Discussion Forum

Let   $P=\begin{bmatrix}3 & -1 &-2\\2 & 0 &\alpha\\3&-5&0\end{bmatrix}$  , where  $\alpha \epsilon R$ .Suppose Q= [$q_{ij}$] is a matrix such that  PQ=kl, where  $k\epsilon R, k\neq 0$ and l is the identity matrix of order 3. If  $q_{23}=-\frac{k}{8}$  and $det(Q)=\frac{k^{2}}{2}$  then

Answer:

Explanation:

Here,  $P=\begin{bmatrix}3 & -1 &-2\\2 & 0 &\alpha\\3&-5&0\end{bmatrix} $

 Now,   $|P|=3(5\alpha)+1(-3\alpha)-2(-10)=12\alpha+20$         ...........(i)

$\therefore$    adj(P)=   $\begin{bmatrix}5\alpha & 2\alpha &-10\\-10 & 6 &12\\\alpha&-(3\alpha+4)&2\end{bmatrix} ^{T}$

 =  $\begin{bmatrix}5\alpha & -10 &-\alpha\\2\alpha & 6 &-3\alpha-4\\-10&12&2\end{bmatrix}$   .....(ii)

 As,  PQ=kI

 $\Rightarrow$    $|P||Q|=|kI|$

 $\Rightarrow$    $|P||Q|=k^{3}$

 $\Rightarrow$    $|P|\left(\frac{k^{2}}{2}\right)=k^{3}$     $\left[ given, |Q|=\frac{k^{2}}{2}\right]$

 $\Rightarrow$  |P|=2k    ..............(iii)

$\because$             PQ=ki

$\because$           $Q= kp^{-1}I$

$=k.\frac{adjP}{|P|}=\frac{k(adjP)}{2k}$   [ from Eq. (iii)]

 $=\frac{adjP}{2}$

 $=\frac{1}{2}\begin{bmatrix}5\alpha & -10 &-\alpha\\2\alpha & 6 &-3\alpha-4\\-10&12&2\end{bmatrix}$

 $\therefore$    $q_{23}=\frac{-3\alpha-4}{2} \left[ given, q_{23}=-\frac{k}{8}\right]$

$\Rightarrow$              $-\frac{(3\alpha+4)}{2}=-\frac{k}{8}$

 $\Rightarrow$    $(3\alpha+4)\times 4=k$

$\Rightarrow$   $12\alpha+16=k$       .......(iv)

From Eq .(iii) . |P|=2k

 $\Rightarrow$   $12\alpha +20=2k$