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Mathematics Question on Transpose of a Matrix

If kk is one of the roots of the equation x2−25x+24=0x^2 - 25x + 24 = 0 such that A=[121 323 11k;]A = \begin{bmatrix}1&2&1\\\ 3&2&3\\\ 1&1&k;\end{bmatrix} is a non-singular matrix, then A−1A^{-1} =

A

−146[90−948 −138460 22−8]- \frac{1}{46} \begin{bmatrix}90&-94&8\\\ -138&46&0\\\ 2&2&-8\end{bmatrix}

B

−192[45−474 −69230 114]- \frac{1}{92} \begin{bmatrix}45&-47&4\\\ -69&23&0\\\ 1&1&4\end{bmatrix}

C

−146[45−474 −69230 11−4]- \frac{1}{46} \begin{bmatrix}45&-47&4\\\ -69&23&0\\\ 1&1&-4\end{bmatrix}

D

−192[90−948 −138460 22−8]- \frac{1}{92} \begin{bmatrix}90&-94&8\\\ -138&46&0\\\ 2&2&-8\end{bmatrix}

Answer

−192[45−474 −69230 114]- \frac{1}{92} \begin{bmatrix}45&-47&4\\\ -69&23&0\\\ 1&1&4\end{bmatrix}

Explanation

Solution

x2−25x+24=0…x^{2}-25 x+24=0 \dots(i)
x2−x−24x+24=0x^{2}-x-24 x+24 = 0
=x(x−1)−24(x−1)=0= x(x-1)-24(x-1)= 0
=(x−1)(x−24)=0= (x-1)(x-24) =0
⇒x=1,24\Rightarrow x=1,24
∵k\because k is one of the root of the E (i),
∴k=1,24\therefore k=1, 24
If k=1k = 1,
∴A=[121 323 111]\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)|A| =1(2-3)-2(3-3)+1(3-2)
=−1−0+1=0=-1-0+1=0
∣A∣=0|A| =0
k=1k=1, not possible, because given matrix AA is singular.
Now, k=24k=24,
∴A=[121 323 1124]\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)|A| =1(48-3)-2(72-3)+1(3-2)
=45−138+1=−92≠0=45-138+1=-92 \neq 0
adj A=[45−691 −47231 40−4]1A =\begin{bmatrix} 45 & -69 & 1 \\\ -47 & 23 & 1 \\\ 4 & 0 & -4\end{bmatrix}^{1} =[45−474 −69230 11−4]=\begin{bmatrix}45 & -47 & 4 \\\ -69 & 23 & 0 \\\ 1 & 1 & -4\end{bmatrix}
A−1=1∣A∣⋅adjAA^{-1} =\frac{1}{|A|} \cdot \text{adj} A
=−192[45−474 −69230 11−4]=-\frac{1}{92}\begin{bmatrix}45 & -47 & 4 \\\ -69 & 23 & 0 \\\ 1 & 1 & -4\end{bmatrix}