AIEEE 2012

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MCQ (Single Correct Answer)

Let $$P$$ and $$Q$$ be $$3 \times 3$$ matrices $$P \ne Q.$$ If $${P^3} = {Q^3}$$ and
$${P^2}Q = {Q^2}P$$ then determinant of $$\left( {{P^2} + {Q^2}} \right)$$ is equal to :

$$-2$$

$$1$$

$$0$$

$$-1$$

Explanation

Given

$${P^3} = {q^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

$${P^2}Q = {Q^2}p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$

Subtracting $$(1)$$ and $$(2)$$, we get

$${P^3} - {P^2}Q = {Q^3} - {Q^2}P$$

$$ \Rightarrow {P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0$$

$$ \Rightarrow \left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0$$

$$ \Rightarrow \left| {{p^2} + {Q^2}} \right| = 0$$

as $$P \ne Q$$

AIEEE 2012

MCQ (Single Correct Answer)

Let $$A = \left( {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right)$$. If $${u_1}$$ and $${u_2}$$ are column matrices such
that $$A{u_1} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right)$$ and $$A{u_2} = \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right),$$ then $${u_1} + {u_2}$$ is equal to :

$$\left( {\matrix{ -1 \cr 1 \cr 0 \cr } } \right)$$

$$\left( {\matrix{ -1 \cr 1 \cr -1 \cr } } \right)$$

$$\left( {\matrix{ -1 \cr -1 \cr 0 \cr } } \right)$$

$$\left( {\matrix{ 1 \cr -1 \cr -1 \cr } } \right)$$

Explanation

Let $$A{u_1} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right)\,\,\,\,\,\,A{u_2} = \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right)$$

Then, $$A{u_1} + A{u_2} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right) + \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right)$$

$$ \Rightarrow A\left( {{u_1} + {u_2}} \right) = \left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$

Also, $$A = \left( {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right)$$

$$ \Rightarrow \left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1$$

We know,

$${A^{ - 1}} = {1 \over {\left| A \right|}}\,adjA \Rightarrow {A^{ - 1}} = adj\left( A \right)$$

( as $$\left| A \right| = 1$$ )

Now, from equation $$(1)$$, we have

$${u_1} + {u_2} = {A^{ - 1}}\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$$

$$ = \left[ {\matrix{ 1 & 0 & 0 \cr { - 2} & 1 & 0 \cr 1 & { - 2} & 1 \cr } } \right]\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$$

$$ = \left[ {\matrix{ 1 \cr { - 1} \cr { - 1} \cr } } \right]$$

AIEEE 2011

MCQ (Single Correct Answer)

The number of values of $$k$$ for which the linear equations
$$4x + ky + 2z = 0,kx + 4y + z = 0$$ and $$2x+2y+z=0$$ possess a non-zero solution is

$$2$$

$$1$$

zero

$$3$$

Explanation

$$\Delta = 0 \Rightarrow \left| {\matrix{ 4 & k & 2 \cr k & 4 & 1 \cr 2 & 2 & 1 \cr } } \right| = 0$$

$$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0$$

$$ \Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0$$

$$ \Rightarrow {k^2} - 6k + 8 = 0$$

$$ \Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2$$

AIEEE 2011

MCQ (Single Correct Answer)

Let $$A$$ and $$B$$ be two symmetric matrices of order $$3$$.
Statement - 1: $$A(BA)$$ and $$(AB)$$$$A$$ are symmetric matrices.
Statement - 2: $$AB$$ is symmetric matrix if matrix multiplication of $$A$$ with $$B$$ is commutative.

statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1.

statement - 1 is true, statement - 2 is false.

statement - 1 is false, statement -2 is true

statement -1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1.

Explanation

$$\therefore$$ $$\,\,\,\,\,A' = A,B' = B$$

Now $$\,\,\,\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A'$$

$$ = \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right)$$

Similarly $$\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A$$

So, $$A\left( {BA} \right)\,\,\,\,$$ and $$A\left( {BA} \right)\,\,\,\,$$ are symmetric matrices.

Again $$\left( {AB} \right)' = B'A' = BA$$

Now if $$BA=AB$$, then $$AB$$ is symmetric matrix.