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

Hint: 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 by\[adjA = {\left[ {\begin{array}{*{20}{c}}
  {{A_{11}}}&{{A_{12}}}&{{A_{13}}} \\
  {{A_{21}}}&{{A_{22}}}&{{A_{23}}} \\
  {{A_{31}}}&{{A_{32}}}&{{A_{33}}}
\end{array}} \right]^T}\], 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 matrix\[A = \left[ {\begin{array}{*{20}{c}}
  1&2&1 \\
  3&0&1 \\
  0&2&1
\end{array}} \right]\]
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
\[ \Rightarrow adjA = {\left[ {\begin{array}{*{20}{c}}
  {{{( - 1)}^{1 + 1}}( - 2)}&{{{( - 1)}^{1 + 2}}(3)}&{{{( - 1)}^{1 + 3}}(6)} \\
  {{{( - 1)}^{2 + 1}}(0)}&{{{( - 1)}^{2 + 2}}(1)}&{{{( - 1)}^{2 + 3}}(2)} \\
  {{{( - 1)}^{3 + 1}}(2)}&{{{( - 1)}^{3 + 2}}( - 2)}&{{{( - 1)}^{3 + 3}}( - 6)}
\end{array}} \right]^T}\]
\[ \Rightarrow adjA = {\left[ {\begin{array}{*{20}{c}}
  {{{( - 1)}^2}( - 2)}&{{{( - 1)}^3}(3)}&{{{( - 1)}^4}(6)} \\
  {{{( - 1)}^3}(0)}&{{{( - 1)}^4}(1)}&{{{( - 1)}^5}(2)} \\
  {{{( - 1)}^4}(2)}&{{{( - 1)}^5}( - 2)}&{{{( - 1)}^6}( - 6)}
\end{array}} \right]^T}\]
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.