Question

If $A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3}&5 \\
  3&2&{ - 4} \\
  1&1&{ - 2}
\end{array}} \right]$ and find ${A^{ - 1}}$ and solve the system of linear equation $2x - 3y + 5z = 11,3x + 2y - 4z = - 5$ and $x + y - 2z = - 3.$

Answer 1

Hint: We have a square matrix $A$ , which is nonsingular then there exist an $n \times n$ matrix ${A^{ - 1}}$ which is called the inverse of $A$ such that $A{A^{ - 1}} = {A^{ - 1}}A = I$ , where $I$ is the identity matrix.

Complete step-by-step solution:

Given:
$
  2x - 3y + 5z = 11 \\
  3x + 2y - 4z = - 5 \\
  x + y - 2z = - 3 \\
 $
System of linear equation $A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3}&5 \\
  3&2&{ - 4} \\
  1&1&{ - 2}
\end{array}} \right]$
$A = \left[ {\begin{array}{*{20}{c}}
  2&{ - 3}&5 \\
  3&2&{ - 4} \\
  1&1&{ - 2}
\end{array}} \right]$

[Solving determinant, we get]

$
  \left| A \right| = \left| {\begin{array}{*{20}{c}}
  2&{ - 3}&5 \\
  3&2&{ - 4} \\
  1&1&{ - 2}
\end{array}} \right| \\
   = 2( - 4 + 4) + 3( - 6 + 4) + 5(3 - 2) \\
   = 0 - 6 + 5 \\
   = - 1 \\
 $
Let $cij$ be the co-factors of the element $aij$ in $A = [aij]$ . Then,
\[
  {C_{11}} = {( - 1)^{1 + 1}}\left| {\begin{array}{*{20}{c}}
  2&{ - 4} \\
  1&{ - 2}
\end{array}} \right| = 0,{C_{12}} = {( - 1)^{1 + 2}}\left| {\begin{array}{*{20}{c}}
  3&{ - 4} \\
  1&{ - 2}
\end{array}} \right| = 2,{C_{13}} = {( - 1)^{1 + 3}}\left| {\begin{array}{*{20}{c}}
  3&2 \\
  1&1
\end{array}} \right| = 1, \\
  {C_{21}} = {( - 1)^{2 + 1}}\left| {\begin{array}{*{20}{c}}
  { - 3}&5 \\
  1&{ - 2}
\end{array}} \right| = - 1,{C_{22}} = {( - 1)^{2 + 2}}\left| {\begin{array}{*{20}{c}}
  2&5 \\
  1&{ - 2}
\end{array}} \right| = - 9,{C_{23}} = {( - 1)^{2 + 3}}\left| {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  1&1
\end{array}} \right| = - 5, \\
  {C_{31}} = {( - 1)^{3 + 1}}\left| {\begin{array}{*{20}{c}}
  { - 2}&5 \\
  2&4
\end{array}} \right| = 2,{C_{32}} = {( - 1)^{3 + 2}}\left| {\begin{array}{*{20}{c}}
  2&5 \\
  3&{ - 4}
\end{array}} \right| = 23,{C_{33}} = {( - 1)^{3 + 3}}\left| {\begin{array}{*{20}{c}}
  2&{ - 3} \\
  3&2
\end{array}} \right| = 13. \\
 \]
Now, finding the $adj(A)$
$
  adj = {\left[ {\begin{array}{*{20}{c}}
  0&2&1 \\
  { - 1}&{ - 9}&5 \\
  2&{23}&{13}
\end{array}} \right]^T} \\
        = \left[ {\begin{array}{*{20}{c}}
  0&{ - 1}&2 \\
  2&{ - 9}&{23} \\
  1&{ - 5}&{13}
\end{array}} \right] \\
 $
We knew that the formula of ${A^{ - 1}}$ is
$
   \Rightarrow {A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adjA \\
  {A^{ - 1}} = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}
  0&{ - 1}&2 \\
  2&{ - 9}&{23} \\
  1&{ - 5}&{13}
\end{array}} \right] \\
 $
Now, the given system of equation can be written in matrix form as follows,
$
  \left[ {\begin{array}{*{20}{c}}
  2&{ - 3}&5 \\
  3&2&{ - 4} \\
  1&1&{ - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {11} \\
  { - 5} \\
  { - 3}
\end{array}} \right] \\
  X = {A^{ - 1}}B \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}
  0&{ - 1}&2 \\
  2&{ - 9}&{23} \\
  1&{ - 5}&{13}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  {11} \\
  { - 5} \\
  { - 3}
\end{array}} \right] \\
 $
Multiplying ${A^{ - 1}}B$ we get,
\[
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}
  {0 + 5 - 6} \\
  {22 + 45 - 69} \\
  {11 + 25 - 39}
\end{array}} \right] \\
   \Rightarrow \left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}}
  { - 1} \\
  { - 2} \\
  { - 3}
\end{array}} \right] \\
   \Rightarrow x = \dfrac{{ - 1}}{{ - 1}},y = \dfrac{{ - 2}}{1},z = \dfrac{{ - 3}}{{ - 1}} \\
  \therefore x = 1,y = - 2,z = 3 \\
 \]

Note: In this type of question students often make mistakes while finding the cofactor so keep your mind focused while doing so. Also ${A^{ - 1}} \ne \dfrac{1}{A}\left| A \right|$ as many people make this mistake. Remember transpose of a matrix is nothing but interchanging the row into column and vice-versa.