Question

For a matrix A = \left[ {\begin{array}{*{20}{c}} 1&1&2 \\\ 3&0&2 \\\ 1&0&3 \end{array}} \right] verify $A\left( {{\text{adj }}A} \right) = \left( {{\text{adj }}A} \right)A = \left| A \right|I$.

Answer 1

First find the cofactor matrix of A. Determine the cofactors of all elements of matrix A. Then transpose the cofactor matrix to get the adjoint of matrix A, i.e. ${\text{adj}}A$. Multiply the adjoint matrix with A in different order and find the determinant value of A to prove the above result.

Complete step-by-step solution:
According to the question, the given matrix is A = \left[ {\begin{array}{*{20}{c}} 1&1&2 \\\ 3&0&2 \\\ 1&0&3 \end{array}} \right].
To determine the adjoint of the matrix, first we have to find out the cofactors of all of its elements. Let the cofactors are denoted by $a$ then we have:
Cofactors of first row elements:

0&2 \\\ 0&3 \end{array}} \right| = 0$$, $${a_{1,2}} = - \left| {\begin{array}{*{20}{c}} 3&2 \\\ 1&3 \end{array}} \right| = - 7$$ and $${a_{1,3}} = \left| {\begin{array}{*{20}{c}} 3&0 \\\ 1&0 \end{array}} \right| = 0$$ Co-factors of second row elements: $${a_{2,1}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\\ 0&3 \end{array}} \right| = - 3$$, $${a_{2,2}} = \left| {\begin{array}{*{20}{c}} 1&2 \\\ 1&3 \end{array}} \right| = 1$$ and $${a_{2,3}} = - \left| {\begin{array}{*{20}{c}} 1&1 \\\ 1&0 \end{array}} \right| = 1$$ Co-factors of third row elements: $${a_{3,1}} = \left| {\begin{array}{*{20}{c}} 1&2 \\\ 0&2 \end{array}} \right| = 2$$, $${a_{3,2}} = - \left| {\begin{array}{*{20}{c}} 1&2 \\\ 3&2 \end{array}} \right| = 4$$ and $${a_{3,3}} = \left| {\begin{array}{*{20}{c}} 1&1 \\\ 3&0 \end{array}} \right| = - 3$$ Thus the cofactor matrix is obtained by putting the co-factors in a matrix in order. So we have the cofactor matrix as: $ \Rightarrow {C_A} = \left[ {\begin{array}{*{20}{c}} 0&{ - 7}&0 \\\ { - 3}&1&1 \\\ 2&4&{ - 3} \end{array}} \right]$ We know from definition that the adjoint of a matrix is the transpose of its cofactor matrix. So we have the adjoint of matrix A as: $ \Rightarrow {\text{adj}}A = \left[ {\begin{array}{*{20}{c}} 0&{ - 3}&2 \\\ { - 7}&1&4 \\\ 0&1&{ - 3} \end{array}} \right]$ To get the value of matrix $A\left( {{\text{adj }}A} \right)$, we will multiply the two matrices, we’ll get:

\Rightarrow A\left( {{\text{adj }}A} \right) = \left[ {\begin{array}{{20}{c}}
1&1&2 \\
3&0&2 \\
1&0&3
\end{array}} \right] \times \left[ {\begin{array}{{20}{c}}
0&{ - 3}&2 \\
{ - 7}&1&4 \\
0&1&{ - 3}
\end{array}} \right] \\
\Rightarrow A\left( {{\text{adj }}A} \right) = \left[ {\begin{array}{{20}{c}}
{0 - 7 + 0}&{ - 3 + 1 + 2}&{2 + 4 - 6} \\
{0 - 0 + 0}&{ - 9 + 0 + 2}&{6 + 0 - 6} \\
{0 - 0 + 0}&{ - 3 + 0 + 3}&{2 + 0 - 9}
\end{array}} \right] = \left[ {\begin{array}{{20}{c}}
{ - 7}&0&0 \\
0&{ - 7}&0 \\
0&0&{ - 7}
\end{array}} \right] \\
\Rightarrow A\left( {{\text{adj }}A} \right) = - 7\left[ {\begin{array}{*{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right]{\text{ }}.....{\text{(1)}} \\

Similarly to determine the matrix $\left( {{\text{adj }}A} \right)A$, we’ll multiply the matrices in opposite order. We’ll get: $ \Rightarrow \left( {{\text{adj }}A} \right)A = \left[ {\begin{array}{*{20}{c}} 0&{ - 3}&2 \\\ { - 7}&1&4 \\\ 0&1&{ - 3} \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} 1&1&2 \\\ 3&0&2 \\\ 1&0&3 \end{array}} \right] \\\ \Rightarrow \left( {{\text{adj }}A} \right)A = \left[ {\begin{array}{*{20}{c}} {0 - 9 + 2}&{0 - 0 + 0}&{0 - 6 + 6} \\\ { - 7 + 3 + 4}&{ - 7 + 0 + 0}&{ - 14 + 2 + 12} \\\ {0 + 3 - 3}&{0 + 0 - 0}&{0 + 2 - 9} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 7}&0&0 \\\ 0&{ - 7}&0 \\\ 0&0&{ - 7} \end{array}} \right] \\\ \Rightarrow \left( {{\text{adj }}A} \right)A = - 7\left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]{\text{ }}.....{\text{(2)}} \\\ $ Further, the determinant value of matrix A is given as: $ \Rightarrow \left| A \right| = 1\left( {0 - 0} \right) - 1\left( {9 - 2} \right) + 2\left( {0 - 0} \right) \\\ \Rightarrow \left| A \right| = 0 - 7 + 0 = - 7 \\\ $ So the value of matrix $\left| A \right|I$ is: $ \Rightarrow \left| A \right|I - 7\left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]{\text{ }}.....{\text{(3)}}$ Here $I = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]$ is the identity matrix of order $3 \times 3$. Now, comparing equations (1), (2) and (3), we can observe that: $ \Rightarrow A\left( {{\text{adj }}A} \right) = \left( {{\text{adj }}A} \right)A = \left| A \right|I$ **Note:** This adjoint method is also used to determine the inverse of any matrix. The formula to determine the inverse of any matrix A is given as: $ \Rightarrow {A^{ - 1}} = \dfrac{1}{{\left| A \right|}}\left( {{\text{adj}}A} \right)$ Thus when all the elements of the adjoint matrix are divided by the determinant value of the original matrix, we get the inverse matrix.