If k is one of the roots of the equation x^2- | vedclass

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(B) Given the quadratic equation $x^2-25x+24=0$. Factoring the equation: $x^2-x-24x+24=0 \Rightarrow x(x-1)-24(x-1)=0 \Rightarrow (x-1)(x-24)=0$. Thus

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$-\frac{1}{92}\left[\begin{array}{ccc}45 & -47 & 4 \ -69 & 23 & 0 \ 1 & 1 & -4\end{array}ight]$

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(B) Given the quadratic equation $x^2-25x+24=0$. Factoring the equation: $x^2-x-24x+24=0 \Rightarrow x(x-1)-24(x-1)=0 \Rightarrow (x-1)(x-24)=0$. Thus,the roots are $x=1$ and $x=24$. Since $A$ is a non-singular matrix,$|A| 
eq 0$. Case $1$: If $k=1$,then $A=\left[\begin{array}{lll}1 & 2 & 1 \ 3 & 2 & 3 \ 1 & 1 & 1\end{array}ight]$. $|A| = 1(2-3) - 2(3-3) + 1(3-2) = -1 - 0 + 1 = 0$. Since $|A|=0$,$k=1$ is not possible. Case $2$: If $k=24$,then $A=\left[\begin{array}{ccc}1 & 2 & 1 \ 3 & 2 & 3 \ 1 & 1 & 24\end{array}ight]$. $|A| = 1(48-3) - 2(72-3) + 1(3-2) = 45 - 138 + 1 = -92$. Now,find the adjoint of $A$: $C_{11} = 45, C_{12} = -69, C_{13} = 1$ $C_{21} = -47, C_{22} = 23, C_{23} = 1$ $C_{31} = 4, C_{32} = 0, C_{33} = -4$ $	ext{adj } A = \left[\begin{array}{ccc}45 & -47 & 4 \ -69 & 23 & 0 \ 1 & 1 & -4\end{array}ight]$. Therefore,$A^{-1} = \frac{1}{|A|} 	ext{adj } A = \frac{1}{-92} \left[\begin{array}{ccc}45 & -47 & 4 \ -69 & 23 & 0 \ 1 & 1 & -4\end{array}ight] = -\frac{1}{92} \left[\begin{array}{ccc}45 & -47 & 4 \ -69 & 23 & 0 \ 1 & 1 & -4\end{array}ight]$.

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

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