Question

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

Answer 1

Hint: In this question first convert the system of equation into matrix format, then apply the formula of A inverse which is ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)$, and later on use the concept of matrix multiplication, so use these concepts to get the solution of the question.

Complete step-by-step answer:
Given system of equation are
$2x - 3y + 5z = 11 \\\ 3x + 2y - 4z = 5 \\\ x + y - 2z = 3 \\\$
First convert the system of equations into matrix format we have,

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]$$ Now as we see that left most part of above equation is equal to given matrix (A) $A = \left[ {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\\ 3&2&{ - 4} \\\ 1&1&{ - 2} \end{array}} \right]$, Let $X = \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right],{\text{ B}} = \left[ {\begin{array}{*{20}{c}} {11} \\\ 5 \\\ 3 \end{array}} \right]$ $ \Rightarrow AX = B$ So the solution of the given system of equations is $X = {A^{ - 1}}B$……………….. (1) So, first calculate $A$ inverse As we know ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right)$ Where $adj\left( A \right) = {\left[ {\begin{array}{*{20}{c}} {{c_{11}}}&{{c_{12}}}&{{c_{13}}} \\\ {{c_{21}}}&{{c_{22}}}&{{c_{23}}} \\\ {{c_{31}}}&{{c_{32}}}&{{c_{33}}} \end{array}} \right]^T}$ Where T is the transpose of matrix, so apply transpose of matrix

\Rightarrow adj\left( A \right) = \left[ {\begin{array}{{20}{c}}
{{c_{11}}}&{{c_{21}}}&{{c_{31}}} \\
{{c_{12}}}&{{c_{22}}}&{{c_{32}}} \\
{{c_{13}}}&{{c_{23}}}&{{c_{33}}}
\end{array}} \right] \\
{A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right) = \dfrac{1}{{\left| A \right|}}\left[ {\begin{array}{{20}{c}}
{{c_{11}}}&{{c_{21}}}&{{c_{31}}} \\
{{c_{12}}}&{{c_{22}}}&{{c_{32}}} \\
{{c_{13}}}&{{c_{23}}}&{{c_{33}}}
\end{array}} \right]...........\left( 2 \right) \\

Now, first calculate determinant of $A$ $ \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\\ 3&2&{ - 4} \\\ 1&1&{ - 2} \end{array}} \right|$ Now, expand the determinant $ \Rightarrow \left| A \right| = \left| {\begin{array}{*{20}{c}} 2&{ - 3}&5 \\\ 3&2&{ - 4} \\\ 1&1&{ - 2} \end{array}} \right| = 2\left| {\begin{array}{*{20}{c}} 2&{ - 4} \\\ 1&{ - 2} \end{array}} \right| - \left( { - 3} \right)\left| {\begin{array}{*{20}{c}} 3&{ - 4} \\\ 1&{ - 2} \end{array}} \right| + 5\left| {\begin{array}{*{20}{c}} 3&2 \\\ 1&1 \end{array}} \right| \\\ = 2\left( { - 4 - \left( { - 4} \right)} \right) + 3\left( { - 6 - \left( { - 4} \right)} \right) + 5\left( {3 - 2} \right) = 0 - 6 + 5 = - 1 \\\ $ Now calculate $adj\left( A \right)$ $ \Rightarrow adj\left( A \right) = \left[ {\begin{array}{*{20}{c}} {{c_{11}}}&{{c_{21}}}&{{c_{31}}} \\\ {{c_{12}}}&{{c_{22}}}&{{c_{32}}} \\\ {{c_{13}}}&{{c_{23}}}&{{c_{33}}} \end{array}} \right]$ So, calculate its internal elements i.e. its cofactors

{c_{11}} = + 1\left| {\begin{array}{{20}{c}}
2&{ - 4} \\
1&{ - 2}
\end{array}} \right| = 1\left( { - 4 - \left( { - 4} \right)} \right) = 0,{\text{ }}{{\text{c}}_{21}} = - 1\left| {\begin{array}{{20}{c}}
{ - 3}&5 \\
1&{ - 2}
\end{array}} \right| = - 1\left( {6 - 5} \right) = - 1,{\text{ }}{{\text{c}}{31}} = + 1\left| {\begin{array}{*{20}{c}}
{ - 3}&5 \\
2&{ - 4}
\end{array}} \right| = 1\left( {12 - 10} \right) = 2 \\
{c{12}} = - 1\left| {\begin{array}{{20}{c}}
3&{ - 4} \\
1&{ - 2}
\end{array}} \right| = - 1\left( { - 6 - \left( { - 4} \right)} \right) = 2,{\text{ }}{{\text{c}}_{22}} = + 1\left| {\begin{array}{{20}{c}}
2&5 \\
1&{ - 2}
\end{array}} \right| = 1\left( { - 4 - 5} \right) = - 9,{\text{ }}{{\text{c}}{32}} = - 1\left| {\begin{array}{*{20}{c}}
2&5 \\
3&{ - 4}
\end{array}} \right| = - 1\left( { - 8 - 15} \right) = 23 \\
{c{13}} = + 1\left| {\begin{array}{{20}{c}}
3&2 \\
1&1
\end{array}} \right| = 1\left( {3 - 2} \right) = 1,{\text{ }}{{\text{c}}_{23}} = - 1\left| {\begin{array}{{20}{c}}
2&{ - 3} \\
1&1
\end{array}} \right| = - 1\left( {2 + 3} \right) = - 5,{\text{ }}{{\text{c}}_{33}} = + 1\left| {\begin{array}{*{20}{c}}
2&{ - 3} \\
3&2
\end{array}} \right| = 1\left( {4 - \left( { - 9} \right)} \right) = 13 \\

$ \Rightarrow adj\left( A \right) = \left[ {\begin{array}{*{20}{c}} 0&{ - 1}&2 \\\ 2&{ - 9}&{23} \\\ 1&{ - 5}&{13} \end{array}} \right]$ Now, from equation 2 ${A^{ - 1}} = \dfrac{1}{{\left| A \right|}}adj\left( A \right) = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}} 0&{ - 1}&2 \\\ 2&{ - 9}&{23} \\\ 1&{ - 5}&{13} \end{array}} \right]$ So, this is the required ${A^{ - 1}}$. Now from equation 1 $ \Rightarrow X = {A^{ - 1}}B = \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]$ Now apply matrix multiplication $ \Rightarrow X = {A^{ - 1}}B = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}} {0 \times 11 - 1 \times 5 + 2 \times 3} \\\ {2 \times 11 - 9 \times 5 + 23 \times 3} \\\ {1 \times 11 - 5 \times 5 + 13 \times 3} \end{array}} \right] = \dfrac{1}{{ - 1}}\left[ {\begin{array}{*{20}{c}} 1 \\\ {46} \\\ {25} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1} \\\ { - 46} \\\ { - 25} \end{array}} \right]$ Hence $X = \left[ {\begin{array}{*{20}{c}} { - 1} \\\ { - 46} \\\ { - 25} \end{array}} \right]$ Now, $X = \left[ {\begin{array}{*{20}{c}} x \\\ y \\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1} \\\ { - 46} \\\ { - 25} \end{array}} \right]$, so on comparing we have, $x = - 1,{\text{ }}y = - 46,{\text{ }}z = - 25$ So, this is the required solution. Note: In such types of questions convert the system of equation into matrix format in the form $AX = B$, so the solution of the system of equations is $X = {A^{ - 1}}B$ so, first calculate the determinant value of $A$ then calculate the value of ${A^{ - 1}}$ using the formula which is stated above, then apply matrix multiplication we will get the required solution of $X$.