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Question: Answered & Verified by Expert
If $A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right], 10 B=\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]$ a is the inverse of $A$, then the value of $\alpha$ is
MathematicsMatricesCOMEDKCOMEDK 2020
Options:
  • A 0
  • B 2
  • C 4
  • D 5
Solution:
2338 Upvotes Verified Answer
The correct answer is: 5
We have,
$A=\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]$ and $B=\frac{1}{10}\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 0 \\ 1 & -2 & 3\end{array}\right]$
Since, $B$ is inverse of $A$, then
$A B=I$
$\Rightarrow \frac{1}{10}\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\left[\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right]=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
$\Rightarrow \quad\left[\begin{array}{ccc}10 & 0 & 5-\alpha \\ 0 & 10 & \alpha-5 \\ 0 & 0 & 5+\alpha\end{array}\right]=\left[\begin{array}{ccc}10 & 0 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10\end{array}\right]$
$\therefore 5-\alpha=0$ or $\alpha-5=0$ or $5+\alpha=10$
$\Rightarrow \quad a=5$

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