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1

### WB JEE 2009

If A and B are square matrices of the same order and AB = 3I, then A$$-$$1 is equal to

A
3B
B
$${1 \over 3}$$B
C
3B$$-$$1
D
$${1 \over 3}$$B$$-$$1

## Explanation

Given AB = 3I

$${A^{ - 1}}(AB) = {A^{ - 1}}(3I)$$ pre-multiplication by A$$-$$1

$$\Rightarrow {A^{ - 1}}AB = 3{A^{ - 1}}I$$

$$\Rightarrow IB = 3{A^{ - 1}}$$ ($$\because$$ $${A^{ - 1}}A = I$$)

$$\Rightarrow B = 3{A^{ - 1}} \Rightarrow {A^{ - 1}} = {1 \over 3}B$$

2

### WB JEE 2009

If A2 $$-$$ A + I = 0, then the inverse of the matrix A is

A
A $$-$$ I
B
I $$-$$ A
C
A + I
D
A

## Explanation

A(A $$-$$ I) = $$-$$I

$$\Rightarrow$$ A(I $$-$$ A) = I $$\Rightarrow$$ A$$-$$1 = I $$-$$ A.

3

### WB JEE 2009

If A is a square matrix. Then

A
A + AT is symmetric
B
AAT is skew-symmetric
C
AT + A is skew-symmetric
D
ATA is skew-symmetric

## Explanation

Let B = A + AT

$$\therefore$$ BT = (A + AT)T = AT + (AT)T = AT + A ($$\because$$ (A + B)T = BT + AT, (AT)T = A)

= A + AT = B (If AT = A, then A is symmetric)

$$\therefore$$ A + AT is symmetric.

4

### WB JEE 2008

If the matrix $$\left[ {\matrix{ a & b \cr c & d \cr } } \right]$$ is commutative with the matrix $$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]$$ then

A
a = 0, b = c
B
b = 0, c = d
C
c = 0, d = a
D
d = 0, a = b

## Explanation

Matrix $$\left[ {\matrix{ a & b \cr c & d \cr } } \right]$$ is commutative with the matrix $$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]$$

If we consider two matrices A and B are commutative on product then AB = BA

$$\Rightarrow \left[ {\matrix{ a & b \cr c & d \cr } } \right]\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ a & b \cr c & d \cr } } \right]$$

$$\Rightarrow \left[ {\matrix{ a & {a + b} \cr c & {c + d} \cr } } \right] = \left[ {\matrix{ {a + c} & {b + d} \cr c & d \cr } } \right]$$

Comparing corresponding elements of two matrices, we get

$$\Rightarrow$$ a + c = a, a + b = b + d and c + d = d

$$\Rightarrow$$ c = 0, a = d and c = 0.

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