Question

Find the inverse of the matrix\left[ {\begin{array}{*{20}{c}} 1&2&1 \\\ 3&0&1 \\\ 0&2&1 \end{array}} \right] using the method of adjoint.

Answer 1

We find the determinant of the matrix. Use the method of minors and cofactors to find the adjoint of the given matrix. Use the formula of inverse of a matrix using adjoint of matrix and inverse of matrix.

• Determinant of a matrix \left[ {\begin{array}{*{20}{c}} a&b;&c; \\\ d&e;&f; \\\ g&h;&i; \end{array}} \right] = a(ei - hf) - b(di - fg) + c(dh - eg)
• Adjoint of a matrix A is given byadjA = {\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]^T}, where T stands for transpose.
  Each element ${A_{ij}}$is given by calculating the determinant of the matrix obtained by removing $i^{th}$ row and $j^{th}$ column from the matrix. Also, the sign of the element ${A_{ij}}$is given by ${( - 1)^{i + j}}$.
• Inverse of a matrix A is given by ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adjA$

Complete step by step solution:
Let us assume the matrixA = \left[ {\begin{array}{*{20}{c}} 1&2&1 \\\ 3&0&1 \\\ 0&2&1 \end{array}} \right]
We find the determinant of the matrix A using the formula of determinant.
$\Rightarrow \left| A \right| = 1\left( {(0 \times 1) - (2 \times 1)} \right) - 2\left( {(3 \times 1) - (0 \times 1)} \right) + 1\left( {(3 \times 2) - (0 \times 0)} \right)$
Calculate the products in the brackets
$\Rightarrow \left| A \right| = 1\left( {0 - 2} \right) - 2\left( {3 - 0} \right) + 1\left( {6 - 0} \right)$
$\Rightarrow \left| A \right| = 1 \times ( - 2) - 2 \times (3) + 1 \times (6)$
Calculate the products
$\Rightarrow \left| A \right| = - 2 - 6 + 6$
Cancel same terms with opposite signs
$\Rightarrow \left| A \right| = - 2$.................… (1)
Since the determinant of the matrix is non-negative, the inverse of the matrix exists.
Now we find the adjoint of matrix A

{{{( - 1)}^{1 + 1}}( - 2)}&{{{( - 1)}^{1 + 2}}(3)}&{{{( - 1)}^{1 + 3}}(6)} \\\ {{{( - 1)}^{2 + 1}}(0)}&{{{( - 1)}^{2 + 2}}(1)}&{{{( - 1)}^{2 + 3}}(2)} \\\ {{{( - 1)}^{3 + 1}}(2)}&{{{( - 1)}^{3 + 2}}( - 2)}&{{{( - 1)}^{3 + 3}}( - 6)} \end{array}} \right]^T}$$ $$ \Rightarrow adjA = {\left[ {\begin{array}{*{20}{c}} {{{( - 1)}^2}( - 2)}&{{{( - 1)}^3}(3)}&{{{( - 1)}^4}(6)} \\\ {{{( - 1)}^3}(0)}&{{{( - 1)}^4}(1)}&{{{( - 1)}^5}(2)} \\\ {{{( - 1)}^4}(2)}&{{{( - 1)}^5}( - 2)}&{{{( - 1)}^6}( - 6)} \end{array}} \right]^T}$$ Write even powers of -1 equal to 1 and odd powers of -1 equal to -1 $$ \Rightarrow adjA = {\left[ {\begin{array}{*{20}{c}} {(1)( - 2)}&{( - 1)(3)}&{(1)(6)} \\\ {( - 1)(0)}&{(1)(1)}&{( - 1)(2)} \\\ {(1)(2)}&{( - 1)( - 2)}&{(1)( - 6)} \end{array}} \right]^T}$$ Multiply the negative signs $$ \Rightarrow adjA = {\left[ {\begin{array}{*{20}{c}} { - 2}&{ - 3}&6 \\\ 0&1&{ - 2} \\\ 2&2&{ - 6} \end{array}} \right]^T}$$ Now take transpose of the matrix in RHS, i.e. write columns in place of rows and rows in place of columns. $$ \Rightarrow adjA = \left[ {\begin{array}{*{20}{c}} { - 2}&0&2 \\\ { - 3}&1&2 \\\ 6&{ - 2}&{ - 6} \end{array}} \right]$$....................… (2) Substitute the values of determinant A and adjoint A in the formula of inverse i.e.$${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adjA$$ Substitute values from equation (1) and (2) $$ \Rightarrow {A^{ - 1}} = \dfrac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} { - 2}&0&2 \\\ { - 3}&1&2 \\\ 6&{ - 2}&{ - 6} \end{array}} \right]$$ $$\therefore $$The inverse of matrix$$\left[ {\begin{array}{*{20}{c}} 1&2&1 \\\ 3&0&1 \\\ 0&2&1 \end{array}} \right]$$ is $$\dfrac{1}{{ - 2}}\left[ {\begin{array}{*{20}{c}} { - 2}&0&2 \\\ { - 3}&1&2 \\\ 6&{ - 2}&{ - 6} \end{array}} \right]$$. **Note:** Students are likely to make the mistake of writing the value of determinant as positive because there is a modulus sign along with it, but keep in mind that sign is the sign for the matrix name, the value of determinant need not be positive.