ELEMENTARY TRANSFORMATIONS TO FIND THE INVERSE OF A MATRIX

To find the inverse by elementary row transformations, we need to follow the following steps:

clip_image001

\left[ {\begin{array}{ccccccccccccccc}
{1\left( {\rm{I}} \right)}&{0\left( {{\rm{IV}}} \right)}\\
{0\left( {{\rm{II}}} \right)}&{1\left( {{\rm{III}}} \right)}
\end{array}} \right]

(OR)

\left[ {\begin{array}{ccccccccccccccc}
{1\left( {\rm{I}} \right)}&{0\left( {\rm{V}} \right)}&{0\left( {{\rm{VIII}}} \right)}\\
{0\left( {{\rm{II}}} \right)}&{1\left( {{\rm{IV}}} \right)}&{0\left( {{\rm{IX}}} \right)}\\
{0\left( {{\rm{III}}} \right)}&{0\left( {{\rm{VI}}} \right)}&{1\left( {{\rm{VII}}} \right)}
\end{array}} \right]

To find the inverse of the matrix \left[ {\begin{array}{ccccccccccccccc}
2&{ - 1}&4\\
4&0&2\\
3&{ - 2}&7
\end{array}} \right]

Let A = \left[ {\begin{array}{ccccccccccccccc}
2&{ - 1}&4\\
4&0&2\\
3&{ - 2}&7
\end{array}} \right]

A = I A

 \Rightarrow \,\,\,\,\left[ {\begin{array}{ccccccccccccccc}
2&{ - 1}&4\\
4&0&2\\
3&{ - 2}&7
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right]\,A

{R_1}\, \to \,\frac{1}{2}{R_1}

 \Rightarrow \,\,\,\,\left[ {\begin{array}{ccccccccccccccc}
1&{ - \frac{1}{2}}&2\\
4&0&2\\
3&{ - 2}&7
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
{\frac{1}{2}}&0&0\\
0&1&0\\
0&0&1
\end{array}} \right]A

{R_2}\, \to \,\,{R_2} - 4{R_1}

 \Rightarrow \,\,\,\,\left[ {\begin{array}{ccccccccccccccc}
1&{ - \frac{1}{2}}&2\\
0&2&{ - 6}\\
3&{ - 2}&7
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
{\frac{1}{2}}&0&0\\
{ - 2}&1&0\\
0&0&1
\end{array}} \right]A 

\,{R_3}\, \to \,\,{R_3}\, - \,3{R_1}

 \Rightarrow \,\,\,\left[ {\begin{array}{ccccccccccccccc}
1&{ - \frac{1}{2}}&2\\
0&2&{ - 6}\\
0&{ - \frac{1}{2}}&1
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
{ - \frac{1}{2}}&0&0\\
{ - 2}&1&0\\
{ - 3}&0&1
\end{array}} \right]A

{R_2}\, \to \,\,{R_2} + 2{R_3}

 \Rightarrow \,\,\,\left[ {\begin{array}{ccccccccccccccc}
1&{ - \frac{1}{2}}&2\\
0&1&{ - 4}\\
0&{ - \frac{1}{2}}&1
\end{array}} \right]\, = \,\left[ {\begin{array}{ccccccccccccccc}
{\frac{1}{2}}&0&0\\
{ - 5}&1&2\\
{ - \frac{3}{2}}&0&1
\end{array}} \right]A

{R_1} \to \,\,{R_1}\, - \,{R_3}

 \Rightarrow \,\,\,\left[ {\begin{array}{ccccccccccccccc}
1&0&1\\
0&1&{ - 4}\\
0&{ - \frac{1}{2}}&1
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
2&0&{ - 1}\\
{ - 5}&1&2\\
{ - \frac{3}{2}}&0&1
\end{array}} \right]A 

\,\,\,{R_3} \to \,{R_3} + \frac{1}{2}{R_2}

 \Rightarrow \,\,\,\left[ {\begin{array}{ccccccccccccccc}
1&0&1\\
0&1&{ - 4}\\
0&0&1
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
2&0&{ - 1}\\
{ - 5}&1&2\\
{ - 4}&{\frac{1}{2}}&2
\end{array}} \right]A

{R_3}\, \to \,\left( { - 1} \right){R_3}

 \Rightarrow \,\,\,\left[ {\begin{array}{ccccccccccccccc}
1&0&1\\
0&1&{ - 4}\\
0&0&1
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
2&0&{ - 1}\\
{ - 5}&1&2\\
4&{ - \frac{1}{2}}&{ - 2}
\end{array}} \right]A

{R_1}\, \to \,\,{R_1}\, - {R_3}

 \Rightarrow \,\,\,\left[ {\begin{array}{ccccccccccccccc}
1&0&0\\
0&1&{ - 4}\\
0&0&1
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
{ - 2}&{\frac{1}{2}}&1\\
{ - 5}&1&2\\
4&{ - \frac{1}{2}}&{ - 2}
\end{array}} \right]A 

{R_2}\,\, \to \,\,{R_2}\, + 4{R_3}

 \Rightarrow \,\left[ {\begin{array}{ccccccccccccccc}
1&0&0\\
0&1&0\\
0&0&1
\end{array}} \right] = \left[ {\begin{array}{ccccccccccccccc}
{ - 2}&{\frac{1}{2}}&1\\
{11}&{ - 1}&{ - 6}\\
4&{ - \frac{1}{2}}&2
\end{array}} \right]A

I = \,\left[ {\begin{array}{ccccccccccccccc}
{ - 2}&{\frac{1}{2}}&1\\
{11}&{ - 1}&{ - 6}\\
4&{ - \frac{1}{2}}&2
\end{array}} \right]A

\therefore \,\,\,{A^{ - 1}} = \left[ {\begin{array}{ccccccccccccccc}
{ - 2}&{\frac{1}{2}}&1\\
{11}&{ - 1}&{ - 6}\\
4&{ - \frac{1}{2}}&2
\end{array}} \right] 

Post Author: E-Maths