Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 3 Matrices Ex 3.2 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 3 Matrices Ex 3.2

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]), C = (left[begin{array}{ll} -2 & 5 \ 3 & 4 end{array}right]).
Find each of the following:
(i) A + B
(ii) A-B
(iii) 3A-C
(iv) AB
(v) BA
Solution:

Question 2.
Compute the following:
(i) (left[begin{array}{cc} a & b \ -b & a end{array}right]) + (left[begin{array}{cc} a & b \ b & a end{array}right])
(ii) (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])
(iii) (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])
(iv) (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])
Solution:

Question 3.
Compute the indicated products:

Solution:

Question 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 (A + B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.
Solution:

Question 5.
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}{5} & frac{4}{5} \ frac{7}{5} & frac{6}{5} & frac{2}{5} end{array}right]), then compute 3A – 5B.
Solution:

Question 6.
Simplify : cos θ (left[begin{array}{cc} cos theta & sin theta \ -sin theta & cos theta end{array}right]) + sin θ(left[begin{array}{cc} sin theta & -cos theta \ cos theta & sin theta end{array}right])
Solution:

Question 7.
Find X and Y, if
(i) 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])
(ii) 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:

Question 8.
Find X, if Y = (left[begin{array}{ll} 3 & 2 \ 1 & 4 end{array}right]) and 2X + Y = (left[begin{array}{ll} 1 & 0 \ -3 & 2 end{array}right]).
Solution:
2X + Y = (left[begin{array}{ll} 1 & 0 \ -3 & 2 end{array}right])
∴ 2X = (left[begin{array}{ll} 1 & 0 \ -3 & 2 end{array}right]) – Y = (left[begin{array}{ll} 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 = (left[begin{array}{ll} -1 & -1 \ -2 & -1 end{array}right]).

Question 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]).
Solution:

Equating the corresponding elements, we get:
2 + y = 5. ∴ y = 5 – 2 = 3.
2x + 2 = 8. ∴ 2x = 8 – 2 = 6. or x = 3
Hence, x = 3 and y = 3.

Question 10.
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}{ll} 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}{ll} 1 & -1 \ 0 & 2 end{array}right]) = 3(left[begin{array}{ll} 3 & 5 \ 4 & 6 end{array}right])
or (left[begin{array}{ll} 2x & 2z \ 2y & 2t end{array}right])+(left[begin{array}{ll} 3 & -3 \ 0 & 6 end{array}right]) = (left[begin{array}{ll} 9 & 15 \ 12 & 18 end{array}right])
or (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]).
Equating corresponding elements, we get:
2x + 3 = 9 or 2x = 9 – 3 = 6. ∴ x = 3.
2y = 12 ∴ y = 6.
2z – 3 = 15 or 2z – 15 + 3 = 18 ∴ z = 9.
2t + 6 = 18 or 2t = 18 – 6 = 12 ∴ t = 6.
Thus, x = 3, y = 6, z = 9 and t = 6.

Question 11.
If x(left[begin{array}{l} 2 \ 3 end{array}right]) + y(left[begin{array}{l} -1 \ 1 end{array}right]) = (left[begin{array}{l} 10 \ 5 end{array}right]), then find the values of x and y.
Solution:
x(left[begin{array}{l} 2 \ 3 end{array}right]) + y(left[begin{array}{l} -1 \ 1 end{array}right]) = (left[begin{array}{l} 10 \ 5 end{array}right]).
or (left[begin{array}{l} 2x \ 3x end{array}right]) + (left[begin{array}{l} -y \ y end{array}right]) = (left[begin{array}{l} 10 \ 5 end{array}right])
or (left[begin{array}{l} 2 x-y \ 3 x+y end{array}right]) = (left[begin{array}{l} 10 \ 5 end{array}right])
Equating the corresponding elements, we get:
2x – y = 10 … (1)
3x + y = 5 … (2)
Adding (1) and (2), we get
5x – 15 ⇒ x = (frac { 15 }{ 5 }) = 3.
Putting value of x in (1), we get
2x – y = 2 x 3 – y = 10
∴ y = 6 – 10 = – 4.
Hence, x = 3, and y = – 4.

Question 12.
Given 3 (left[begin{array}{ll} x & y \ z & w end{array}right]) = (left[begin{array}{ll} x & 6 \ -1 & 2w end{array}right]) + (left[begin{array}{ll} 4 & x+y \ z+w & 3 end{array}right]), find the values of x, y, z and w.
Solution:

Question 13.
If F(x) = (left[begin{array}{ccc} cos x & -sin x & 0 \ sin x & cos x & 0 \ 0 & 0 & 1 end{array}right]), then show that
F(x)F(y) = F(x + y).
Solution:
L.H.S = F(x)F(y)

= F(x+y) = R.H.S
Hence, F(x)F(y) = F(x+y)

Question 14.
Show that:

Solution:

Question 15.
Find A² – 5A + 6I, if A = (left[begin{array}{ccc} 2 & 0 & 1 \ 2 & 1 & 3 \ 1 & -1 & 0 end{array}right])
Solution:

Question 16.
If A = (left[begin{array}{ccc} 1 & 0 & 2 \ 0 & 2 & 1 \ 2 & 0 & 3 end{array}right]), prove that A³ – 6A² + 7A + 2I = 0.
Solution:

Question 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]), then find k such that A² = kA – 2I.
Solution:

Question 18.
If A = (left[begin{array}{rr} 0 & -tan frac{alpha}{2} \ tan frac{alpha}{2} & 0 end{array}right]) and I is the identity matrix of order 2, show that I + A = (I – A)(left[begin{array}{cc} cos alpha & -sin alpha \ sin alpha & cos alpha end{array}right]).
Solution:

Hence, the result.

Question 19.
A trust has ₹ 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹ 30,000 among the two types of bond, if the trust fund obtains an annual total interest of :
(a) ₹ 1,800
(b) ₹ 2,000.
Solution:
Let ₹ 30,000 be divided into two parts.
viz. ₹ x and ₹ (30000 – x).
Let it be represented by 1 x 2 matrix [x (30000 – x)]
Rate of interest is 0.05 and 0.07 per rupee. Let it be denoted a matrix R of order 2 x 1 as shown below :

Thus, the two parts are ₹ 15000 each.

(b) Here, the total interest is ₹ 2000.
Proceeding as above, we have :

Question 20.
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books and 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60, ₹ 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 be represented by matrix A of order 1 x 3 as shown below :
A = [10 dozen 8 dozen 10 dozen]
= [120 96 120].
The cost of each book of chemistry, physics and economics are ₹ 80, ₹ 60 and ₹ 40 respectively. They are represented by the matrix R of order 3 x 1, i.e.,
R = (left[begin{array}{l} 80 \ 60 \ 40 end{array}right])
Total amount (in ₹) received by selling all the books is given
AR = [120 96 120] (left[begin{array}{l} 80 \ 60 \ 40 end{array}right])
= [120 x 80 + 96 x 60 + 120 x 40]
= [9600 + 5760 + 4800]
= [20160].
∴ Total amount received = ₹ 20160.
Assume that X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. Now answer the following questions 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:
The order of matrix P is p x k
and the order of matrix Y is 3 x k.
PY is defined, when
No. of columns in P = No. of rows in Y.
⇒ k = 3
∴ The order of PY is p x 3.
Now, the order of W is n x 3.
The order of Y is 3 x k.
WY is defined, when
No. of columns in W = No. of rows in Y = 3.
So, WY is defined.
The order of WY = n x 3. .
Matrix PY + WY is defined when PY and WY are of the same order.
But they are of the order p x 3 and nx 3 respectively.
⇒ PY + WY is defined when p = n.
Hence, PY + WY is defined when k = 3 and p = n.
∴ Part (A) is the correct answer.

Question 22.
If n = p, then the order of the matrix 7X – 5Z is
(A) p x 2
(B) 2 x n
(C) n x 3
(D) p x n
Solution:
The order of matrix X is 2 x n.
The order of matrix Z is 2 x p.
7X – 5Z is defined when X and Z are of the same order.
⇒ n-p (given)
Thus, the order of Z is also 2 x n.
Hence, order of 7X – 5Z is 2 x n.
∴ Part (B) is the correct answer.