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

Question

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

Aicharya · Accepted Answer

$ \begin{aligned} x^2-25 x+24 & =0 \ x^2-x-24 x+24 & =0 \ x(x-1)-24(x-1) & =0 \ (x-1)(x-24) & =0 \Rightarrow x=1,24 \end{aligned} $ $\because k$ is one of the root of the Eq. (i), $ \begin{array}{rlrl} 	herefore & k & =1,24 \ & & k & =1, \ 	herefore & \quad A & =\left[\begin{array}{lll} 1 & 2 & 1 \ 3 & 2 & 3 \ 1 & 1 & 1 \end{array}ight] \ |A| & =1(2-3)-2(3-3)+1(3-2) \ & =-1-0+1=0 \ |A| & =0 \end{array} $ $k=1$, not possible, because given matrix $A$ is singular. Now, $ \begin{aligned} & k=24 	ext {, } \ & 	herefore \quad A=\left[\begin{array}{ccc} 1 & 2 & 1 \ 3 & 2 & 3 \ 1 & 1 & 24 \end{array}ight] \ & |A|=1(48-3)-2(72-3)+1(3-2) \ & =45-138+1=-92 
eq 0 \ & \operatorname{adj} A=\left[\begin{array}{ccc} 45 & -69 & 1 \ -47 & 23 & 1 \ 4 & 0 & -4 \end{array}ight]^1 \ & =\left[\begin{array}{ccc} 45 & -47 & 4 \ -69 & 23 & 0 \ 1 & 1 & -4 \end{array}ight] \ & 	herefore \quad A^{-1}=\frac{1}{|A|} \cdot \operatorname{adj} A \ & =-\frac{1}{92}\left[\begin{array}{ccc} 45 & -47 & 4 \ -69 & 23 & 0 \ 1 & 1 & -4 \end{array}ight] \ & \end{aligned} $

If $k$ is one of the roots of the equation $x^2-25 x+24=0$ such that $A=\left[\begin{array}{lll}1 & 2 & 1 \\…

Solution