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1

WB JEE 2010

MCQ (Single Correct Answer)

If $$\omega$$ is an imaginary cube root of unity and $$\left| {\matrix{ {x + {\omega ^2}} & \omega & 1 \cr \omega & {{\omega ^2}} & {1 + x} \cr 1 & {x + \omega } & {{\omega ^2}} \cr } } \right| = 0$$, then one of the values of x is

A
1
B
0
C
$$-$$1
D
2

Explanation

$$\left| {\matrix{ {x + 1 + \omega + {\omega ^2}} & \omega & 1 \cr {x + 1 + \omega + {\omega ^2}} & {{\omega ^2}} & {1 + x} \cr {x + 1 + \omega + {\omega ^2}} & {x + \omega } & {{\omega ^2}} \cr } } \right| = 0\,[{C_1} \to {C_1} + ({C_2} + {C_3})]$$

$$ \Rightarrow \left| {\matrix{ x & \omega & 1 \cr x & {{\omega ^2}} & {1 + x} \cr x & {x + \omega } & {{\omega ^2}} \cr } } \right| = 0$$ [$$\because$$ $$1 + \omega + {\omega ^2} = 0$$]

$$ \Rightarrow \left| {\matrix{ x & \omega & 1 \cr 0 & {{\omega ^2} - \omega } & x \cr 0 & x & {{\omega ^2} - 1} \cr } } \right| = 0\left[ {\matrix{ {{R_2} \to {R_2} - {R_1}} \cr {{R_3} \to {R_3} - {R_1}} \cr } } \right]$$

$$ \Rightarrow x\left| {\matrix{ {{\omega ^2} - \omega } & x \cr x & {{\omega ^2} - 1} \cr } } \right| = 0 \Rightarrow x\{ ({\omega ^2} - \omega )({\omega ^2} - 1) - {x^2}\} = 0$$

$$ \Rightarrow x = 0$$ or $$({\omega ^4} - {\omega ^3} - {\omega ^2} + \omega - {x^2}) = 0$$

$$ \Rightarrow x = 0$$ or $${x^2} = 3\omega $$ (complex)

$$\therefore$$ $$x = 0$$ is only solution.

2

WB JEE 2010

MCQ (Single Correct Answer)

If the matrices $$A = \left[ {\matrix{ 2 & 1 & 3 \cr 4 & 1 & 0 \cr } } \right]$$ and $$B = \left[ {\matrix{ 1 & { - 1} \cr 0 & 2 \cr 5 & 0 \cr } } \right]$$, then AB will be

A
$$\left[ {\matrix{ {17} & 0 \cr 4 & { - 2} \cr } } \right]$$
B
$$\left[ {\matrix{ 4 & 0 \cr 0 & 4 \cr } } \right]$$
C
$$\left[ {\matrix{ {17} & 4 \cr 0 & { - 2} \cr } } \right]$$
D
$$\left[ {\matrix{ 0 & 0 \cr 0 & 0 \cr } } \right]$$

Explanation

$$AB = \left( {\matrix{ 2 & 1 & 3 \cr 4 & 1 & 0 \cr } } \right)\left( {\matrix{ 1 & { - 1} \cr 0 & 2 \cr 5 & 0 \cr } } \right) = \left( {\matrix{ {17} & 0 \cr 4 & { - 2} \cr } } \right)$$

3

WB JEE 2009

MCQ (Single Correct Answer)

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

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

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

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

4

WB JEE 2009

MCQ (Single Correct Answer)

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.

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