Question

How do find the inverse of \[A = ((1,1,2)\,(2,2,2)\,(2,1,1))\] ?

Answer 1

Hint: Here the question is of matrix and we have to find the inverse of the given matrix. To find the inverse of a matrix we have two different methods, one is by using row reduced echelon form and other one by using determinant and cofactors of a matrix.Hence, we can determine the required solution.

Complete step by step solution:
The inverse of a matrix A is a matrix that, when multiplied by A results in the identity. The notation for this inverse matrix is \[{A^{ - 1}}\]. Now consider the matrix.
\[A = \left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  2&2&2 \\
  2&1&1
\end{array}} \right]\]
We find the inverse of a matrix, to find the inverse of the above matrix we use the row reduced echelon form. As we know that \[A = IA\], where \[I\] is the identity matrix of order 3 cross 3.

The matrix is written as
\[\left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  2&2&2 \\
  2&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]A\]
Now we have to convert the LHS matrix into an identity matrix, for the conversion we use row reduced echelon form and apply row transformation to the matrix.

Now apply the row transformation \[{R_2} \to {R_2} \leftrightarrow {R_3}\]
\[\left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  2&1&1 \\
  2&2&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&0&1 \\
  0&1&0
\end{array}} \right]A\]

Now apply the row transformation \[{R_2} \to {R_2} - {R_1}\]
\[\left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  1&0&{ - 1} \\
  2&2&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  { - 1}&0&1 \\
  0&1&0
\end{array}} \right]A\]

Now apply the row transformation \[{R_3} \to \dfrac{{{R_3}}}{2}\]
\[\left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  1&0&{ - 1} \\
  1&1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  { - 1}&0&1 \\
  0&{\dfrac{1}{2}}&0
\end{array}} \right]A\]

Now apply the row transformation \[{R_3} \to {R_3} - {R_1}\]
\[\left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  1&0&{ - 1} \\
  0&0&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  { - 1}&0&1 \\
  { - 1}&{\dfrac{1}{2}}&0
\end{array}} \right]A\]

Now apply the row transformation \[{R_2} \to {R_2} - {R_3}\]
\[\left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  1&0&0 \\
  0&0&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&{ - \dfrac{1}{2}}&1 \\
  { - 1}&{\dfrac{1}{2}}&0
\end{array}} \right]A\]

Now apply the row transformation \[{R_2} \to {R_2} \leftrightarrow {R_1}\]
\[\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  1&1&2 \\
  0&0&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - \dfrac{1}{2}}&1 \\
  1&0&0 \\
  { - 1}&{\dfrac{1}{2}}&0
\end{array}} \right]A\]

Now apply the row transformation \[{R_2} \to {R_2} - {R_1}\]
\[\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&2 \\
  0&0&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - \dfrac{1}{2}}&1 \\
  1&{\dfrac{1}{2}}&1 \\
  { - 1}&{\dfrac{1}{2}}&0
\end{array}} \right]A\]

Now apply the row transformation \[{R_2} \to {R_2} + 2{R_3}\]
\[\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - \dfrac{1}{2}}&1 \\
  { - 1}&{\dfrac{3}{2}}&1 \\
  { - 1}&{\dfrac{1}{2}}&0
\end{array}} \right]A\]

Now apply the row transformation \[{R_3} \to {R_3} \times - 1\]
\[\left[ {\begin{array}{*{20}{c}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - \dfrac{1}{2}}&1 \\
  { - 1}&{\dfrac{3}{2}}&1 \\
  1&{ - \dfrac{1}{2}}&0
\end{array}} \right]A\]

Hence we have obtained \[I = {A^{ - 1}}A\]

Therefore the inverse of \[A = \left[ {\begin{array}{*{20}{c}}
  1&1&2 \\
  2&2&2 \\
  2&1&1
\end{array}} \right]\] is \[{A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  0&{ - \dfrac{1}{2}}&1 \\
  { - 1}&{\dfrac{3}{2}}&1 \\
  1&{ - \dfrac{1}{2}}&0
\end{array}} \right]\]

Note: When we use row reduced echelon form either we must use row transformation or column transformation. We can’t use both the transformation at a time to the matrix. While using transformation we must take care of signs and sometimes we alter the rows for easier simplification.