Let A=[1−1121−3111] and (10B)=[422−50α123] , If B is the inverse of matrix A, then α is

AB=I ⇒ [1−1121−3111]110 [422−50α1−23] =[100010001] ⇒ 110 [1005−α010−5+α005+α]=[100010001] α=5

Let A=[1−1121−3111] and (10B)=[422−50α123] , If B is the inverse of matrix A, then α is

AB=I ⇒ [1−1121−3111]110 [422−50α1−23] =[100010001] ⇒ 110 [1005−α010−5+α005+α]=[100010001] α=5

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Let A= $[\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix}]$ and $(10 B) = [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & α \\ 1 & 2 & 3 \end{matrix}]$ , If B is the inverse of matrix A, then $α$ is

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Detailed Solution

$A B = I$

$\Rightarrow [\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix}] \frac{1}{10} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & α \\ 1 & - 2 & 3 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$\Rightarrow \frac{1}{10} [\begin{matrix} 10 & 0 & 5 - α \\ 0 & 10 & - 5 + α \\ 0 & 0 & 5 + α \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$