Let P- begin bmatrix 3 -1 -22 - 0 - lalpha 38-5 - 0lendbmatrix - where - R- Suppose Q - q ij - is a matrix such that PQ - kI- where k - R - k -0 and I is the identity matrix of order 3 - If q 23- - k -8 and det Q - k 2-2 thenA- -0- k -8B- 4 - k -8-0C- detP adjQ-29D- det Q adj P -213

The correct options are B 4α k+8=0 C det(P adj (Q))=2^9Here, P=[ 3 amp; 1 amp; 2; 2 amp;0 amp;α; 3 amp; 5 amp;0 ] Now, |P|=3(5 α) + 1( 3α) ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

Let P= begin ...

Question

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

α = 0, k = 8

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4 α - k + 8 = 0

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

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

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Solution

The correct options are
B $4 α - k + 8 = 0$
C $d e t (P a d j (Q)) = 2^{9}$
Here, $P = ⎡ ⎢ ⎣ \begin{matrix} 3 & - 1 & - 2 2 & 0 & α 3 & - 5 & 0 \end{matrix} ⎤ ⎥ ⎦$
Now, $| P | = 3 (5 α) + 1 (- 3 α) - 2 (- 10)$
$= 12 α + 20$ . . . (i)
$∴ a d j (P) = {⎡ ⎢ ⎣ \begin{matrix} 5 α & 3 α & - 10 - 10 & 6 & 12 - α & - (3 α + 4) & 2 \end{matrix} ⎤ ⎥ ⎦}^{T}$
$= ⎡ ⎢ ⎣ \begin{matrix} 5 α & 10 & - α 3 α & 6 & - 3 α - 4 - 10 & 12 & 2 \end{matrix} ⎤ ⎥ ⎦$ . . . (ii)
A PQ = kI
$\Rightarrow$ |P||Q| = |kI|
$\Rightarrow$ |P||Q| = $k^{3}$
$\Rightarrow | P | (\frac{k^{2}}{2}) = k^{3} [g i v e n, | Q | = \frac{k^{2}}{2}]$
$\Rightarrow$ |P| = 2k . . . (iii)
$∵$ PQ = ki
$∴$ Q $= k p^{- 1} I$
$= k . \frac{a d j P}{| P |} = \frac{k (a d j P)}{2 k}$ [from Eq. (iii)]
$= \frac{a d j P}{2} = \frac{1}{2} ⎡ ⎢ ⎣ \begin{matrix} 5 α & - 10 & - α 3 α & 6 & - 3 α - 4 - 10 & 12 & 2 \end{matrix} ⎤ ⎥ ⎦ ∴ q_{23} = \frac{- 3 α - 4}{2} [g i v e n, q_{23} = - \frac{k}{8}]$
$\Rightarrow - \frac{(3 α + 4)}{2} = - \frac{k}{8} \Rightarrow (3 α + 4) \times 4 = k$
$\Rightarrow 12 α + 16 = k$ . . . (iv)
From Eq. (iii), |P| = 2k
$\Rightarrow$ $4 α$ - k + 8 = - 4 - 4 + 8 = 0
$∴$ Option (b) is correct.
Now, |P adj (Q)| = |P||adj Q|
$= 2 k {(\frac{k^{2}}{2})}^{2} = \frac{k^{5}}{2} = \frac{2^{10}}{2} = 2^{9}$
$∴$ Option (c) is correct.

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Similar questions

Q. Let

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

, where

α \in R

. Suppose

Q = [q_{i j}]

is a matrix such that

P Q = k I

, where

k \in R, k \neq 0

and

I

the identity matrix of order

3

. If

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

and

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

, then:

Q. Let

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

, where

α \in R

. Suppose

Q = [q_{i j}]

is matrix such that

P Q = k I

, where

k \in R, k \neq 0

and

I

is the identity matrix of order

3

. If

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

and

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

, then

Q. Let

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

, where

α ϵ R

. Suppose

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}

then

Q. 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.

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

and

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

then

Q. Let

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

, where

α \in R

. Suppose

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}

and

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

, then

α^{2} + k^{2}

is equal to

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Let P=⎡⎢⎣3−1−220α3−50⎤⎥⎦, where αϵR. Suppose Q=[qij] is a matrix such that PQ = kI, where kϵR, k≠0 and I is the identity matrix of order 3. If q23=−k8 and det(Q)=k22 then