Question

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

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

Answer 1

Answer: (B)

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