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# 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$2x - 3y + 5z = 11 \\ 3x + 2y - 4z = 5 \\ x + y - 2z = 3 \\$

Last updated date: 19th Mar 2023
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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.

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 $X$.