Question

Using elementary row transformation, find the inverse of the matrix
$A = \left[ {\begin{array}{*{20}{l}}
  1&2&3 \\
  2&5&7 \\
  { - 2}&{ - 4}&{ - 5}
\end{array}} \right]$

Answer 1

Hint: In ERT method we first consider given matrix as product of identity matrix and given matrix and then using different row transformation we will convert given matrix to identity matrix and applying same transformation in identity matrix and matrix so obtained is called inverse of given matrix.

Complete step-by-step answer:
Given matrix is
$A = \left[ {\begin{array}{*{20}{l}}
  1&2&3 \\
  2&5&7 \\
  { - 2}&{ - 4}&{ - 5}
\end{array}} \right]$
To find its inverse by ERT method.
In this method we equate given matrix as product of identity matrix and given matrix and then using row transformation to convert given matrix into identity and using same transformation applying in identity matrix taken as a product and finally when given matrix converted or become identity matrix then corresponding matrix obtained on right hand side will be inverse of given matrix.
Considering
$A = IA$
Substituting value of matrix A on left side and value of identity matrix on right side. We have,
$\left[ {\begin{array}{*{20}{l}}
  1&2&3 \\
  2&5&7 \\
  { - 2}&{ - 4}&{ - 5}
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right]A$
Using transformation
${R_2} = {R_2} - 2{R_1}\,\,and\,\,\,{R_3} = {R_3} + 2{R_1}$to make elements ${a_{21}}\,\,and\,\,\,\,{a_{31}}$equal to zero. Also, applying same row operation in identity matrix to have its new corresponding elements.

$\left[ {\begin{array}{*{20}{l}}
  1&2&3 \\
  0&1&1 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  1&0&0 \\
  { - 2}&1&0 \\
  2&0&1
\end{array}} \right]A$
Now, making an element ${a_{22}}$as one if not.
If it is one then using it makes the other element of second columns zero.
${R_1} = {R_1} - 2{R_2}$Using this row transformation. We have
$\left[ {\begin{array}{*{20}{l}}
  1&0&1 \\
  0&1&1 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  5&{ - 2}&0 \\
  { - 2}&1&0 \\
  2&0&1
\end{array}} \right]A$
Now, making an element ${a_{33}}$ as one if not.
If it is one then using it make other element of second columns zero.
${R_1} = {R_1} - {R_3}\,\,and\,\,\,{R_2} = {R_2} - {R_3}$ Using this row transformation. We have
$\left[ {\begin{array}{*{20}{l}}
  1&0&0 \\
  0&1&0 \\
  0&0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{l}}
  3&{ - 2}&{ - 1} \\
  { - 4}&1&{ - 1} \\
  2&0&1
\end{array}} \right]A$
In above we see that the given matrix is converted into a unit matrix.
Therefore, from the right hand side we see that the matrix so formed in the right hand side with product of matrix A will be the inverse of the given matrix.
Hence, inverse of given matrix $A = \left[ {\begin{array}{*{20}{l}}
  1&2&3 \\
  2&5&7 \\
  { - 2}&{ - 4}&{ - 5}
\end{array}} \right]$by ERT method is$\left[ {\begin{array}{*{20}{l}}
  3&{ - 2}&{ - 1} \\
  { - 4}&1&{ - 1} \\
  2&0&1
\end{array}} \right]$.

Note: In ERT method only row transformation will be applied irrespective to manner in which any one can apply but final answer will always remain same or we can say that the inverse obtained in every case will be same whether student apply different transformation on different rows to get it.