# Using row operations to solve the system. -10x+15y+6z=-123 \\ -4x+9y+2x=-63 \\ -2x+3y+z=-24

## Question:

Using row operations to solve the system.

$$-10x+15y+6z=-123 \\ -4x+9y+2x=-63 \\ -2x+3y+z=-24 $$

## System Of Equations :

Consider a system of equations given by

{eq}\eqalign{ & {a_1}x + {b_1}y + {c_1}z = {d_1} \cr & {a_2}x + {b_2}y + {c_2}z = {d_2} \cr & {a_3}x + {b_3}y + {c_3}z = {d_3} \cr} {/eq}

Then, the coefficient matrix is given by

{eq}A = \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \\ {{a_3}}&{{b_3}}&{{c_3}} \end{array}} \right] {/eq}

If the determinant of this matrix is non-zero, then the system of equations given by {eq}AX=B {/eq}, has a unique solution given by {eq}X=A^{-1} B {/eq}

## Answer and Explanation:

Given, a system of equations with three unknowns.

{eq}\eqalign{ &-10x + 15y + 6z = -123 \cr & -4x+9y +2z = - 63 \cr & -2x+3y + z = - 24 \cr} {/eq}

We need to find the unknown variables {eq}x,y {/eq} and {eq}z {/eq}

This system in matrix form can be written as

{eq}AX=B {/eq}

where A is the coefficient matrix, X is the unknown vector and B is the right-hand side vector.

{eq}\left[ {\begin{array}{*{20}{c}} { - 10}&{15}&6 \\ 4&9&2 \\ { - 2}&3&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 123} \\ { - 63} \\ { - 24} \end{array}} \right] {/eq}

To find {eq}X, {/eq} we have {eq}X=A^{-1}B {/eq}

Now, we find {eq}A^{-1} {/eq} by elementary row transformations.

{eq}\left[ {\begin{array}{*{20}{c}} { - 10}&{15}&6 \\ 4&9&2 \\ { - 2}&3&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]A {/eq}

{eq}{R_1} \to \dfrac{{{R_1}}}{{ - 10}} {/eq}

{eq}\left[ {\begin{array}{*{20}{c}} 1&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{5}} \\ 4&9&2 \\ { - 2}&3&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{10}}}&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right]A {/eq}

{eq}{R_2} \to {R_2} - 4{R_1},{R_3} \to {R_3} + 2{R_1} {/eq}

{eq}\left[ {\begin{array}{*{20}{c}} 1&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{5}} \\ 0&{15}&{\frac{{22}}{5}} \\ 0&0&{\frac{{ - 1}}{5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{10}}}&0&0 \\ {\frac{2}{5}}&1&0 \\ {\frac{{ - 1}}{5}}&0&1 \end{array}} \right]A {/eq}

{eq}{R_2} \to \dfrac{{{R_2}}}{{15}} {/eq}

{eq}\left[ {\begin{array}{*{20}{c}} 1&{\frac{{ - 3}}{2}}&{\frac{{ - 3}}{5}} \\ 0&1&{\frac{{22}}{{75}}} \\ 0&0&{\frac{{ - 1}}{5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{ - 1}}{{10}}}&0&0 \\ {\frac{2}{{75}}}&{\frac{1}{{15}}}&0 \\ {\frac{{ - 1}}{5}}&0&1 \end{array}} \right]A {/eq}

{eq}{R_1} \to {R_1} + \dfrac{3}{2}{R_2} {/eq}

{eq}\left[ {\begin{array}{*{20}{c}} 1&0&{\frac{{ - 4}}{{25}}} \\ 0&1&{\frac{{22}}{{75}}} \\ 0&0&{\frac{{ - 1}}{5}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{ - 3}}{{50}}}&{\frac{1}{{10}}}&0 \\ {\frac{2}{{75}}}&{\frac{1}{{15}}}&0 \\ {\frac{{ - 1}}{5}}&0&1 \end{array}} \right]A {/eq}

{eq}{R_3} \to - 5\left( {{R_3}} \right) {/eq}

{eq}\left[ {\begin{array}{*{20}{c}} 1&0&{\frac{{ - 4}}{{25}}} \\ 0&1&{\frac{{22}}{{75}}} \\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{{ - 3}}{{50}}}&{\frac{1}{{10}}}&0 \\ {\frac{2}{{75}}}&{\frac{1}{{15}}}&0 \\ 1&0&{ - 5} \end{array}} \right]A {/eq}

{eq}{R_1} \to {R_1} + \dfrac{4}{{25}}{R_3},{R_2} \to {R_2} - \dfrac{{22}}{{75}}{R_3} {/eq}

{eq}\left[ {\begin{array}{*{20}{c}} 1&0&0 \\ 0&1&0 \\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{1}{{10}}}&{\frac{1}{{10}}}&{\frac{{ - 4}}{5}} \\ {\frac{{ - 4}}{{15}}}&{\frac{1}{{15}}}&{\frac{{22}}{{15}}} \\ 1&0&{ - 5} \end{array}} \right]A {/eq}

The matrix thus obtained is the inverse of matrix A

The solution of the system can be obtained by

{eq}\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{1}{{10}}}&{\dfrac{1}{{10}}}&{\dfrac{{ - 4}}{5}} \\ {\dfrac{{ - 4}}{{15}}}&{\dfrac{1}{{15}}}&{\dfrac{{22}}{{15}}} \\ 1&0&{ - 5} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} { - 123} \\ { - 63} \\ { - 24} \end{array}} \right] {/eq}

Hence,

{eq}\left[ {\begin{array}{*{20}{c}} x \\ y \\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{{ - 123}}{{10}} - \dfrac{{63}}{{10}} + \dfrac{{96}}{5}} \\ {\dfrac{{492}}{{15}} - \dfrac{{63}}{{15}} - \dfrac{{528}}{{15}}} \\ { - 123 + 120} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\dfrac{3}{5}} \\ {\dfrac{{ - 33}}{5}} \\ { - 3} \end{array}} \right] {/eq}

Hence, the required values are

{eq}\color{red}{x = \dfrac{3}{5},y = \dfrac{{ - 33}}{5},z = - 3} {/eq}

#### Learn more about this topic:

from High School Algebra II: Homework Help Resource

Chapter 8 / Lesson 8