Q44 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
If $\mathrm{A}=\left[\begin{array}{lll}x & y & y \\ y & x & y \\ y & y & x\end{array}\right]$ is a matrix such that $5 A^{-1}=\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3\end{array}\right]$ then $A^2-4 A=$
- A$5 \mathrm{~A}^{-1}$
- Correct answerB5 I
- C0
- DI
Solution
Given that $5 A^{-1}=\left[\begin{array}{ccc}-3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3\end{array}\right]$
Post multiply by $A$ on both sides $\begin{aligned} & 5 I=\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right]\left[\begin{array}{lll} x & y & y \\ y & x & y \\ y & y & x \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]=\left[\begin{array}{ccc} -3 x+4 y & 2 x-y & 2 x-y \\ 2 x-y & -3 x+4 y & 2 x-y \\ 2 x-y & 2 x-y & -3 x+4 y \end{array}\right] \\ & \therefore-3 x+4 y=5...(i) \\ & 2 x-y=0....(ii) \end{aligned}$
On solving equation (i) and (ii), we get $x=1$ and $y=2$ $\begin{aligned} & \therefore A=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \\ & 5 A^{-1}=\left[\begin{array}{ccc} 1-4 & 2 & 2 \\ 2 & 1-4 & 2 \\ 2 & 2 & 1-4 \end{array}\right]=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]-\left[\begin{array}{lll} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] \\ & \therefore 5 A^{-1}=A-4 I \end{aligned}$
Post multiply by $A$, we get $5 I=A^2-4 A$.
Post multiply by $A$ on both sides $\begin{aligned} & 5 I=\left[\begin{array}{ccc} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{array}\right]\left[\begin{array}{lll} x & y & y \\ y & x & y \\ y & y & x \end{array}\right] \\ & \Rightarrow\left[\begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{array}\right]=\left[\begin{array}{ccc} -3 x+4 y & 2 x-y & 2 x-y \\ 2 x-y & -3 x+4 y & 2 x-y \\ 2 x-y & 2 x-y & -3 x+4 y \end{array}\right] \\ & \therefore-3 x+4 y=5...(i) \\ & 2 x-y=0....(ii) \end{aligned}$
On solving equation (i) and (ii), we get $x=1$ and $y=2$ $\begin{aligned} & \therefore A=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right] \\ & 5 A^{-1}=\left[\begin{array}{ccc} 1-4 & 2 & 2 \\ 2 & 1-4 & 2 \\ 2 & 2 & 1-4 \end{array}\right]=\left[\begin{array}{lll} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{array}\right]-\left[\begin{array}{lll} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array}\right] \\ & \therefore 5 A^{-1}=A-4 I \end{aligned}$
Post multiply by $A$, we get $5 I=A^2-4 A$.
Hence, the correct answer is option: 'B'.