Matrices: 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 VideoCorrect
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 FormFirst, 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 FormStart with simplifying the two rowsSimplifying `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 FormRemember that each row in this matrix corresponds to a term in an equationConvert `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` -
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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 VideoCorrect
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 FormFirst, 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 FormStart 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 FormRemember that each row in this matrix corresponds to a term in an equationConvert `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` -
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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 VideoCorrect
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 FormFirst, 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 FormStart 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 columnTransforming 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 FormRemember that each row in this matrix corresponds to a term in an equationConvert `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` -
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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 VideoCorrect
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 FormFirst, 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 FormStart 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 FormRemember that each row in this matrix corresponds to a term in an equationConvert `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
- Add & Subtract Matrices 1
- Add & Subtract Matrices 2
- Add & Subtract Matrices 3
- Multiply Matrices 1
- Multiply Matrices 2
- Matrices: Multiplication Word Problems
- Determinant of a Matrix
- Inverse of a Matrix
- Matrices: Systems of Equations 1
- Matrices: Systems of Equations 2
- Gauss Jordan Elimination
- Cramer’s Rule