RBSE Class 12 Maths Solutions Chapter 3 Matrices Ex 3.2

RBSE Class 12 Maths Solutions Chapter 3 Matrices Ex 3.2

Question 1.

Let
A = $\left[\begin{array}{cc}2& 4\\ 3& 2\end{array}\right],$ 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:
(i) A + B
Answer:

(ii) A - B
Answer:

(iii) 3A - C
Answer:

(iv) AB
Answer:

(v) BA
Answer:

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

(ii) $\left[\begin{array}{cc}{a}^2+b^2& b^2+c^2\\ a^2+c^2& a^2+b^2\end{array}\right]+\left[\begin{array}{cc}2ab& 2bc\\ -2ac& -2ab\end{array}\right]$
Answer:

(iii) $\left[\begin{array}{ccc}-1& 4& -6\\ 8& 5& 16\\ 2& 8& 5\end{array}\right]+\left[\begin{array}{ccc}12& 7& 6\\ 8& 0& 5\\ 3& 2& 4\end{array}\right]$
Answer:

(iv) $\left[\begin{array}{cc}{\cos }^{2}x& {\sin }^{2}x\\ {\sin }^{2}x& {\cos }^{2}x\end{array}\right]+\left[\begin{array}{cc}{\sin }^{2}x& {\cos }^{2}x\\ {\cos }^{2}x& {\sin }^{2}x\end{array}\right]$
Answer:

Question 3.
Compute the indicated products:
(i) $\left[\begin{array}{cc}a& b\\ -b& a\end{array}\right]\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$
Answer:

(ii) $\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]\left[\begin{array}{ccc}2& 3& 4\end{array}\right]$
Answer:

(iii) $\left[\begin{array}{cc}1& -2\\ 2& 3\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right]$
Answer:

(iv) $\left[\begin{array}{ccc}2& 3& 4\\ 3& 4& 5\\ 4& 5& 6\end{array}\right]\left[\begin{array}{ccc}1& -3& 5\\ 0& 2& 4\\ 3& 0& 5\end{array}\right]$
Answer:

(v) $\left[\begin{array}{cc}2& 1\\ 3& 2\\ -1& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 1\\ -1& 2& 1\end{array}\right]$
Answer:

(vi) $\left[\begin{array}{ccc}3& -1& 3\\ -1& 0& 2\end{array}\right]\left[\begin{array}{cc}2& -3\\ 1& 0\\ 3& 1\end{array}\right]$
Answer:

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.
Answer:

Question 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 = $\left[\begin{array}{ccc}\frac{2}{5}& \frac{3}{5}& 1\\ \frac{1}{5}& \frac{2}{5}& \frac{4}{5}\\ \frac{7}{5}& \frac{6}{5}& \frac{2}{5}\end{array}\right]$ then compute 3A - 5B.
Answer:

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

Question 7.
(i) X + Y = $\left[\begin{array}{cc}7& 0\\ 2& 5\end{array}\right]$ and X - Y = $\left[\begin{array}{cc}3& 0\\ 0& 3\end{array}\right]$
Answer:

(ii) 2X + 3Y = $\left[\begin{array}{cc}2& 3\\ 4& 0\end{array}\right]$ and 3X + 2Y = $\left[\begin{array}{cc}2& -2\\ -1& 5\end{array}\right]$
Answer:

Question 8.
Find X if Y = $\left[\begin{array}{cc}3& 2\\ 1& 4\end{array}\right]$ and 2X + Y = $\left[\begin{array}{cc}1& 0\\ -3& 2\end{array}\right]$.
Answer:

Question 9.
Find x and y if
2$\left[\begin{array}{cc}1& 3\\ 0& x\end{array}\right]+\left[\begin{array}{cc}y& 0\\ 1& 2\end{array}\right]=\left[\begin{array}{cc}5& 6\\ 1& 8\end{array}\right]$
Answer:

Comparing corresponding elements, we get
2 + y = 5,
⇒ y = 5 - 2,
⇒ y = 3,
∴ y = 3,

2x + 2 = 8
⇒ 2x = 8 - 2
⇒ 2x = 6
∴ x = 3

Question 10.
Solve the equation for x, y, z and t, if
2$\left[\begin{array}{cc}x& z\\ y& t\end{array}\right]+3\left[\begin{array}{cc}1& -1\\ 0& 2\end{array}\right]=3\left[\begin{array}{cc}3& 5\\ 4& 6\end{array}\right]$
Answer:

Comparing corresponding elements, we get
2x + 3 = 9, 2z - 3 = 15, 2y = 12, 2t + 6 = 18
⇒ 2x = 9 - 3, 2z = 15 + 3, y = $\frac{12}{2}$, 2t = 18 - 6
⇒ 2x = 6, 2z = 18, y = 6, 2t = 12
∴ x = 3, z = 9, y = 6, t = 6

Question 11.
If x$\left[\begin{array}{c}2\\ 3\end{array}\right]$+ y$\left[\begin{array}{c}-1\\ 1\end{array}\right]$ = $\left[\begin{array}{c}10\\ 5\end{array}\right]$, then find the values of x and y.
Answer:

⇒ 2x - y = 10 ........ (i)
3x + y = 5 .......... (ii)
Adding equation (i) and (ii), we get

Putting value of x in euqation (i), we get
2 × 3 - y = 10 ⇒ 6 - y = 10
⇒ - y = 10 - 6 ⇒ y = 4
⇒ y = - 4
Thus, x = 3, y = - 4

Question 12.
Given
3$\left[\begin{array}{cc}x& y\\ z& w\end{array}\right]=\left[\begin{array}{cc}x& 6\\ -1& 2w\end{array}\right]+\left[\begin{array}{cc}4& x+y\\ z+w& 3\end{array}\right]$
find the value of x, y, z and w.
Answer:

Comparing corresponding elements, we get
3x = x + 4,
3z = - 1 + z + w,
⇒ 3x - x = 4
⇒ 2x = 4,
3z - z = w - 1,
2z = w - 1,

3y = 6 + x + y
3w = 2w + 3
3y - y = 6 + x
2y = 6 + x
3w - 2w = 3
w = 3

∴ x = 2,
2y = 6 + 2, ⇒ 2y = 8,
y = 4

And w = 3, then 2z = w - 1
⇒ 2z = 3 - 1
⇒ 2z = 2
∴ z = 1
Thus, x = 2, y = 4, z = 1, w = 3

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

Question 14.
Show that
(i) $\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]\left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\ne \left[\begin{array}{cc}2& 1\\ 3& 4\end{array}\right]\left[\begin{array}{cc}5& -1\\ 6& 7\end{array}\right]$
Answer:

(ii)
$\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]$ ≠ $\left[\begin{array}{ccc}-1& 1& 0\\ 0& -1& 1\\ 2& 3& 4\end{array}\right]\left[\begin{array}{ccc}1& 2& 3\\ 0& 1& 0\\ 1& 1& 0\end{array}\right]$
Answer:

Question 15.
If A = $\left[\begin{array}{ccc}2& 0& 1\\ 2& 1& 3\\ 1& -1& 0\end{array}\right]$, then find the value of A2 - 5A + 6I
Answer:
Here, A2 = AA

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

Question 17.
If A = $\left[\begin{array}{cc}3& -2\\ 4& -2\end{array}\right]$ and I = $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ and A2 = KA - 2I, then find k.
Answer:

⇒ 1 = 3k - 2, - 2 = - 2k
4 = 4k, - 4 = - 2k - 2
⇒ 3k = 2 + 1, 2k = 2
4k = 4, - 2k = - 4 + 2
⇒ 3k = 3, 2k = 2
4k = 4, - 2k = - 2
⇒ k = 1, k = 1

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

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) ₹ 1,800
(b) ₹ 2,000
Answer:
Let x and (30000 - x) are two parts of ₹ 30,000.
Then matrix A = [x (30000 - x)]
Rate of interest are 5% and 7% or 0.05 and 0.07
Let matrix of rate R = $\left[\begin{array}{c}0.05\\ 0.07\end{array}\right]$

(a) We have, total interest = ₹ 1800
= [x (30000 - x)] $\left[\begin{array}{c}0.05\\ 0.07\end{array}\right]$ = [1800]
⇒ [x × 0.05 + (30000 - x) × 0.07] - [1800]
⇒ [0.05x + (30000 × 0.07) - 0 07 × x] = [1800]
⇒ 0.05x + 2100 - 0.07× = 1800
⇒ 0.05x - 0.07x = 1800 - 2100
⇒ - 0.02x = - 300
⇒ x = $\frac{-300}{-0.02}$ = $\frac{30000}{2}$ = 15000
First bond = ₹ 15,000
Now 30000 - x = 30000 -15000
∴ Second bond = ₹ 15,000
Thus, each bond should be bought for ₹ 15,000 so that interest of ₹ 1800 is received.

(b) For interest ₹ 2,000, matrix equation will be of the following type
[x (3000 - x)] $\left[\begin{array}{c}0.05\\ 0.07\end{array}\right]$ = [2000]
⇒ [x × 0.05 + (30000 - x) × 0.07] = [2000]
⇒ [0.05x + (30000 × 0.07) - (x × 0.07]) = [2000]
⇒ 0.05x + 2100.00 - 0.07x = 2000
⇒ 0.05x - 0.07x = 2000 - 2100
⇒ - 0.02x = - 100
⇒ x = $\frac{-100}{-0.02}$ = $\frac{10000}{2}$ = 5000
∴ First bond = ₹ 5,000
Thus 30000 - x = 30000 - 5000 = 25000
∴ Second bond = ₹ 25,000
To get ₹ 2,000 as interest two bonds should be divided into ₹ 5,000 and ₹ 25,000.

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 ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the book shop will receive from selling all the books using matrix algebra.
Answer:
Number of books in school is as follows:

Subject	Number of books
Chemistry	10 dozen = 10 × 12 = 120
Physics	8 dozen = 8 × 12 = 96
Economics	10 dozen = 10 × 12 = 120

Let number of books are represented by A
∴ A = [120 96 120]
And selling prices are represented by R then
∴ R = $\left[\begin{array}{c}80\\ 60\\ 40\end{array}\right]$
Money obtained by selling the books
= AR = [120 96 120] $\left[\begin{array}{c}80\\ 60\\ 40\end{array}\right]$
= [(120 × 80) + (96 × 60) + (120 × 40)]
= [9600 + 5760 + 4800] = [20160]
Total money obtained = ₹ 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 defined are:
(A) k = 3, p = n
(B) k is arbitary, p = 2
(C) p is arbitrary, k = 3
(D) k = 2, p = 3
Answer:
Order P matrix P = p × k
Order of martix Y = 3 × k
Order of matrix W = n × 3
Then for PY defined
Number of columns in P = number of rows in Y
k =3
and WY defined
Number of columns in W = number of rows in Y
(w = 3)
Now PY + WY defined
order of PY = p × k = p × 3
Then order of WY = n × k
Thus, PY + WY is defined if order of PY and WY are same
∴ p = n and k = 3
Thus (A) is correct.

Question 22.
If n = p, then order of matrix 7X - 5Z.
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n
Answer:
X Order of X = 2 × n
Z Order of Z = 2 × p
∵ p = n
Thus, order of 7X - 5Z = 2 × n,
Hence, choice (B) is correct.

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