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,
\[\left[ {\begin{array}{*{20}{c}}
  2&{ - 3}&5 \\
  3&2&{ - 4} \\
  1&1&{ - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  x \\
  y \\
  z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  {11} \\
  5 \\
  3
\end{array}} \right]\]
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 $X$.