Question

If matrix \[A = \left[ {\begin{array}{*{20}{c}}3&2&4\\1&2&{ - 1}\\0&1&1\end{array}} \right]\] and \[{A^{ - 1}} = \dfrac{1}{K}adj\left( A \right)\]. Then what is the value of \[K\]?
A. 7
B. \[ - 7\]
C. \[\dfrac{1}{7}\]
D. \[11\]

Answer 1

Hint: First, compare the given value of the inverse matrix with the standard formula of the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]. Then calculate the determinant of the matrix \[A\] to the required value of \[K\].

Formula used:
Inverse matrix: \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]
The determinant of a \[3 \times 3\] matrix \[A = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\] is: \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{22}}} \right)\]

Complete step by step solution:
The given matrix is \[A = \left[ {\begin{array}{*{20}{c}}3&2&4\\1&2&{ - 1}\\0&1&1\end{array}} \right]\] and the inverse of the matrix is \[{A^{ - 1}} = \dfrac{1}{K}adj\left( A \right)\]

Compare the value of the given inverse matrix with the standard formula of the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\].
We get,
\[\dfrac{1}{{\left| A \right|}}adj\left( A \right) = \dfrac{1}{K}adj\left( A \right)\]
After comparing both sides, we get
\[ \Rightarrow \left| A \right| = K\] \[.....\left( 1 \right)\]

Now calculate the determinant of the given \[3 \times 3\] matrix.
Apply the formula of the determinant of a \[3 \times 3\] matrix \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{32}}{a_{23}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{31}}{a_{23}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{31}}{a_{23}}} \right)\].
We get,
\[\left| A \right| = 3\left( {2 \times 1 - 1 \times \left( { - 1} \right)} \right) - 2\left( {1 \times 1 - 0 \times \left( { - 1} \right)} \right) + 4\left( {1 \times 1 - 0 \times 2} \right)\]
\[ \Rightarrow \left| A \right| = 3\left( {2 + 1} \right) - 2\left( {1 - 0} \right) + 4\left( {1 - 0} \right)\]
\[ \Rightarrow \left| A \right| = 3\left( 3 \right) - 2\left( 1 \right) + 4\left( 1 \right)\]
\[ \Rightarrow \left| A \right| = 9 - 2 + 4\]
\[ \Rightarrow \left| A \right| = 11\]
Comparing with the equation \[\left( 1 \right)\], we get
\[K = 11\]
Hence the correct option is D.

Note: Students often get confused and calculate the inverse matrix of the matrix by calculating the co-factors, adjoint matrix, and determinant using the formula of the inverse matrix. This method is correct but too lengthy for this type of question.