Solving Systems of Equations 2

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Year 12

Matrices

Solving Systems of Equations

Solving Systems of Equations 2

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Question 1 of 4

1. Question

Solve for `x` and `y`

`6x-3y=-9`

`-4x-2y=26`

`x=` (-4)

`y=` (-5)

Hint

Help Video

Correct

Great Work!

Incorrect

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

2. Question
Solve for `x,y` and `z`

`5x+3z=-3`

`y+4z=15`

`2x-4y-z=-6`
- `x=` (-3)
  
  `y=` (-1)
  
  `z=` (4)
Hint

Help Video

Correct

Correct!

Incorrect

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

3. Question

Solve for `x,y` and `z`

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

`x-3y=-11`

`2x+y-2z=1`

`x=` (-2)

`y=` (3)

`z=` (-1)

Hint

Help Video

Correct

Keep Going!

Incorrect

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)
Hint

Help Video

Correct

Well Done!

Incorrect

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`

Quizzes

Adding & Subtracting Matrices 1
Adding & Subtracting Matrices 2
Adding & Subtracting Matrices 3
Multiplying Matrices 1
Multiplying Matrices 2
Multiplication Word Problems
Determinant of a Matrix
Inverse of a Matrix
Solving Systems of Equations 1
Solving Systems of Equations 2
Gauss Jordan Elimination
Cramer’s Rule

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

`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`

`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`

Solving 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