# 2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise

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## Karnataka 2nd PUC Maths Question Bank Chapter 3 Matrices Miscellaneous Exercise

Question 1.
Let $$A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$$, Show that (al +bA)n + nan – 1 ba where 1 is the identity matrix of order 2 and n ∈ N.

Question 2.
If$$A=\left[\begin{array}{lll}{1} & {1} & {1} \\{1} & {1} & {1} \\{1} & {1} & {1}\end{array}\right]$$
,prove that
An = $$\left[\begin{array}{lll}{3^{n-1}} & {3^{n-1}} & {3^{n-1}} \\{3^{n-1}} & {3^{n-1}} & {3^{-1}} \\{3^{n-1}} & {3^{n-1}} & {3^{n-1}}\end{array}\right], n \in N$$

a

Question 3.
$$\mathbf{A}=\left[\begin{array}{ll}{\mathbf{3}} & {-4} \\{\mathbf{1}} & {\mathbf{1}}\end{array}\right]$$
then prove that
An = $$\left[\begin{array}{cc}{1+2 n} & {-4 n} \\{n} & {1-2 n}\end{array}\right]$$ where n is any positive integer.

∴ P (k +1) is true
P(n) is true for all the values of +ve n.

Question 4.
If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
Since A &B are symmetric matrices A = a’ & B = B’
To prove (AB – BA)’ = – (AB – BA)
L.H.S = (AB – BA)’
(AB)’ – (BA)’
= B’A’-A’B’
= BA – AB
= – (AB – BA) = R.H.S
Thus proved.

Question 5.
Show that the matrix g’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Question 6.
find the values of x, y, z if the matrix
A = $$\left[\begin{array}{rrr}{0} & {2 y} & {x} \\{x} & {y} & {-z} \\{x} & {-y} & {z}\end{array}\right]$$ satisfy the equation A’ A = 1

Question 7.
For what values of x:
$$\left[\begin{array}{lll}{1} & {2} & {1}\end{array}\right]\left[\begin{array}{rrr}{1} & {2} & {0} \\{2} & {0} & {1} \\{1} & {0} & {2}\end{array}\right]\left[\begin{array}{l}{0} \\{2} \\{x}\end{array}\right]=0 ?$$

Question 8.
$$\mathbf{A}=\left[\begin{array}{rr}{3} & {1} \\{-1} & {2}\end{array}\right]$$ show that A2 – 5A + 7I = 0

Question 9.
Find x, if
$$\left[\begin{array}{ccc}{\mathbf{x}} & {-5} & {-1}\end{array}\right]\left[\begin{array}{ccc}{\mathbf{1}} & {\mathbf{0}} & {\mathbf{2}} \\{\mathbf{0}} & {\mathbf{2}} &{\mathbf{1}} \\{\mathbf{2}} & {\mathbf{0}} &{\mathbf{3}}\end{array}\right]\left[\begin{array}{l}{\mathbf{x}} \\{\mathbf{4}} \\ {\mathbf{1}}\end{array}\right]=\mathbf{0}$$

Question 10.
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are indicated below:

(a) If unit sale prices of, y and z are ₹2. 50, ₹1.50 and ₹1.00, respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the costs of the above three commodities are ₹ 2.00, ₹ 1.00 and 50 paise respectively. Find the gross profit.
Matrix representing the sales is

Profit in market I is
₹ 46,000 – ₹ 31,000
= ₹ 15,000

Profit in market 11 is
₹ 53,000 – ₹36,000
= ₹ 17,000

Gross profit
= ₹ 15,000 + ₹17,000
= ₹ 32,000.

Question 11.
Find the matrix X so that
$$\mathbf{X}\left[\begin{array}{lll}{1} & {2} & {3} \\{4} & {5} & {6}\end{array}\right]_{2 \times 3}=\left[\begin{array}{rrr}{-7} & {-8} & {-9} \\{2} & {4} & {6}\end{array}\right]_{2 \times 3}$$

Question 12.
If A and B are square matrices of the same order such that AB = BA, then prove by induction that ABn = BnA. Further, prove that (AB)n = An Bn for all n ∈ N.
(i) Let P(n) be the statement
P(n) – ⇒ Ab n = BnA
P(1) ⇒ AB = BA
∴ P( 1) is true
Let P(k) be true
P(k) ⇒ ABk = BkA
P(k +1) ⇒ ABk+1
= A(BkB) = (ABk)B
∵ matrix multiplication is associative.
⇒ (BkA) B ⇒ Bk (AB) = Bk (BA)
= (BkB) A = Bk+1 A
Hence P(k + 1) is true.
∴ P(n) is true for all the values of n ∈ N

(ii) Let P(n) be the statement
P(n) ⇒ (AB)n = AnBn
P(1) = (AB)1 = AB
∴ P(1) is true Let P(k) be true
P(k) ⇒ (AB)k
= AkBk P(k+1)
⇒ (AB)k+1 = (AB)kAB = (AkBk) AB
= Ak (BkA)B = Ak (ABk) B
(∵ ABn = BnA whenever Ab = BA)
= Ak+1 Bk+1
∴ P (k+1) is true
Hence P(n) is true for all the natures of n ∈ N

Choose correct answer in the following questions:

Question 13.
If $$\mathbf{A}=\left[\begin{array}{ll}{\alpha} & {\beta} \\{\gamma} & {-\alpha}\end{array}\right]$$
is such that A2= I ,then
(A) $$1+\alpha^{2}+\beta \gamma=0$$
(B) $$1-\alpha^{2}+\beta \gamma=0$$
(C) $$1-\alpha^{2}-\beta \gamma=0$$
(D) $$1+\alpha^{2}-\beta \gamma=0$$

Question 14.
If the matrix A is both symmetric and skew symmetric, then
(A) A is a diagonal martix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Let A be a square matrix such that A is both symmetric & skew-symmetric
⇒ A’ = A – (i) & A’ = – A -(ii)
⇒ A + A = 0  from (i) & (ii)
2A = 0
⇒ A = o
∴ (b) is correct

Question 15.
If A is square matrix such that A2 = A, then (I + A)3 – 7A is equal to
(A) A
(B) I – A
(C) I
(D) 3A
A2 = A
(I + A)3 – 7A = (I + A) (I+A)2 – 7A
= (I + A) {(I + A) (I+A)} -7A
= (I + A) (I2 + 2A+A2) – 7A
[v IA = AI = A]
= (I + A) (I+2A+A) – 7A (A2= A)
= (I + A) (I + 3A) – 7A
= I2 + AI + 3IA + 3A2 – 7A
= I + A + 3A + 3A – 7A
[∵ A2 = A & IA = AI = A]
=1
∴ (C) is correct.

### 2nd PUC Maths Matrices Miscellaneous Exercise Additional Questions and Answers

One Mark Questions:

Question 1.
If $$\mathbf{A}=\left(\begin{array}{ll}{\mathbf{0}} & {\mathbf{1}} \\{\mathbf{1}} & {\mathbf{0}}\end{array}\right), \mathbf{F} \text { ind } \mathbf{A}^{4}$$
(UP CET 2010)

Question 2.
If
$$\mathbf{A}=\left(\begin{array}{rr}{0} & {2} \\{3} & {-4}\end{array}\right), \mathbf{K A}=\left(\begin{array}{ll}{0} & {3 \mathbf{a}} \\{2 \mathbf{b}} & {24}\end{array}\right)$$,find k, a and b.

Question 3.
Find X and Y
$$2 \mathbf{X}-\mathbf{Y}=\left[\begin{array}{rrr}{6} & {-6} & {0} \\{-4} & {2} & {1} \end{array}\right]$$
$$\mathbf{X}+2 \mathbf{Y}=\left[\begin{array}{rrr}{3} & {2} & {5} \\{-2} & {1} & {-7}\end{array}\right]$$
$$\left[\begin{array}{rrr}{1} & {1} & {1} \\{1} & {-2} & {-2} \\{1} & {3} &{2}\end{array}\right]\left[\begin{array}{l}{x} \\{y} \\{z}\end{array}\right]=\left[\begin{array}{l}{0} \\{3} \\{4}\end{array}\right], \text { then }\left[\begin{array}{l}{x} \\{y} \\{z}\end{array}\right]$$(I I T 2006)
$$\mathbf{A}=\left[\begin{array}{lll}{6} & {8} & {7} \\{4} & {2} & {3} \\{9} & {7} & {1}\end{array}\right]$$