If \(Q\) is the inverse of \(A\), when \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 &…

If \(Q\) is the inverse of \(A\), when \(A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\) and \(10 \times Q=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3\end{array}\right]\), find \(x=\)
  1. 2
  2. 3
  3. 4
  4. 5

Solution

Given, \(Q=A^{-1}\) \(\begin{aligned} A & =\left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{array}\right] \\ |A| & =1(1+3)+1(2+3)+1(2-1)=4+5+1 \\ |A| & =10 \end{aligned}\) \(\operatorname{Adj} A=\left[\begin{array}{ccc}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{array}\right]^{\mathrm{T}}\) Adj \(A=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\) \(\begin{aligned} & A^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{Adj} A \\ & A^{-1}=\frac{1}{10}\left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right] \\ & 10 Q=\left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3 \end{array}\right] \\ & 10 A^{-1}=\left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3 \end{array}\right] \\ & {\left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3 \end{array}\right]=\left[\begin{array}{rrr} 4 & 2 & 2 \\ -5 & 0 & x \\ 1 & -2 & 3 \end{array}\right]} \\ & \therefore \quad x=5 \\ \end{aligned}\) Hence, option (d) is correct.

Asked in: AP EAMCET 2020 (18 Sep Shift 2)

Practice more Matrices questions on Aicharya