Question

if $P=\left|\begin{array}{ccc}1& \alpha & 3\\ 1& 3& 3\\ 2& 4& 4\end{array}\right|$ is the adjoint of $a3\times 3$ matrix $A$ and $|A|$$=4$, then $\alpha$ is equal to:

• A. $4$
• B. $11$`
• C. $5$
• D. $0$

Answer 1

The correct option is B $11$`

Solution:-

$P=\left|\begin{array}{ccc}1& \alpha & 3\\ 1& 3& 3\\ 2& 4& 4\end{array}\right|$

Solving the determinant, we have

$\left|\begin{array}{c}P\end{array}\right|=1(12-12)-\alpha (4-6)+3(4-6)$

$\Rightarrow \left|\begin{array}{c}P\end{array}\right|=0-(-2\alpha )+3(-2)$

$\left|\begin{array}{c}P\end{array}\right|=2\alpha -6$ $.....\left(1\right)$

$\because P=adj\left(A\right)\left(\text{Given}\right)$

$\therefore \left|\begin{array}{c}P\end{array}\right|=\left|\begin{array}{c}adj\left(A\right)\end{array}\right|$

$\Rightarrow \left|\begin{array}{c}P\end{array}\right|={\left|\begin{array}{c}A\end{array}\right|}^{(3-1)}=4^2(\because \left|\begin{array}{c}adj\left(A\right)\end{array}\right|={\left|\begin{array}{c}A\end{array}\right|}^{(n-1)};\text{whereas n is the order of matrix A})$

$\left|\begin{array}{c}P\end{array}\right|=16.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$2\alpha -6=16$

$2\alpha =22$

$\Rightarrow \alpha =11$

Hence the value of $\alpha$ is $11$.