Question

Let two matrices are given as A \[=\left( \begin{matrix}
   1 & -1 & 1 \\
   2 & 1 & -3 \\
   1 & 1 & 1 \\
\end{matrix} \right)\] and 10B \[=\left( \begin{matrix}
   4 & 2 & 2 \\
   -5 & 0 & \alpha \\
   1 & -2 & 3 \\
\end{matrix} \right)\] . If B is the inverse of A, then find the value of \[\alpha \] .

Answer 1

HINT: If two matrices say X and Y, of the same order are inverse of each other, then the following result or property can be used directly which is
 \[X\cdot Y=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)\]
That is, if two matrices of the same order are inverse of each other, then the product of those two matrices is an identity matrix of the same order.

Complete step-by-step solution -
As mentioned in the question, we have to find the value of \[\alpha \] , so, for that, on following the information provided in the hint, we get
\[A\cdot B=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)\]
Now, we have to multiply A and B as follows
\[\begin{align}
  & A\cdot B=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right) \\
 & \left( \begin{matrix}
   1 & -1 & 1 \\
   2 & 1 & -3 \\
   1 & 1 & 1 \\
\end{matrix} \right)\cdot \dfrac{1}{10}\left( \begin{matrix}
   4 & 2 & 2 \\
   -5 & 0 & \alpha \\
   1 & -2 & 3 \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right) \\
 & \dfrac{1}{10}\left( \begin{matrix}
   1\cdot 4+\left( -1 \right)\cdot (-5)+1\cdot 1 & 1\cdot 2+(-1)\cdot 0+1\cdot (-2) & 1\cdot 2+(-1)\cdot \alpha +1\cdot 3 \\
   2\cdot 4+1\cdot (-5)+(-3)\cdot 1 & 2\cdot 2+1\cdot 0+(-3)\cdot (-2) & 2\cdot 2+1\cdot \alpha +(-3)\cdot 3 \\
   1\cdot 4+1\cdot (-5)+1\cdot 1 & 1\cdot 2+1\cdot 0+1\cdot (-2) & 1\cdot 2+1\cdot \alpha +1\cdot 3 \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right) \\
 & \dfrac{1}{10}\left( \begin{matrix}
   4+5+1 & 2-2 & 2-\alpha +3 \\
   8-5-3 & 4+6 & 4+\alpha -9 \\
   4-5+1 & 2-2 & 2+\alpha +3 \\
\end{matrix} \right)=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right) \\
 & \left( \begin{matrix}
   10 & 0 & 5-\alpha \\
   0 & 10 & -5+\alpha \\
   0 & 0 & 5+\alpha \\
\end{matrix} \right)=\left( \begin{matrix}
   10 & 0 & 0 \\
   0 & 10 & 0 \\
   0 & 0 & 10 \\
\end{matrix} \right) \\
\end{align}\]
On comparing LHS and RHS, we get that
\[\begin{align}
  & 5-\alpha =0 \\
 & -5+\alpha =0 \\
 & 5+\alpha = 10 \\
\end{align}\]
From the above three equations, we can deduce that the value of \[\alpha \] is 5.

NOTE: - The students can make an error if they don’t know the fact that is mentioned in the hint which is that if two matrices say X and Y, of the same order are inverse of each other, then the following result or property can be used directly which is
 \[X\cdot Y=\left( \begin{matrix}
   1 & 0 & 0 \\
   0 & 1 & 0 \\
   0 & 0 & 1 \\
\end{matrix} \right)\]
That is, if two matrices of the same order are inverse of each other, then the product of those two matrices is an identity matrix of the same order.
Solving the question would not be possible if the above fact is not known beforehand.