Question

Find the value of the following using transformation method.$\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  { - 1}&2
\end{array}} \right]$

Answer 1

Hint: Linear transformation can be represented by matrices. If $T$is a linear transformation mapping ${R^n}\,\,to\,\,{R^m}$and $\vec x$ is a column vector with $n$entries , then $T(\vec x) = A\vec x$
For some \[m \times n\] matrix $A$called the transformation matrix of $T$.


Complete step by step solution:
Let $A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  1&2
\end{array}} \right]$
$A = IA$, here $I$ is identity matrix
So, $I = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
Then $A = IA$
$\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  1&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]A$
Now, ${R_2} \to {R_2} + {R_1}$
$\left[ {\begin{array}{*{20}{c}}
  1&{ - 3} \\
  { - 1 + 2}&{2 + ( - 3)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  {0 + 1}&{1 + 0}
\end{array}} \right]A$
$\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  1&{2 - 3}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right]A$
$\left[ {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  1&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  1&1
\end{array}} \right]A$
${R_1} \to {R_2} - {R_2}$
$\left[ {\begin{array}{*{20}{c}}
  {2 - 1}&{ - 3 - ( - 1)} \\
  1&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {1 - 1}&{0 - 1} \\
  1&1
\end{array}} \right]A$
$\left[ {\begin{array}{*{20}{c}}
  1&{ - 3 + 1} \\
  1&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&1
\end{array}} \right]A$
$\left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  1&{ - 1}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&1
\end{array}} \right]A$
Now ${R_2} \to {R_1} - {R_2}$
$\left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  {1 - 1}&{ - 1 - (2)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  {0 - 1}&{ - 1 - 1}
\end{array}} \right]A$
$\left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  0&{ - 1 + 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  1&{ - 2}
\end{array}} \right]A$
$\left[ {\begin{array}{*{20}{c}}
  1&{ - 2} \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  0&{ - 1} \\
  { - 1}&{ - 2}
\end{array}} \right]A$
${R_1} \to {R_1} + 2{R_2}$
\[\left[ {\begin{array}{*{20}{c}}
  {1 + 0}&{ - 2 + 2(1)} \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {0 + 2 \times ( - 1)}&{ - 1 + 2( - 2)} \\
  { - 1}&{ - 2}
\end{array}} \right]\]
$\left[ {\begin{array}{*{20}{c}}
  1&{ - 2 + 2} \\
  0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  { - 2}&{ - 1 - 4} \\
  { - 1}&{ - 2}
\end{array}} \right]A$
$\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  { - 2}&{ - 5} \\
  { - 1}&{ - 2}
\end{array}} \right]A$
$\therefore {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  { - 2}&{ - 5} \\
  { - 1}&{ - 2}
\end{array}} \right]$


Note: Students must do equating the rows and columns carefully if you add or subtract and also the operations on rows and columns should be done in such a way so that we will get the required answer