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Using elementary transformation, find the inverse of the matrix $\left[ {\begin{array}{*{20}{c}}
  3&{10} \\
  2&7
\end{array}} \right]$.

Answer Verified Verified
Hint – In this question we have to find the inverse of the given matrix using elementary transformations. Elementary transformations means that we start with a row or a column and by applying various transformations on the chosen entity either rows or columns we try to make maximum possible zeroes. Use this concept along with A=IA where I is the identity matrix, to get the inverse.

Complete step-by-step answer:
Given matrix is
$\left[ {\begin{array}{*{20}{c}}
  3&{10} \\
  2&7
\end{array}} \right]$
Now we have to find out the inverse of this matrix using elementary transformations,
Let,
$A = \left[ {\begin{array}{*{20}{c}}
  3&{10} \\
  2&7
\end{array}} \right]$
As we know A is also written as $A = IA$, where I is an identity matrix.
$I = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
$
   \Rightarrow IA = \left[ {\begin{array}{*{20}{c}}
  3&{10} \\
  2&7
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  3&{10} \\
  2&7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]A \\
 $
Now, apply row transformations so that the L.H.S matrix become identity matrix therefore we apply,
${R_1} \to {R_1} - {R_2}$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {3 - 2}&{10 - 7} \\
  2&7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {1 - 0}&{0 - 1} \\
  0&1
\end{array}} \right]A$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&7
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  0&1
\end{array}} \right]A$
Now apply ${R_2} \to {R_2} - 2{R_1}$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  {2 - 2}&{7 - 6}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  {0 - 2}&{1 + 2}
\end{array}} \right]A$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  1&{ - 1} \\
  { - 2}&3
\end{array}} \right]A$
Now apply ${R_1} \to {R_1} - 3{R_2}$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {1 - 0}&{3 - 3} \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {1 + 6}&{ - 1 - 9} \\
  { - 2}&3
\end{array}} \right]A$
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  7&{ - 10} \\
  { - 2}&3
\end{array}} \right]A$
Now shift A to L.H.S
$ \Rightarrow {A^{ - 1}}\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  7&{ - 10} \\
  { - 2}&3
\end{array}} \right]$
$ \Rightarrow {A^{ - 1}}I = \left[ {\begin{array}{*{20}{c}}
  7&{ - 10} \\
  { - 2}&3
\end{array}} \right]$
$ \Rightarrow {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
  7&{ - 10} \\
  { - 2}&3
\end{array}} \right]$
So, this is the required A inverse using elementary transformations.
So, this is the required answer.

Note – Whenever we face such type of problems the key concept is to apply the row or column transformations accurately, however this problem could also be solved using other direct formula technique to find inverse of the matrix that is ${{\text{A}}^{ - 1}} = \dfrac{{}}{{\left| A \right|}}adj(A)$, but in this problem only elementary transformation is being asked.

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