Question

Let \[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right]\] , and \[\left( {10} \right)B = \left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right]\]. If \[B\] is the inverse of matrix \[A\], then what is the value of \[\alpha \]?
A. 5
B. \[ - 1\]
C. 2
D. \[ - 2\]

Answer 1

Hint: First, use the given information \[B\] is the inverse of matrix \[A\] and calculate the value of \[10{A^{ - 1}}\]. Then multiply both sides by the matrix \[A\] and simplify the equation. After that apply the identity \[A{A^{ - 1}} = A{A^{ - 1}} = I\] and further simplify the equation. In the end, substitute the values of the matrices in the equation and solve it to get the required answer

Formula used:
For a non-singular square matrix \[A\], \[A{A^{ - 1}} = A{A^{ - 1}} = I\]

Complete step by step solution:
The given matrices are
\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right]\] \[.....\left( 1 \right)\]
\[\left( {10} \right)B = \left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right]\] \[.....\left( 2 \right)\]
Let’s apply the given information \[B\] is the inverse of matrix \[A\].
We get,
\[{A^{ - 1}} = B\]

Substitute the values in the equation \[\left( 2 \right)\].
\[\left( {10} \right){A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right]\]
Now multiply both sides by the matrix \[A\].
\[\left( {10} \right){A^{ - 1}}A = \left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right]A\]
Use the identity \[A{A^{ - 1}} = A{A^{ - 1}} = I\].
\[\left( {10} \right)I = \left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right]A\]
Now substitute the values in the above equation.
\[\left( {10} \right)\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}4&2&2\\{ - 5}&0&\alpha \\1&{ - 2}&3\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&{ - 1}&1\\2&1&{ - 3}\\1&1&1\end{array}} \right]\]
Apply the scalar and matrix multiplication properties of the matrices.
\[\left[ {\begin{array}{*{20}{c}}{10}&0&0\\0&{10}&0\\0&0&{10}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{\left( {4 \times 1} \right) + \left( {2 \times 1} \right) + \left( {2 \times 1} \right)}&{\left( {4 \times - 1} \right) + \left( {2 \times 1} \right) + \left( {2 \times 1} \right)}&{\left( {4 \times 1} \right) + \left( {2 \times - 3} \right) + \left( {2 \times 1} \right)}\\{\left( { - 5 \times 1} \right) + \left( {0 \times 2} \right) + \left( {\alpha \times 1} \right)}&{\left( { - 5 \times - 1} \right) + \left( {0 \times 1} \right) + \left( {\alpha \times 1} \right)}&{\left( { - 5 \times 1} \right) + \left( {0 \times - 3} \right) + \left( {\alpha \times 1} \right)}\\{\left( {1 \times 1} \right) + \left( { - 2 \times 2} \right) + \left( {3 \times 1} \right)}&{\left( {1 \times - 1} \right) + \left( { - 2 \times 1} \right) + \left( {3 \times 1} \right)}&{\left( {1 \times 1} \right) + \left( { - 2 \times - 3} \right) + \left( {3 \times 1} \right)}\end{array}} \right]\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}{10}&0&0\\0&{10}&0\\0&0&{10}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{4 + 2 + 2}&{ - 4 + 2 + 2}&{4 - 6 + 2}\\{ - 5 + \alpha }&{5 + \alpha }&{ - 5 + \alpha }\\{1 - 4 + 3}&{ - 1 - 2 + 3}&{1 + 6 + 3}\end{array}} \right]\]
\[ \Rightarrow \left[ {\begin{array}{*{20}{c}}{10}&0&0\\0&{10}&0\\0&0&{10}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{10}&0&0\\{\alpha - 5}&{\alpha + 5}&{\alpha - 5}\\0&0&{10}\end{array}} \right]\]
Now compare both sides.
We get,
\[\alpha - 5 = 0\]
\[ \Rightarrow \alpha = 5\]
Hence the correct option is A.

Note: Students often get confused about the matrix multiplication. The product of two matrices is defined if the number of columns of the first matrix is equal to the number of rows of the second matrix. We can equate the matrices only if both the matrices have the same number of rows and the same number of columns, only then the corresponding elements of both matrices will be equal.