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Expansion of (x-y)^2

Let P = [ [...

Question

Let $P = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 \end{matrix} ⎤ ⎥ ⎦,$ where $α \in R .$ Suppose $Q = [q_{n}]$ is a matrix such that $P Q = k l,$ where $k \in R,$ $k \neq 0$ and $l$ is the identity matrix of order $3.$ if $q_{23} = - \frac{k}{8}$ and $d e t (Q) = \frac{k^{2}}{2},$ then

A

α = 0, k = 8

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B

4 a - k + 8 = 0

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C

d e t (P

a d j (Q)) = 2^{9}

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D

d e t (Q

a d j (P)) = 2^{13}

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Solution

The correct options are
B $4 a - k + 8 = 0$
C $d e t (P$ $a d j (Q)) = 2^{9}$

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Similar questions

P = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 \end{matrix} ⎤ ⎥ ⎦

α ϵ R

Q = [q_{i j}]

is a matrix such that PQ = kI, where

k ϵ R, k \neq 0

and I is the identity matrix of order 3. If

q_{23} = - \frac{k}{8} a n d d e t (Q) = \frac{k^{2}}{2}

P = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 \end{matrix} ⎤ ⎥ ⎦

α ϵ R

Q = [q_{i j}]

is a matrix such that PQ = kI, where

k ϵ R, k \neq 0

and I is the identity matrix of order 3. If

q_{23} = - \frac{k}{8} a n d d e t (Q) = \frac{k^{2}}{2}

P = ⎡ ⎢ ⎣ \begin{matrix} 323 \end{matrix} \begin{matrix} - 1 0 - 5 \end{matrix} \begin{matrix} - 2 α 0 \end{matrix} ⎤ ⎥ ⎦

α \in R

Q = [q_{i j}]

is matrix such that

P Q = k I

k \in R, k \neq 0

I

is the identity matrix of order

3

q_{23} = - \frac{k}{8}

d e t (Q) = \frac{k^{2}}{2}

P = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 \end{matrix} ⎤ ⎥ ⎦

α \in R

Q = [q_{i j}]

is a matrix such that

P Q = k I

k \in R, k \neq 0

I

the identity matrix of order

3

q_{23} = - \frac{k}{8}

det (Q) = \frac{k^{2}}{2}

P = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 \end{matrix} ⎤ ⎥ ⎦

α \in R

Q = [q_{i j}]

is a matrix satisfying

P Q = k I_{3}

for some non-zero

k \in R .

q_{23} = - \frac{k}{8}

| Q | = \frac{k^{2}}{2}

α^{2} + k^{2}

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