Referring to the AP Inter 2nd Year Maths Study Material Chapter 3 Matrices Exercise 3b Solutions makes it easier to understand complex problems.
AP Inter 2nd Year Maths Matrices Solutions Exercise 3b
I.
Question 1.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\). Find A + B
Solution:
A + B
= \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) + \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\) = \(\left[\begin{array}{ll}
2+1 & 4+3 \\
3-2 & 2+5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3 & 7 \\
1 & 7
\end{array}\right]\)
Question 2.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\). Find A – B
Solution:
A – B
= \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) – \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\) = \(\left[\begin{array}{ll}
2-1 & 4-3 \\
3+2 & 2-5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 1\\
5 & -3
\end{array}\right]\)
Question 3.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), C = \(\left[\begin{array}{ll}
-2 & 5 \\
3 & 4
\end{array}\right]\). Find 3A – C
Solution:
3A – C
= 3\(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\) – \(\left[\begin{array}{ll}
-2 & 5 \\
3 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
3 \times 2 & 3 \times 4 \\
3 \times 3 & 3 \times 2
\end{array}\right]-\left[\begin{array}{cc}
-2 & 5 \\
3 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
6+2 & 12-5 \\
9-3 & 6-4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
8 & 7 \\
6 & 2
\end{array}\right]\)
Question 4.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\). Find AB
Solution:
AB
= \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\left[\begin{array}{cc}
1 & 3 \\
-2 & 5
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2(1)+4(-2) & 2(3)+4(5) \\
3(1)+2(-2) & 3(3)+2(5)
\end{array}\right]\)
= \(\left[\begin{array}{ll}
2-8 & 6+20 \\
3-4 & 9+10
\end{array}\right]\) = \(\left[\begin{array}{ll}
-6 & 26 \\
-1 & 19
\end{array}\right]\)
Question 5.
Let A = \(\left[\begin{array}{ll}
2 & 4 \\
3 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\). Find BA
Solution:
BA
= \(\left[\begin{array}{ll}
1 & 3 \\
-2 & 5
\end{array}\right]\left[\begin{array}{cc}
2 & 4 \\
3 & 2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
1(2)+3(3) & 1(4)+3(2) \\
-2(2)+5(3) & -2(4)+5(2)
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2+9 & 4+6 \\
-4+15 & -8+10
\end{array}\right]\) = \(\left[\begin{array}{cc}
11 & 10 \\
11 & 2
\end{array}\right]\)
Question 6.
Compute \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\) + \(\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\) + \(\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\) = \(\left[\begin{array}{cc}
\mathrm{a}+\mathrm{a} & \mathrm{~b}+\mathrm{b} \\
-\mathrm{b}+\mathrm{b} & \mathrm{a}+\mathrm{a}
\end{array}\right]\)
= \(\left[\begin{array}{cc}
2 \mathrm{a} & 2 \mathrm{~b} \\
0 & 2 \mathrm{a}
\end{array}\right]\)
Question 7.
Compute \(\left[\begin{array}{ll}
a^2+b^2 & b^2+c^2 \\
a^2+c^2 & a^2+b^2
\end{array}\right]\) + \(\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ll}
a^2+b^2 & b^2+c^2 \\
a^2+c^2 & a^2+b^2
\end{array}\right]\) + \(\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]\)
= \(\left[\begin{array}{ll}
a^2+b^2+2 a b & b^2+c^2+2 b c \\
a^2+c^2-2 a c & a^2+b^2-2 a b
\end{array}\right]\)
= \(\left[\begin{array}{ll}
(a+b)^2 & (b+c)^2 \\
(a-c)^2 & (a-b)^2
\end{array}\right]\)
Question 8.
Compute \(\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]\) + \(\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]\) + \(\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-1+12 & 4+7 & -6+6 \\
8+8 & 5+0 & 16+5 \\
2+3 & 8+2 & 5+4
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
11 & 11 & 0 \\
16 & 5 & 21 \\
5 & 10 & 9
\end{array}\right]\)
Question 9.
Compute \(\left[\begin{array}{ll}
\cos ^2 x & \sin ^2 x \\
\sin ^2 x & \cos ^2 x
\end{array}\right]\) + \(\left[\begin{array}{ll}
\sin ^2 x & \cos ^2 x \\
\cos ^2 x & \sin ^2 x
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ll}
\cos ^2 x & \sin ^2 x \\
\sin ^2 x & \cos ^2 x
\end{array}\right]\) + \(\left[\begin{array}{ll}
\sin ^2 x & \cos ^2 x \\
\cos ^2 x & \sin ^2 x
\end{array}\right]\)
= \(\left[\begin{array}{ll}
\cos ^2 x+\sin ^2 x & \sin ^2 x+\cos ^2 x \\
\sin ^2 x+\cos ^2 x & \cos ^2 x+\sin ^2 x
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Question 10.
Compute \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right]\) = \(\left[\begin{array}{cc}
\mathrm{a}(\mathrm{a})+\mathrm{b}(\mathrm{~b}) & \mathrm{a}(-\mathrm{b})+\mathrm{b}(\mathrm{a}) \\
-\mathrm{b}(\mathrm{a})+\mathrm{a}(\mathrm{~b}) & -\mathrm{b}(-\mathrm{b})+\mathrm{a}(\mathrm{a})
\end{array}\right]\)
= \(\left[\begin{array}{cc}
a^2+b^2 & -a b+a b \\
-a b+a b & b^2+a^2
\end{array}\right]\)
= \(\left[\begin{array}{cc}
a^2+b^2 & 0 \\
0 & a^2+b^2
\end{array}\right]\)
Question 11.
Compute \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]|233|\)
Solution:
\(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]|233|\) = \(\left[\begin{array}{lll}
1(2) & 1(3) & 1(4) \\
2(2) & 2(3) & 2(4) \\
3(2) & 3(3) & 3(4)
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
2 & 3 & 4 \\
4 & 6 & 8 \\
6 & 9 & 12
\end{array}\right]\)
Question 12.
Compute \(\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
1 & -2 \\
2 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{lll}
1(1)-2(2) & 1(2)-2(3) & 1(3)-2(1) \\
2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1)
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
-3 & -4 & 1 \\
8 & 13 & 9
\end{array}\right]\)
Question 13.
Compute \(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]\)
Solution:
\(\left[\begin{array}{lll}
2 & 3 & 4 \\
3 & 4 & 5 \\
4 & 5 & 6
\end{array}\right]\left[\begin{array}{ccc}
1 & -3 & 5 \\
0 & 2 & 4 \\
3 & 0 & 5
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\
3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\
4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5)
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
14 & 0 & 42 \\
18 & -1 & 56 \\
22 & -2 & 70
\end{array}\right]\)
Question 14.
Compute \(\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
1 & 2 & 1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
2 & 1 \\
3 & 2 \\
1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
1 & 2 & 1
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\
3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\
-1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1)
\end{array}\right]\)
= \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
1 & 4 & 5 \\
-2 & 2 & 0
\end{array}\right]\)
Question 15.
Compute \(\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & 3 \\
1 & 0 \\
3 & 1
\end{array}\right]\)
Solution:
\(\left[\begin{array}{ccc}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{cc}
2 & 3 \\
1 & 0 \\
3 & 1
\end{array}\right]\)
= \(\left[\begin{array}{cc}
3(2)-1(1)+3(3) & 3(-3)-1(0)+3(1) \\
-1(2)+0(1)+2(3) & -1(-3)+0(0)+2(1)
\end{array}\right]\)
= \(\left[\begin{array}{cc}
14 & -6 \\
4 & 5
\end{array}\right]\)
Question 16.
If A = \(\left[\begin{array}{lll}
\frac{2}{3} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
\frac{2}{5} & \frac{3}{5} & 1 \\
\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\
\frac{7}{5} & \frac{6}{5} & \frac{2}{5}
\end{array}\right]\), then compute 3A – 5B
Solution:
3A – 5B = 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\(\left[\begin{array}{ccc}
\frac{2}{5} & \frac{3}{5} & 1 \\
\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\
\frac{7}{5} & \frac{6}{5} & \frac{2}{5}
\end{array}\right]\)
= \(\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{array}\right]\) – \(\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{array}\right]\)
= \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\)
Question 17.
Simplify cosθ \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) + sinθ \(\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
Solution:
cosθ \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) + sinθ \(\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 \theta & \cos \theta \sin \theta \\
-\sin \theta \cos \theta & \cos ^2 \theta
\end{array}\right]\) + \(\left[\begin{array}{cc}
\sin ^2 \theta & -\sin \theta \cos \theta \\
\sin \theta \cos \theta & \sin ^2 \theta
\end{array}\right]\)
= \(\left[\begin{array}{cc}
\cos ^2 \theta+\sin ^2 \theta & \sin \theta \cos \theta-\sin \theta \cos \theta \\
-\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^2 \theta+\sin ^2 \theta
\end{array}\right]\)
= \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Question 18.
Find X and Y, if X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)
Solution:
Given X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) …………… (1)
X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\) …………… (2)
(1) + (2) ⇒ 2X = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) + \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\) = \(\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right]\)
⇒ X = \(\frac{1}{2}\left[\begin{array}{cc}
10 & 0 \\
2 & 8
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]\)
⇒ X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\)
⇒ Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) – \(\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]\)
⇒ Y = \(\left[\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right]\)
Question 19.
Find X and Y, if 2X + 3Y = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) and 3X + 2Y = \(\left[\begin{array}{cc}
2 & -2 \\
-1 & 5
\end{array}\right]\)
Solution:
Given 2X + 3Y = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) ………….. (1)
3X + 2Y = \(\left[\begin{array}{cc}
2 & -2 \\
-1 & 5
\end{array}\right]\) ……….. (2)
Multiplying equation(1) by 2, we have 2(2X +3Y) = 2\(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) ⇒ 4X + 6Y = \(\left[\begin{array}{ll}
4 & 6 \\
8 & 0
\end{array}\right]\) …..(3)
MultIplying equation (2) by 3, we have 3(3X + 2Y) = 3 \(\left[\begin{array}{cc}
2 & -2 \\
-1 & 5
\end{array}\right]\) ⇒ 9X + 6Y = \(\left[\begin{array}{cc}
6 & -6 \\
-3 & 15
\end{array}\right]\) …………. (4)
From (3) and(4), we have (4X + 6Y)(9X + 6Y) = \(\left[\begin{array}{ll}
4 & 6 \\
8 & 0
\end{array}\right]\) – \(\left[\begin{array}{cc}
6 & -6 \\
-3 & 15
\end{array}\right]\)
Question 20.
Find X, if Y = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) and 2X + Y = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\)
Solution:
2X + Y = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\) ⇒ 2X + \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\)
⇒ 2X = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\) – \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\)
= \(\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right]\)
⇒ X = \(\frac{1}{2}\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right]\) = \(\left[\begin{array}{ll}
-1 & -1 \\
-2 & -1
\end{array}\right]\)
Question 21.
Find x and y, if 2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]\) + \(\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
Solution:
Given 2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]\) + \(\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 & 6 \\
0 & 2 x
\end{array}\right]\) + \(\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2+y & 6 \\
1 & 2 x+2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
Equating the corresponding elements of these two matrices, 2 + y = 5 ⇒ y = 3
2x + 2 = 8 ⇒ x = 3
∴ x = 3, y = 3
Question 22.
Solve the equation for x, y, 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\(\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
Solution:
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\(\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 \mathrm{x} & 2 \mathrm{z} \\
2 \mathrm{y} & 2 \mathrm{t}
\end{array}\right]\) + \(\left[\begin{array}{cc}
3 & -3 \\
0 & 6
\end{array}\right]\) = \(\left[\begin{array}{cc}
9 & 15 \\
12 & 18
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2 \mathrm{x}+3 & 2 \mathrm{z}-3 \\
2 \mathrm{y} & 2 \mathrm{t}+6
\end{array}\right]=\) = \(\left[\begin{array}{cc}
9 & 15 \\
12 & 18
\end{array}\right]\)
Equating the corresponding elements of these two matrIces, 2x + 3 = 9 ⇒ 2x = 6 ⇒ x = 3
2y = 12 ⇒ y = 6
2z – 3 = 15 ⇒ 2z = 18 ⇒ z = 9
2t + 6 = 18 ⇒ 2t = 12 ⇒ t = 6;
∴ x = 3, y = 6, z = 9, t = 6
Question 23.
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.
Solution:
\(x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{c}
-1 \\
1
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{c}
2 \mathrm{x} \\
3 \mathrm{x}
\end{array}\right]+\left[\begin{array}{c}
-\mathrm{y} \\
\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\)
⇒ \(\left[\begin{array}{l}
2 \mathrm{x}-\mathrm{y} \\
3 \mathrm{x}+\mathrm{y}
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\)
Equating the corresponding elements of these two matrices,
2x – y = 10 ……….. (1)
3x + y = 5 ……….. (2)
By adding these two equations, we get 5x = 15 ⇒ x = 3
Now putting this value in (2)
3x + y = 5 ⇒ y = 5 – 3x
⇒ y = 5 – 3(3) ⇒ y = 5 – 9
⇒ y = -4
∴ x = 3, y = -4
Question 24.
Given, 3\(\left[\begin{array}{cc}
\mathrm{x} & \mathrm{y} \\
\mathrm{z} & \mathrm{w}
\end{array}\right]\) = \(\left[\begin{array}{cc}
x & 6 \\
-1 & 2 w
\end{array}\right]\) + \(\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right]\) find the values of x, y, z and w.
Solution:
3\(\left[\begin{array}{cc}
\mathrm{x} & \mathrm{y} \\
\mathrm{z} & \mathrm{w}
\end{array}\right]\) = \(\left[\begin{array}{cc}
x & 6 \\
-1 & 2 w
\end{array}\right]\) + \(\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
3 \mathrm{x} & 3 \mathrm{y} \\
3 \mathrm{z} & 3 \mathrm{w}
\end{array}\right]\) = \(\left[\begin{array}{cc}
x+4 & 6+x+y \\
-1+z+w & 2 w+3
\end{array}\right]\)
Equating the corresponding elements of these two matrices,
3x = x + 4 ⇒ 2x = 4 ⇒ x = 2
3y = 6 + x + y ⇒ 2y = 6 + x ⇒ 2y = 6 + 2 ⇒ 2y = 8 ⇒ y = 4
3w = 2w + 3 ⇒ w = 3
3z = -1 + z + w ⇒ 2z = w – 1 ⇒ 2z = 3 – 1 ⇒ 2z = 2 ⇒ z = 1
∴ x = 2, y = 4, z = 1, w = 3 .
Question 25.
Show that \(\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{cc}
2 & 1 \\
3 & 4
\end{array}\right]\) ≠ \(\left[\begin{array}{cc}
2 & 1 \\
3 & 4
\end{array}\right]\left[\begin{array}{cc}
5 & -1 \\
6 & 7
\end{array}\right]\)
Solution:
II.
Question 1.
If F(x) = \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\) show that F(x) F(y) = F(x + y)
Solution:
Question 2.
Show that \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\) ≠ \(\left[\begin{array}{ccc}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\)
Solution:
Question 3.
Find A2 – 5A + 6I, if A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\)
Solution:
Question 4.
If A = \(\), prove that A3 – 6A2 + 7A + 2I = 0
Solution:
Hence, A3 – 6A2 + 7A + 2I = 0
Question 5.
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 A2 = kA – 2I
Solution:
Equating the corresponding elements, we have 3k – 2 = 1
⇒ 3k = 3
⇒ k = 1
∴ k = 1
Question 6.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{u}{2} \\
\tan \frac{u}{2} & 0
\end{array}\right]\) and I is the identify matrix of order 2, show that I + A – (I – A) \(\left[\begin{array}{cc}
0 & -\tan \frac{u}{2} \\
\tan \frac{u}{2} & 0
\end{array}\right]\)
Solution:
Question 7.
A trust fund has ₹ 30,000 that must be invested in two different types of bonds. – The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) ₹ 1800 (b) ₹ 2000
Solution:
(a) Let be invested in the first bond.
Then, the sum of money invested in the second bond will be ₹(3000 – x)
It is given that the first bond pays 5% interest per year and the second bond pays 7% interest per year.
Now in order to obtain an annual total interest of ₹ 1800, we have:
\(\left[\begin{array}{ll}
\mathrm{x} & (30000-\mathrm{x})
\end{array}\right]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]\) = 1800
\(\left[\text { S.I for } 1 \text { year }=\frac{\text { Principal } \text { × } \text { Rate }}{100}\right]\) ⇒ \(\frac{5 x}{100}+\frac{7(30000-x)}{100}\) = 1800
⇒ 5x + 210000 – 7x = 180000
⇒ 210000 – 2x = 180000 .
⇒ -2x = 210000 -180000 ⇒ 2x = 30000
⇒ x = 15000
Thus, 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.
(b) Let ₹ x be invested in the first bond.
Then, the sum of money invested in the second bond will be ₹ (3000 – x)
Now in order to obtain art annual total interest of ₹ 2000, we have:
\(\left[\begin{array}{ll}
\mathrm{x} & (30000-\mathrm{x})
\end{array}\right]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]\) = 2000
\(\left[\text { S.I for } 1 \text { year }=\frac{\text { Principal × Rate }}{100}\right]\) ⇒ \(\frac{5 x}{100}+\frac{7(30000-x)}{100}\) = 2000
⇒ 5x + 210000 – 7x = 200000 ⇒ 210000 – 2x = 200000
⇒ 2x = 210000- 200000 ⇒ 2x = 10000
⇒ x = 5000
Thus, in order to obtain an annual total interest of ₹ 2000, the trust fund should invest ₹ 5000 in the first bond and the remaining ₹ 25000 in the second bond.
Question 8.
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution:
The total amount of money that will be received from the sale of all these books can be represented in D the matrix form as:
\(12\left[\begin{array}{lll}
10 & 8 & 10
\end{array}\right]\left[\begin{array}{l}
80 \\
60 \\
40
\end{array}\right]\) = 12[10(80) + 8(60) + 10(40)]
= 12(800 + 480 + 400) = 12(1680) = 20160
Thus, the book shop receives ₹ 20160 from the sale of all these books.
Question 9.
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 (A+B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Solution:
