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

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.