Matrices: Systems of Equations 2

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Algebra 2

Matrices

Matrices: Systems of Equations

Matrices: Systems of Equations 2

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Question 1 of 4

Solve for

$x$ and

$y$

$6x-3y=-9$

$-4x-2y=26$

$x=$ (-4)

$y=$ (-5)

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Row-Echelon Form

$[[1,x,x],[0,1,x]]$

A system of equations can be solved by using matrices and applying row operations to achieve the Row-Echelon Form

First, write the system of equations in matrix form by taking the constants

$6x-3y=-9$

$-4x-2y=26$

$\begin{bmatrix} 6 & -3 & | & -9 \\ -4 & -2 & | & 26 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

Use row operations for the matrix to achieve Row-Echelon Form

Start with simplifying the two rows

Simplifying

$R_1$ :

$\begin{bmatrix} 6 & -3 & | & -9 \\ -4 & -2 & | & 26 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

$R_1\div3→R_1$

$\begin{matrix} R_1\div3→ 2 & -1 & | & -3 \end{matrix}$

Simplifying

$R_2$ :

$\begin{bmatrix} 2 & -1 & | & -3 \\ -4 & -2 & | & 26 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

$R_2\div2→R_2$

$\begin{matrix} R_1\div3→ -2 & -1 & | & 13 \end{matrix}$

New matrix:

$\begin{bmatrix} 2 & -1 & | & -3 \\ -2 & -1 & | & 13 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

Proceed with transforming

$R_2$

Transforming the First Element of

$R_2$ :

$\begin{bmatrix} 2 & -1 & | & -3 \\ -2 & -1 & | & 13 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

$R_1+R_2→R_2$

$\begin{matrix} R_1: & 2 & -1 & | & -3 \\ R_2: & -2 & -1 & | & 13 \\ \hline R_2→ & 0 & -2 & | & 10 \end{matrix}$

Transforming the Second Element of

$R_2$ :

$\begin{bmatrix} 2 & -1 & | & -3 \\ 0 & -2 & | & 10 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

$R_2\div-2→R_2$

$\begin{matrix} R_2\div41→ 0 & 1 & | & -5 \end{matrix}$

New matrix:

$\begin{bmatrix} 2 & -1 & | & -3 \\ 0 & 1 & | & -5 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

Next, transform

$R_1$

Transforming the First Element of

$R_1$ :

$\begin{bmatrix} 2 & -1 & | & -3 \\ 0 & 1 & | & -5 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

$R_1\div2→R_1$

$\begin{matrix} R_1\div2→ 1 & -\frac{1}{2} & | & -\frac{3}{2} \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{3}{2} \\[0.3em] 0 & 1 & | & -5 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

This new matrix is in Row-Echelon Form

Remember that each row in this matrix corresponds to a term in an equation

Convert

$R_2$ into an equation

$\begin{matrix} \color{#6F6161}{\:x} & \color{#6F6161}{y} & \: & \color{#6F6161}{\:\:C} \end{matrix}$

$\begin{bmatrix} 0 & 1 & | & -5 \end{bmatrix}$

$\begin{matrix} \:\: & \color{#00880A}{y} & = & \color{#00880A}{-5} \end{matrix}$

This time, convert

$R_1$ into an equation then substitute the value of

$y$ to find

$x$

$\begin{matrix} \color{#6F6161}{\:\;x} & \color{#6F6161}{\:\:y} & \: & \color{#6F6161}{\:\:\:\:C} \end{matrix}$

$\begin{bmatrix} 1 & -\frac{1}{2} & | & -\frac{3}{2} \end{bmatrix}$

$x-1/2y$	$=$	$-3/2$

$x-1/2(-5)$	$=$	$-3/2$	Substitute $y=-5$

$x+5/2$	$=$	$-3/2$

$x+5/2$ $-5/2$	$=$	$-3/2$ $-5/2$	Subtract $5/2$ from both sides

$x$	$=$	$-8/2$

$x$	$=$	$-4$

$x=-4$

$y=-5$

Question 2 of 4

Solve for

$x,y$ and

$z$

$5x+3z=-3$

$y+4z=15$

$2x-4y-z=-6$

$x=$ (-3)

$y=$ (-1)

$z=$ (4)

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Row-Echelon Form

$[[1,x,x,x],[0,1,x,x],[0,0,1,x]]$

A system of equations can be solved by using matrices and applying row operations to achieve the Row-Echelon Form

First, write the system of equations in matrix form by taking the constants

$5x+3z=-3$

$y+4z=15$

$2x-4y-z=-6$

$\begin{bmatrix} 5 & 0 & 3 & | & -3 \\ 0 & 1 & 4 & | & 15 \\ 2 & -4 & -1 & | & -6 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Use row operations for the matrix to achieve Row-Echelon Form

Start with transforming

$R_3$

Transforming the First Element of

$R_3$ :

$\begin{bmatrix} 5 & 0 & 3 & | & -3 \\ 0 & 1 & 4 & | & 15 \\ 2 & -4 & -1 & | & -6 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$2R_1-5R_3→R_3$

$\begin{matrix} \:\:2R_1: & 10 & 0 & 6 & | & -6 \\ -5R_3: & -10 & 20 & 5 & | & 30 \\ \hline \:R_3→ & 0 & 20 & 11 & | & 24 \end{matrix}$

Transforming the Second Element of

$R_3$ :

$\begin{bmatrix} 5 & 0 & 3 & | & -3 \\ 0 & 1 & 4 & | & 15 \\ 0 & 20 & 11 & | & 24 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$20R_2-R_3→R_3$

$\begin{matrix} 20R_2: & 0 & 20 & 80 & | & 300 \\ -R_3: & 0 & -20 & -11 & | & -24 \\ \hline \:R_3→ & 0 & 0 & 69 & | & 276 \end{matrix}$

Transforming the Third Element of

$R_3$ :

$\begin{bmatrix} 5 & 0 & 3 & | & -3 \\ 0 & 1 & 4 & | & 15 \\ 0 & 0 & 69 & | & 276 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_3\div69→R_3$

$\begin{matrix} R_3\div69→ 0 & 0 & 1 & | & 4 \end{matrix}$

New matrix:

$\begin{bmatrix} 5 & 0 & 3 & | & -3 \\ 0 & 1 & 4 & | & 15 \\ 0 & 0 & 1 & | & 4 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Notice that

$R_2$ is already in Row-Echelon Form. Hence, continue with transforming

$R_1$

Transforming the First Element of

$R_1$ :

$\begin{bmatrix} 5 & 0 & 3 & | & -3 \\ 0 & 1 & 4 & | & 15 \\ 0 & 0 & 1 & | & 4 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_1\div5→R_1$

$\begin{matrix} R_1\div5→ 1 & 0 & \frac{3}{5} & | & \frac{-3}{5} \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & 0 & \frac{3}{5} & | & \frac{-3}{5} \\[0.3em] 0 & 1 & 4 & | & 15 \\ 0 & 0 & 1 & | & 4 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

This new matrix is in Row-Echelon Form

Remember that each row in this matrix corresponds to a term in an equation

Convert

$R_3$ into an equation

$\begin{matrix} \color{#6F6161}{\:x} & \color{#6F6161}{y} & \color{#6F6161}{z} & \: & \color{#6F6161}{C} \end{matrix}$

$\begin{bmatrix} 0 & 0 & 1 & | & 4 \end{bmatrix}$

$\begin{matrix} \:\: & \:\: & \color{#00880A}{z} & = & \color{#00880A}{4} \end{matrix}$

Next, convert

$R_2$ into an equation then substitute the value of

$z$ to find

$y$

$\begin{matrix} \color{#6F6161}{\:x} & \color{#6F6161}{y} & \color{#6F6161}{z} & \: & \color{#6F6161}{\:C} \end{matrix}$

$\begin{bmatrix} 0 & 1 & 4 & | & 15 \end{bmatrix}$

$y+4z$	$=$	$15$
$y+4(4)$	$=$	$15$	Substitute $z=4$
$y+16$	$=$	$15$
$y+16$ $-16$	$=$	$15$ $-16$
$y$	$=$	$-1$

Lastly, convert

$R_1$ into an equation then substitute the value of

$z$ to find

$x$

$\begin{matrix} \color{#6F6161}{\;\:x} & \color{#6F6161}{y} & \color{#6F6161}{z} & \: & \color{#6F6161}{\;C} \end{matrix}$

$\begin{bmatrix} 1 & 0 & \frac{3}{5} & | & \frac{-3}{5} \end{bmatrix}$

$x+3/5z$	$=$	$(-3)/5$

$x+3/5(4)$	$=$	$(-3)/5$	Substitute $z=4$

$x+12/5$	$=$	$(-3)/5$

$x+12/5$ $-12/5$	$=$	$(-3)/5$ $-12/5$

$x$	$=$	$(-15)/5$

$x$	$=$	$-3$

$x=-3$

$y=-1$

$z=4$

Question 3 of 4

Solve for

$x,y$ and

$z$

$4x+2y+z=-3$

$x-3y=-11$

$2x+y-2z=1$

$x=$ (-2)

$y=$ (3)

$z=$ (-1)

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Row-Echelon Form

$[[1,x,x,x],[0,1,x,x],[0,0,1,x]]$

A system of equations can be solved by using matrices and applying row operations to achieve the Row-Echelon Form

First, write the system of equations in matrix form by taking the constants

$4x+2y+z=-3$

$x-3y=-11$

$2x+y-2z=1$

$\begin{bmatrix} 4 & 2 & 1 & | & -3 \\ 1 & -3 & 0 & | & -11 \\ 2 & 1 & -2 & | & 1 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Use row operations for the matrix to achieve Row-Echelon Form

Start with swapping

$R_1$ and

$R_2$ , since the first element of

$R_2$ is already

$1$

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\ 4 & 2 & 1 & | & -3 \\ 2 & 1 & -2 & | & 1 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Proceed with transforming the first column

Transforming the First Element of

$R_3$ :

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\ 4 & 2 & 1 & | & -3 \\ 2 & 1 & -2 & | & 1 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-2R_1+R_3→R_3$

$\begin{matrix} -2R_1: & -2 & 6 & 0 & | & 22 \\ \:R_3: & 2 & 1 & -2 & | & 1 \\ \hline \:R_3→ & 0 & 7 & -2 & | & 23 \end{matrix}$

Transforming the First Element of

$R_2$ :

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\ 4 & 2 & 1 & | & -3 \\ 0 & 7 & -2 & | & 23 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-4R_1+R_2→R_2$

$\begin{matrix} -4R_1: & -4 & 12 & 0 & | & 44 \\ R_2: & 4 & 2 & 1 & | & -3 \\ \hline \:R_3→ & 0 & 14 & 1 & | & 41 \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\ 0 & 14 & 1 & | & 41 \\ 0 & 7 & -2 & | & 23 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Continue with transforming

$R_3$

Transforming the Second Element of

$R_3$ :

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\ 0 & 14 & 1 & | & 41 \\ 0 & 7 & -2 & | & 23 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_2-2R_3→R_3$

$\begin{matrix} R_2: & 0 & 14 & 1 & | & 41 \\ -2R_3: & 0 & -14 & 4 & | & -46 \\ \hline \:R_3→ & 0 & 0 & 5 & | & -5 \end{matrix}$

Transforming the Third Element of

$R_3$ :

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\ 0 & 14 & 1 & | & 41 \\ 0 & 0 & 5 & | & -5 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_3\div5→R_3$

$\begin{matrix} R_3\div5→ 0 & 0 & 1 & | & -1 \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\ 0 & 14 & 1 & | & 41 \\ 0 & 0 & 1 & | & -1 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Next, transform

$R_2$

Transforming the Second Element of

$R_2$ :

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\ 0 & 14 & 1 & | & 41 \\ 0 & 0 & 1 & | & -1 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_2\div14→R_2$

$\begin{matrix} R_2\div14→ 0 & 1 & \frac{1}{14} & | & \frac{41}{14} \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \\[0.3em] 0 & 1 & \frac{1}{14} & | & \frac{41}{14} \\[0.3em] 0 & 0 & 1 & | & -1 \end{bmatrix}$

$\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

This new matrix is in Row-Echelon Form

Remember that each row in this matrix corresponds to a term in an equation

Convert

$R_3$ into an equation

$\begin{matrix} \color{#6F6161}{\:x} & \color{#6F6161}{y} & \color{#6F6161}{z} & \: & \color{#6F6161}{\:\;C} \end{matrix}$

$\begin{bmatrix} 0 & 0 & 1 & | & -1 \end{bmatrix}$

$\begin{matrix} \:\: & \:\: & \color{#00880A}{\:z} & = & \color{#00880A}{-1} \end{matrix}$

Next, convert

$R_2$ into an equation then substitute the value of

$z$ to find

$y$

$\begin{matrix} \color{#6F6161}{\:\,x} & \color{#6F6161}{y} & \color{#6F6161}{\:z} & \: & \color{#6F6161}{\:\:\;C} \end{matrix}$

$\begin{bmatrix} 0 & 1 & \frac{1}{14} & | & \frac{41}{14} \end{bmatrix}$

$y+1/14z$	$=$	$41/14$

$y+1/14(-1)$	$=$	$41/14$	Substitute $z=-1$

$y-1/14$	$=$	$41/14$

$y-1/14$ $+1/14$	$=$	$41/14$ $+1/14$	Add $1/14$ to both sides

$y$	$=$	$42/14$

$y$	$=$	$3$

Lastly, convert

$R_1$ into an equation then substitute the value of

$y$ to find

$x$

$\begin{matrix} \color{#6F6161}{\:x} & \color{#6F6161}{\:\:y} & \color{#6F6161}{\:z} & \: & \color{#6F6161}{\;\:\:C} \end{matrix}$

$\begin{bmatrix} 1 & -3 & 0 & | & -11 \end{bmatrix}$

$x-3y$	$=$	$-11$
$x-3(3)$	$=$	$-11$	Substitute $y=3$
$x-9$	$=$	$-11$
$x-9$ $+9$	$=$	$-11$ $+9$	Add $9$ to both sides
$x$	$=$	$-2$

$x=-2$

$y=3$

$z=-1$

Question 4 of 4

4. Question
Solve for $x,y$ and $z$

$3x-2y=16$

$2x+y+5z=1$

$x+4y-z=-3$
- $x=$ (4)
  
  $y=$ (-2)
  
  $z=$ (-1)
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Row-Echelon Form

$[[1,x,x,x],[0,1,x,x],[0,0,1,x]]$

A system of equations can be solved by using matrices and applying row operations to achieve the Row-Echelon Form

First, write the system of equations in matrix form by taking the constants

$3x-2y=16$

$2x+y+5z=1$

$x+4y-z=-3$

$\begin{bmatrix} 3 & -2 & 0 & | & 16 \\ 2 & 1 & 5 & | & 1 \\ 1 & 4 & -1 & | & -3 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Use row operations for the matrix to achieve Row-Echelon Form

Start with transforming $R_2$

Transforming the First Element of $R_2$ :

$\begin{bmatrix} 3 & -2 & 0 & | & 16 \\ 2 & 1 & 5 & | & 1 \\ 1 & 4 & -1 & | & -3 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-2R_3+R_2→R_2$

$\begin{matrix} -2R_3: & -2 & -8 & 2 & | & 6 \\ \:R_2: & 2 & 1 & 5 & | & 1 \\ \hline \:R_2→ & 0 & -7 & 7 & | & 7 \end{matrix}$

Transforming the Second Element of $R_2$ :

$\begin{bmatrix} 3 & -2 & 0 & | & 16 \\ 0 & -7 & 7 & | & 7 \\ 1 & 4 & -1 & | & -3 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_2\div-7→R_2$

$\begin{matrix} R_2\div-7→ 0 & 1 & -1 & | & -1 \end{matrix}$

New matrix:

$\begin{bmatrix} 3 & -2 & 0 & | & 16 \\ 0 & 1 & -1 & | & -1 \\ 1 & 4 & -1 & | & -3 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Next, swap $R_1$ and $R_3$ , since the first element of $R_3$ is already $1$

$\begin{bmatrix} 1 & 4 & -1 & | & -3 \\ 0 & 1 & -1 & | & -1 \\ 3 & -2 & 0 & | & 16 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

Continue with transforming $R_3$

Transforming the First Element of $R_3$ :

$\begin{bmatrix} 1 & 4 & -1 & | & -3 \\ 0 & 1 & -1 & | & -1 \\ 3 & -2 & 0 & | & 16 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$-3R_1+R_3→R_3$

$\begin{matrix} -3R_1: & -3 & -12 & 3 & | & 9 \\ R_3: & 3 & -2 & 0 & | & 16 \\ \hline \:R_3→ & 0 & -14 & 3 & | & 25 \end{matrix}$

Transforming the Second Element of $R_3$ :

$\begin{bmatrix} 1 & 4 & -1 & | & -3 \\ 0 & 1 & -1 & | & -1 \\ 0 & -14 & 3 & | & 25 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$14R_2+R_3→R_3$

$\begin{matrix} 14R_2: & 0 & 14 & -14 & | & -14 \\ R_3: & 0 & -14 & 3 & | & 25 \\ \hline \:R_3→ & 0 & 0 & -11 & | & 11 \end{matrix}$

Transforming the Third Element of $R_3$ :

$\begin{bmatrix} 1 & 4 & -1 & | & -3 \\ 0 & 1 & -1 & | & -1 \\ 0 & 0 & -11 & | & 11 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

$R_3\div-11→R_3$

$\begin{matrix} R_3\div-11→ 0 & 0 & 1 & | & -1 \end{matrix}$

New matrix:

$\begin{bmatrix} 1 & 4 & -1 & | & -3 \\ 0 & 1 & -1 & | & -1 \\ 0 & 0 & 1 & | & -1 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \\ \color{#6F6161}{R_3} \end{matrix}$

This new matrix is in Row-Echelon Form

Remember that each row in this matrix corresponds to a term in an equation

Convert $R_3$ into an equation

$\begin{matrix} \color{#6F6161}{\:x} & \color{#6F6161}{y} & \color{#6F6161}{z} & \: & \color{#6F6161}{\:\;C} \end{matrix}$
$\begin{bmatrix} 0 & 0 & 1 & | & -1 \end{bmatrix}$

$\begin{matrix} \:\: & \:\: & \color{#00880A}{\:z} & = & \color{#00880A}{-1} \end{matrix}$

Next, convert $R_2$ into an equation then substitute the value of $z$ to find $y$

$\begin{matrix} \color{#6F6161}{\:\,x} & \color{#6F6161}{y} & \color{#6F6161}{\:z} & \: & \color{#6F6161}{\:\:\;C} \end{matrix}$
$\begin{bmatrix} 0 & 1 & -1 & | & -1 \end{bmatrix}$

$y-z$ $=$ $-1$

$y-(-1)$ $=$ $-1$ Substitute $z=-1$

$y+1$ $=$ $-1$

$y+1$ $-1$ $=$ $-1$ $-1$ Subtract $1$ from both sides

$y$ $=$ $-2$

Lastly, convert $R_1$ into an equation then substitute the values of $y$ and $z$ to find $x$

$\begin{matrix} \color{#6F6161}{\:x} & \color{#6F6161}{\:\:y} & \color{#6F6161}{\:z} & \: & \color{#6F6161}{\;\:\:C} \end{matrix}$
$\begin{bmatrix} 1 & 4 & -1 & | & -3 \end{bmatrix}$

$x+4y-z$ $=$ $-3$

$x+4(-2)-(-1)$ $=$ $-3$ Substitute $y=-2$ and $z=-1$

$x-8+1$ $=$ $-3$

$x-7$ $=$ $-3$

$x-7$ $+7$ $=$ $-3$ $+7$ Add $7$ to both sides

$x$ $=$ $4$

$x=4$

$y=-2$

$z=-1$

Matrices: Systems of Equations 2

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Quiz summary

Information

1. Question

Hint

Great Work!

Row-Echelon Form

2. Question

Hint

Correct!

Row-Echelon Form

3. Question

Hint

Keep Going!

Row-Echelon Form

4. Question

Hint

Well Done!

Row-Echelon Form

Quizzes

$y-z$	$=$	$-1$
$y-(-1)$	$=$	$-1$	Substitute $z=-1$
$y+1$	$=$	$-1$
$y+1$ $-1$	$=$	$-1$ $-1$	Subtract $1$ from both sides
$y$	$=$	$-2$

$x+4y-z$	$=$	$-3$
$x+4(-2)-(-1)$	$=$	$-3$	Substitute $y=-2$ and $z=-1$
$x-8+1$	$=$	$-3$
$x-7$	$=$	$-3$
$x-7$ $+7$	$=$	$-3$ $+7$	Add $7$ to both sides
$x$	$=$	$4$