Question

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

Answer 1

If we want to find ${A^{ - 1}}$ using row operations, write $A = IA$ and apply a sequence of row operations on $A = IA$ till we get, $I = BA$. Then, the matrix $B$ will be the inverse of $A$.

Complete step-by-step answer:
Write $A = IA$, i.e.,
\left[ {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\\ 3&2&{ - 4} \\\ 1&1&{ - 2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]A
Applying${R_1} \leftrightarrow {R_3}$,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 1&1&{ - 2} \\\ 3&2&{ - 4} \\\ 2&{ - 3}&5 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ 0&1&0 \\\ 1&0&0 \end{array}} \right]A
Applying${R_2} \to {R_2} - 3{R_1}$,

1&1&{ - 2} \\\ 0&{ - 1}&2 \\\ 2&{ - 3}&5 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ 0&1&{ - 3} \\\ 1&0&0 \end{array}} \right]A$$ Applying$${R_3} \to {R_3} - 2{R_1}$$, $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&1&{ - 2} \\\ 0&1&{ - 2} \\\ 0&{ - 5}&9 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ 0&{ - 1}&3 \\\ 1&0&{ - 2} \end{array}} \right]A$$ $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&1&{ - 2} \\\ 0&{ - 1}&2 \\\ 0&{ - 5}&9 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ 0&1&{ - 3} \\\ 1&0&{ - 2} \end{array}} \right]A$$ Applying$${R_2} \to \left( { - 1} \right){R_2}$$, $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&1&{ - 2} \\\ 0&1&{ - 2} \\\ 0&{ - 5}&9 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ 0&{ - 1}&3 \\\ 1&0&{ - 2} \end{array}} \right]A$$ Applying$${R_3} \to {R_3} + 5{R_2}$$, $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&1&{ - 2} \\\ 0&1&{ - 2} \\\ 0&0&{ - 1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&0&1 \\\ 0&{ - 1}&3 \\\ 1&{ - 5}&{13} \end{array}} \right]A$$ Applying$${R_3} \to \left( { - 1} \right){R_3}$$, $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0&{ - 0} \\\ 0&1&{ - 2} \\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1&{ - 2} \\\ 0&{ - 1}&3 \\\ { - 1}&5&{ - 13} \end{array}} \right]A$$ Applying${R_1} \to {R_1} - {R_2}$, $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&{ - 2} \\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1&{ - 2} \\\ 0&{ - 1}&3 \\\ { - 1}&5&{ - 13} \end{array}} \right]A$$ Applying$${R_2} \to {R_2} + 2{R_3}$$, $$ \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&1&{ - 2} \\\ { - 2}&9&{ - 23} \\\ { - 1}&5&{ - 13} \end{array}} \right]A$$ Since it is of the form $I = BA$, where matrix $B$ will be the inverse of $A$. $\therefore {A^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 0&1&{ - 2} \\\ { - 2}&9&{ - 23} \\\ { - 1}&5&{ - 13} \end{array}} \right]$ ….. (1) Now, the given system of equations is $2x - 3y + 5z = 11$ $3x + 2y - 4z = - 5$ $x + y - 2z = - 3$ The system of linear equations can be written in the form of $AX = B$, where $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\\ 3&2&{ - 4} \\\ 1&1&{ - 2} \end{array}} \right]$, $X = \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}} {11} \\\ { - 5} \\\ { - 3} \end{array}} \right]$ The solution of system of linear equations is given by, $X = {A^{ - 1}}B$ $ \Rightarrow \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \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]$ ….. [Use the value of ${A^{ - 1}}$from (1)] $ \Rightarrow \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \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] = \left[ {\begin{array}{*{20}{c}} 1 \\\ 2 \\\ 3 \end{array}} \right]$ **Hence, $x = 1,y = 2$ and $z = 3$.** **Note:** Matrix is an arrangement of elements or numbers into rows and columns. Determinants is a scalar value which is computed from square matrices. To solve the linear system of equations, firstly write a system of linear equations in the form of $AX = B$ and then find the solution of the system of linear equations by using the formula $X = {A^{ - 1}}B$.