Question

How do you solve $5x - 5y + 10z = - 11$, $10x + 5y - 5z = 1$ and $15x - 15y - 10z = - 1$ using matrices?

Answer 1

Hint: Use the matrix equation ${\mathbf{AX}} = {\mathbf{B}}$ to represent the given system of equations. You need ${{\mathbf{A}}^{ - 1}}$ to solve this equation. Use ${{\mathbf{A}}^{ - 1}} = \dfrac{1}{{\left| {\mathbf{A}} \right|}}Adj{\mathbf{A}}$ to find ${{\mathbf{A}}^{ - 1}}$ and solve the equation after substituting the values.

Complete step by step solution:
The given system of equations has three unknown quantities and three linear equations. So, we can find the particular solution of the system, if it exists. Now, we will use matrices to solve this system.
The given system of linear equations is,
$5x - 5y + 10z = - 11$
$10x + 5y - 5z = 1$
$15x - 15y - 10z = - 1$
We can represent the system in the matrix form as ${\mathbf{AX}} = {\mathbf{B}}$.
Here, \[{\mathbf{A}} = \left[ {\begin{array}{*{20}{c}}
 5&{ - 5}&{10} \\
 {10}&5&{ - 5} \\
 {15}&{ - 15}&{ - 10}
\end{array}} \right]\], ${\mathbf{X}} = \left[ {\begin{array}{*{20}{c}}
 x \\
 y \\
 z
\end{array}} \right]$ and \[{\mathbf{B}} = \left[ {\begin{array}{*{20}{c}}
 { - 11} \\
 1 \\
 { - 1}
\end{array}} \right]\].
Now, we get ${\mathbf{X}} = {{\mathbf{A}}^{ - 1}}{\mathbf{B}}$ from the matrix equation, if the matrix ${{\mathbf{A}}^{ - 1}}$ exists.
To confirm the existence of ${{\mathbf{A}}^{ - 1}}$ we check whether $\left| {\mathbf{A}} \right| \ne 0$.
We have the formula to find $\left| {\mathbf{A}} \right|$ given as, ${a_{11}}({a_{22}}{a_{33}} - {a_{23}}{a_{32}}) - {a_{21}}({a_{12}}{a_{33}} - {a_{13}}{a_{32}}) + {a_{31}}({a_{12}}{a_{23}} - {a_{13}}{a_{22}})$ where ${a_{ij}}$ is the element at the ${i^{th}}$ row and ${j^{th}}$ column of the matrix ${\mathbf{A}}$.
So, we get $\left| {\mathbf{A}} \right| = 5( - 50 - 75) - ( - 5)( - 100 + 75) + 10( - 150 - 75)$.
$ \Rightarrow \left| {\mathbf{A}} \right| = 5( - 125) - ( - 5)( - 25) + 10( - 225)$
$ \Rightarrow \left| {\mathbf{A}} \right| = - 625 - 125 - 2250$
$ \Rightarrow \left| {\mathbf{A}} \right| = - 3000$
Since, $\left| {\mathbf{A}} \right| \ne 0$, we conclude that ${{\mathbf{A}}^{ - 1}}$ exists.
Now we will use the formula ${{\mathbf{A}}^{ - 1}} = \dfrac{1}{{\left| {\mathbf{A}} \right|}}Adj{\mathbf{A}}$ to get ${{\mathbf{A}}^{ - 1}}$. But firstly, we need to find $Adj{\mathbf{A}}$.
\[{\mathbf{A}} = \left[ {\begin{array}{*{20}{c}}
 5&{ - 5}&{10} \\
 {10}&5&{ - 5} \\
 {15}&{ - 15}&{ - 10}
\end{array}} \right]\]
The adjoint of matrix ${\mathbf{A}}$ is the transpose of its cofactor matrix whose elements are the cofactors of each element in ${\mathbf{A}}$.
Cofactor of ${a_{11}} = \left| {\begin{array}{*{20}{c}}
 5&{ - 5} \\
 { - 15}&{ - 10}
\end{array}} \right| = 5( - 10) - ( - 5)( - 15) = - 125$
Cofactor of ${a_{12}} = - \left| {\begin{array}{*{20}{c}}
 {10}&{ - 5} \\
 {15}&{ - 10}
\end{array}} \right| = - \left( {10( - 10) - ( - 5)(15)} \right) = 25$
Cofactor of ${a_{13}} = \left| {\begin{array}{*{20}{c}}
 {10}&5 \\
 {15}&{ - 15}
\end{array}} \right| = 10( - 15) - (5)(15) = - 225$
Cofactor of ${a_{21}} = - \left| {\begin{array}{*{20}{c}}
 { - 5}&{10} \\
 { - 15}&{ - 10}
\end{array}} \right| = - \left( {( - 10)( - 5) - (10)( - 15)} \right) = - 200$
Cofactor of ${a_{22}} = \left| {\begin{array}{*{20}{c}}
 5&{10} \\
 {15}&{ - 10}
\end{array}} \right| = 5( - 10) - (10)(15) = - 200$
Cofactor of ${a_{23}} = - \left| {\begin{array}{*{20}{c}}
 5&{ - 5} \\
 {15}&{ - 15}
\end{array}} \right| = - \left( {( - 15)(5) - ( - 5)(15)} \right) = 0$
Cofactor of ${a_{31}} = \left| {\begin{array}{*{20}{c}}
 { - 5}&{10} \\
 5&{ - 5}
\end{array}} \right| = ( - 5)( - 5) - (10)(5) = - 25$
Cofactor of ${a_{32}} = - \left| {\begin{array}{*{20}{c}}
 5&{10} \\
 {10}&{ - 5}
\end{array}} \right| = - \left( {( - 5)(5) - (10)(10)} \right) = 125$
Cofactor of${a_{33}} = \left| {\begin{array}{*{20}{c}}
 5&{ - 5} \\
 {10}&5
\end{array}} \right| = (5)(5) - (10)( - 5) = 75$.
So, the cofactor matrix is $\left[ {\begin{array}{*{20}{c}}
 { - 125}&{25}&{ - 225} \\
 { - 200}&{ - 200}&0 \\
 { - 25}&{125}&{75}
\end{array}} \right]$.
Now adjoint is the transpose of the co-factor matrix. So, $Adj{\mathbf{A}} = \left[ {\begin{array}{*{20}{c}}
 { - 125}&{ - 200}&{ - 25} \\
 {25}&{ - 200}&{125} \\
 { - 225}&0&{75}
\end{array}} \right]$.
Now, using ${{\mathbf{A}}^{ - 1}} = \dfrac{1}{{\left| {\mathbf{A}} \right|}}Adj{\mathbf{A}}$, we have \[{{\mathbf{A}}^{ - 1}} = \left[ {\begin{array}{*{20}{c}}
 {\dfrac{1}{{24}}}&{\dfrac{1}{{15}}}&{\dfrac{1}{{120}}} \\
 { - \dfrac{1}{{120}}}&{\dfrac{1}{{15}}}&{ - \dfrac{1}{{24}}} \\
 {\dfrac{3}{{40}}}&0&{ - \dfrac{1}{{40}}}
\end{array}} \right]\].
So, using the equation ${\mathbf{X}} = {{\mathbf{A}}^{ - 1}}{\mathbf{B}}$ we get \[{\mathbf{X}} = \left[ {\begin{array}{*{20}{c}}
 {\dfrac{1}{{24}}}&{\dfrac{1}{{15}}}&{\dfrac{1}{{120}}} \\
 { - \dfrac{1}{{120}}}&{\dfrac{1}{{15}}}&{ - \dfrac{1}{{24}}} \\
 {\dfrac{3}{{40}}}&0&{ - \dfrac{1}{{40}}}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
 { - 11} \\
 1 \\
 { - 1}
\end{array}} \right]\]

\[ \Rightarrow {\mathbf{X}} = \left[ {\begin{array}{*{20}{c}}
 { - \dfrac{{11}}{{24}} + \dfrac{1}{{15}} - \dfrac{1}{{120}}} \\
 { + \dfrac{{11}}{{120}} + \dfrac{1}{{15}} + \dfrac{1}{{24}}} \\
 { - \dfrac{{33}}{{40}} + \dfrac{1}{{40}}}
\end{array}} \right]\]
\[ \Rightarrow {\mathbf{X}} = \left[ {\begin{array}{*{20}{c}}
 { - \dfrac{{48}}{{120}}} \\
 {\dfrac{{24}}{{120}}} \\
 { - \dfrac{{32}}{{40}}}
\end{array}} \right]\]
\[ \Rightarrow {\mathbf{X}} = \left[ {\begin{array}{*{20}{c}}
 { - \dfrac{2}{5}} \\
 {\dfrac{1}{5}} \\
 { - \dfrac{4}{5}}
\end{array}} \right]\]
Hence, the solution of the $5x - 5y + 10z = - 11$, $10x + 5y - 5z = 1$ and $15x - 15y - 10z = - 1$ is $x = \dfrac{{ - 2}}{5},{\text{ }}y = \dfrac{1}{5}{\text{ and }}z = \dfrac{{ - 4}}{5}$.

Note:
It is important to be careful while doing calculations in these questions. As, it would be tedious and time-consuming to re-check the calculations, once done. Also, remember that if the coefficient matrix doesn’t have an inverse, then the system either is inconsistent (does not have a solution) or has infinitely many solutions.