshop will receive Rs. 20160 from the sale of all these books.
21. The restriction on \( \mathrm{n}, \mathrm{k} \) and p so that \( \mathrm{PY}+\mathrm{WY} \) will be defined are:
A. \( \mathrm{k}=3, \mathrm{p}=\mathrm{n} \) B. k is arbitrary, \( \mathrm{p}=2 \) C. p is arbitrary, \( \mathrm{k}=3 \) D. \( k=2, p=3 \)
A. \( \mathrm{k}=3, \mathrm{p}=\mathrm{n} \) B. k is arbitrary, \( \mathrm{p}=2 \) C. p is arbitrary, \( \mathrm{k}=3 \) D. \( k=2, p=3 \)
Answer
Matrices P and Y are of the orders \( \mathrm{p} \times \mathrm{k} \) and \( 3 \times \mathrm{k} \) respectively.
Therefore, matrix PY will be defined if \( \mathrm{k}=3 \).
Then, PY will be of the order \( \mathrm{p} \times \mathrm{k} \).
Matrices W and Y are of the orders \( \mathrm{n} \times 3 \) and \( 3 \times \mathrm{k} \) 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 \( \mathrm{n} \times \mathrm{k} \).
Matrices PY and WY can be added only when their orders are the same. Therefore, PY is of the order \( \mathrm{p} \times \mathrm{k} \) and WY is of the order \( \mathrm{n} \times \mathrm{k} \).
Thus, we must have \( \mathrm{p}=\mathrm{n} \).
Therefore, \( \mathrm{k}=3 \) and \( \mathrm{p}=\mathrm{n} \) are the restrictions on \( \mathrm{n}, \mathrm{k} \) and p so that \( \mathrm{PY}+ \) WY will be defined.
22. If \( \mathrm{n}=\mathrm{p} \), then the order of the matrix \( 7 \mathrm{X}-5 \mathrm{Z} \) is:
A. \( \mathrm{p} \times 2 \) B. \( 2 \times \mathrm{n} \) C. \( \mathrm{n} \times 3 \) D. \( \mathrm{p} \times \mathrm{n} \)
A. \( \mathrm{p} \times 2 \) B. \( 2 \times \mathrm{n} \) C. \( \mathrm{n} \times 3 \) D. \( \mathrm{p} \times \mathrm{n} \)
Answer
Matrix X is of the order \( 2 \times \mathrm{n} \).
Therefore, matrix 7X is also of the same order.
Matrix Z is of order \( 2 \times \mathrm{p} \)
\(
\Rightarrow 2 \times \mathrm{n}
\)
\( \Rightarrow \) matrix 5 Z is also of the same order.
Now, both the matrices 7 X and 5 Z are of the order \( 2 \times \mathrm{n} \).
Thus, matrix \( 7 \mathrm{X}-5 \mathrm{Z} \) is well- defined and is of the order \( 2 \times \mathrm{n} \).
