Question

If \[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\] , \[I = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\] and \[{A^{ - 1}} = \dfrac{1}{6}\left[ {{A^2} + cA + dI} \right]\] where \[c,d \in R\]. Then find the value of the pair \[\left( {c,d} \right)\].
A. \[\left( {6, 11} \right)\]
B. \[\left( {6, - 11} \right)\]
C. \[\left( { - 6, 11} \right)\]
D. \[\left( { - 6, - 11} \right)\]

Answer 1

Hint: First, calculate the determinant of the given matrix \[A\]. Check whether the determinant is non-zero or not. Then by using the cofactor matrix find the adjoint matrix of the matrix \[A\]. Substitute the values in the formula of the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)\]. Then use the matrix multiplication method and calculate the value of \[{A^2}\]. After that, use the scalar multiplication method and find the values of \[cA\], and \[dI\]. In the end, substitute the values in the given equation \[{A^{ - 1}} = \dfrac{1}{6}\left[ {{A^2} + cA + dI} \right]\] and solve it to get the required answer.

Formula used:
The 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 matrices are \[A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\] and \[I = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\].
The equation for the inverse matrix is \[{A^{ - 1}} = \dfrac{1}{6}\left[ {{A^2} + cA + dI} \right]\].
Let’s calculate the determinant of the given \[3 \times 3\] matrix \[A\].
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,
\[\left| A \right| = 1\left( {1 \times 4 - 1 \times \left( { - 2} \right)} \right) - 0\left( {0 \times 4 - 0 \times 1} \right) + 0\left( {0 \times \left( { - 2} \right) - 0 \times 1} \right)\]
\[ \Rightarrow \left| A \right| = 1\left( {4 + 2} \right)\]
\[ \Rightarrow \left| A \right| = 6\]

Now find out the adjoint matrix of the given matrix \[A\].
Let’s calculate the co-factors of the matrix \[A\].
\[{A_{11}} = {\left( { - 1} \right)^{1 + 1}}\left( {1 \times 4 - 1 \times \left( { - 2} \right)} \right) = 6\]
\[{A_{12}} = {\left( { - 1} \right)^{1 + 2}}\left( {0 \times 4 - 0 \times 1} \right) = 0\]
\[{A_{13}} = {\left( { - 1} \right)^{1 + 3}}\left( {0 \times \left( { - 2} \right) - 0 \times 1} \right) = 0\]
\[{A_{21}} = {\left( { - 1} \right)^{2 + 1}}\left( {0 \times 4 - 0 \times \left( { - 2} \right)} \right) = 0\]
\[{A_{22}} = {\left( { - 1} \right)^{2 + 2}}\left( {1 \times 4 - 0 \times 0} \right) = 4\]
\[{A_{23}} = {\left( { - 1} \right)^{2 + 3}}\left( {1 \times \left( { - 2} \right) - 0 \times 0} \right) = 2\]
\[{A_{31}} = {\left( { - 1} \right)^{3 + 1}}\left( {0 \times 1 - 0 \times 1} \right) = 0\]
\[{A_{32}} = {\left( { - 1} \right)^{3 + 2}}\left( {1 \times 1 - 0 \times 0} \right) = - 1\]
\[{A_{33}} = {\left( { - 1} \right)^{3 + 3}}\left( {1 \times 1 - 0 \times 0} \right) = 1\]
So, the co-factor matrix of the given matrix is \[\left[ {\begin{array}{*{20}{c}}6&0&0\\0&4&2\\0&{ - 1}&1\end{array}} \right]\]
We know that the cofactor matrix is the transpose of the adjoint matrix.
So, the adjoint matrix of the given matrix is,
\[adj A = \left[ {\begin{array}{*{20}{c}}6&0&0\\0&4&{ - 1}\\0&2&1\end{array}} \right]\]
Now substitute the values of determinant and adjoint matrix in the formula of the inverse matrix \[{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {adj A} \right)\].
We get,
\[{A^{ - 1}} = \dfrac{1}{6}\left[ {\begin{array}{*{20}{c}}6&0&0\\0&4&{ - 1}\\0&2&1\end{array}} \right]\] \[.....\left( 1 \right)\]

Apply the matrix multiplication method to calculate the value of \[{A^2}\].
\[{A^2} = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\]
\[ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}{1 \times 1 + 0 \times 0 + 0 \times 0}&{1 \times 0 + 0 \times 1 + 0 \times - 2}&{1 \times 0 + 0 \times 1 + 0 \times 4}\\{0 \times 1 + 1 \times 0 + 1 \times 0}&{0 \times 0 + 1 \times 1 + 1 \times - 2}&{0 \times 0 + 1 \times 1 + 1 \times 4}\\{0 \times 1 + \left( { - 2} \right) \times 0 + 4 \times 0}&{0 \times 0 + \left( { - 2} \right) \times 1 + 4 \times - 2}&{0 \times 0 + \left( { - 2} \right) \times 1 + 4 \times 4}\end{array}} \right]\]
\[ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&{ - 1}&5\\0&{ - 10}&{14}\end{array}} \right]\] \[.....\left( 2 \right)\]

Apply the scalar multiplication method to calculate the values of \[cA\], and \[dI\].
We get,
\[cA = c\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\]
\[ \Rightarrow cA = \left[ {\begin{array}{*{20}{c}}c&0&0\\0&c&c\\0&{ - 2c}&{4c}\end{array}} \right]\] \[.....\left( 3 \right)\]
And \[dI = d\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}d&0&0\\0&d&0\\0&0&d\end{array}} \right]\] \[.....\left( 4 \right)\]

Now substitute the equations \[\left( 1 \right)\]\[\left( 2 \right)\]\[\left( 3 \right)\], and \[\left( 4 \right)\] in the given equation \[{A^{ - 1}} = \dfrac{1}{6}\left[ {{A^2} + cA + dI} \right]\].
\[\dfrac{1}{6}\left[ {\begin{array}{*{20}{c}}6&0&0\\0&4&{ - 1}\\0&2&1\end{array}} \right] = \dfrac{1}{6}\left[ {\left[ {\begin{array}{*{20}{c}}1&0&0\\0&{ - 1}&5\\0&{ - 10}&{14}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}c&0&0\\0&c&c\\0&{ - 2c}&{4c}\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}d&0&0\\0&d&0\\0&0&d\end{array}} \right]} \right]\]

\[ \Rightarrow \dfrac{1}{6}\left[ {\begin{array}{*{20}{c}}6&0&0\\0&4&{ - 1}\\0&2&1\end{array}} \right] = \dfrac{1}{6}\left[ {\begin{array}{*{20}{c}}{1 + c + d}&0&0\\0&{ - 1 + c + d}&{5 + c}\\0&{ - 10 - 2c}&{14 + 4c + d}\end{array}} \right]\]
Equate both sides
\[5 + c = - 1\] and \[1 + c + d = 6\]
Solving these equations, we get
\[c = - 6\] and \[d = 11\]
Hence the correct option is C.

Note: Students should keep in mind that the product of two matrices is defined only if the number of columns of the first matrix is equal to the number of rows of the second matrix. And the adjoint matrix of any matrix is the transpose of its cofactor matrix. We can equate two matrices only if they have the same order.