Question

If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =

• A. $- \frac{1}{46} \begin{bmatrix}90&-94&8\\ -138&46&0\\ 2&2&-8\end{bmatrix} $
• B. $- \frac{1}{92} \begin{bmatrix}45&-47&4\\ -69&23&0\\ 1&1&4\end{bmatrix} $
• C. $ - \frac{1}{46} \begin{bmatrix}45&-47&4\\ -69&23&0\\ 1&1&-4\end{bmatrix} $
• D. $ - \frac{1}{92} \begin{bmatrix}90&-94&8\\ -138&46&0\\ 2&2&-8\end{bmatrix} $

Answer 1

Answer: (B)

$x^{2}-25 x+24=0 \dots$(i)
$ x^{2}-x-24 x+24 = 0 $
$= x(x-1)-24(x-1)= 0 $
$= (x-1)(x-24) =0$
$ \Rightarrow x=1,24$
$\because k$ is one of the root of the Eq. (i),
$\therefore k=1, 24$
If $k = 1$,
$\therefore A= \begin{bmatrix} 1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 1\end{bmatrix}$
$|A| =1(2-3)-2(3-3)+1(3-2) $
$=-1-0+1=0 $
$|A| =0 $
$k=1$, not possible, because given matrix $A$ is singular.
Now, $k=24$,
$ \therefore A =\begin{bmatrix} 1 & 2 & 1 \\ 3 & 2 & 3 \\ 1 & 1 & 24\end{bmatrix} $
$ |A| =1(48-3)-2(72-3)+1(3-2) $
$=45-138+1=-92 \neq 0 $
adj $A =\begin{bmatrix} 45 & -69 & 1 \\ -47 & 23 & 1 \\ 4 & 0 & -4\end{bmatrix}^{1}$ $=\begin{bmatrix}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{bmatrix}$
$ A^{-1} =\frac{1}{|A|} \cdot \text{adj} A $
$=-\frac{1}{92}\begin{bmatrix}45 & -47 & 4 \\ -69 & 23 & 0 \\ 1 & 1 & -4\end{bmatrix} $