Matrices: Systems of Equations 1
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Question 1 of 4
1. Question
Solve for `x` and `y``x-3y=19``5x+2y=10`-
`x=` (4)`y=` (-5)
Hint
Help VideoCorrect
Fantastic!
Incorrect
Inverse of a `2xx2` Matrix
If \begin{bmatrix}
\color{#007DDC}{a} & \color{#9a00c7}{b} \\
\color{#9a00c7}{c} & \color{#007DDC}{d}
\end{bmatrix} then $$A^{-1}=\frac{1}{|A|}$$\begin{bmatrix}
\color{#007DDC}{d} & -\color{#9a00c7}{b} \\
-\color{#9a00c7}{c} & \color{#007DDC}{a}
\end{bmatrix}Determinant of a `2xx2` Matrix
If \begin{bmatrix}
\color{#007DDC}{a} & \color{#9a00c7}{b} \\
\color{#9a00c7}{c} & \color{#007DDC}{d}
\end{bmatrix} then $$|A|=\color{#007DDC}{ad}-\color{#9a00c7}{bc}$$First, convert the system of equations into a matrix equation`x-3y` `=` `19` `5x+2y` `=` `10` $$\:\:\:\:\:\:\:A\:\:\:\:\:\:\:\:\:\:\:\,X\:=\:\:\:C$$\begin{bmatrix}
1 & -3 \\
5 & 2
\end{bmatrix} \begin{bmatrix}
x \\
y
\end{bmatrix}$$=$$\begin{bmatrix}
\color{#e65021}{19} \\
\color{#e65021}{10}
\end{bmatrix}Note that `A` is the matrix of coefficients, `X` is the matrix of variables, and `C` is the matrix of constantsRemember that we are looking for the values of `x` and `y`, which is contained in the matrix `X``AX` `=` `C` `X` `=` `A^(-1)C` Start finding `A^(-1)` by solving for the determinant`|A|` `=` `ad` `-``bc` Determinant Formula \begin{vmatrix}
\color{#007DDC}{1} & \color{#9a00c7}{-3} \\
\color{#9a00c7}{5} & \color{#007DDC}{2}
\end{vmatrix}`=` `1*2` `-``(-3)*5` Substitute values `=` `2-(-15)` `=` `17` Substitute respective values into the Inverse Formula`A^(-1)` `=` $$\frac{1}{|A|}$$\begin{bmatrix}
\color{#007DDC}{d} & -\color{#9a00c7}{b} \\
-\color{#9a00c7}{c} & \color{#007DDC}{a}
\end{bmatrix}Inverse Formula `=` $$\frac{1}{17}$$\begin{bmatrix}
\color{#007DDC}{2} & -\color{#9a00c7}{(-3)} \\
-\color{#9a00c7}{5} & \color{#007DDC}{1}
\end{bmatrix}Substitute values `A^(-1)` `=` $$\frac{1}{17}$$\begin{bmatrix}
\color{#007DDC}{2} & \color{#9a00c7}{3} \\
\color{#9a00c7}{-5} & \color{#007DDC}{1}
\end{bmatrix}Finally, substitute the matrices into the `X` formula`X` `=` `A^(-1)C` `=` $$\frac{1}{17}$$\begin{bmatrix}
\color{#007DDC}{2} & \color{#9a00c7}{3} \\
\color{#9a00c7}{-5} & \color{#007DDC}{1}
\end{bmatrix}\begin{bmatrix}
\color{#e65021}{19} \\
\color{#e65021}{10}
\end{bmatrix}Substitute values `=` $$\frac{1}{17}$$\begin{bmatrix}
(\color{#007DDC}{2}\cdot\color{#e65021}{19})+(\color{#9a00c7}{3}\cdot\color{#e65021}{10}) \\
(\color{#9a00c7}{-5}\cdot\color{#e65021}{19})+(\color{#007DDC}{1}\cdot\color{#e65021}{10})
\end{bmatrix}`=` $$\frac{1}{17}$$\begin{bmatrix}
38+30 \\
-95+10
\end{bmatrix}`=` $$\frac{1}{17}$$\begin{bmatrix}
68 \\
-85
\end{bmatrix}`X` `=` \begin{bmatrix}
4 \\
-5
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}`=` \begin{bmatrix}
4 \\
-5
\end{bmatrix}`x=4``y=-5` -
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Question 2 of 4
2. Question
Solve for `x` and `y``2x-3y=9``x+5y=-2`-
`x=` (3)`y=` (-1)
Hint
Help VideoCorrect
Excellent!
Incorrect
Inverse of a `2xx2` Matrix
If \begin{bmatrix}
\color{#007DDC}{a} & \color{#9a00c7}{b} \\
\color{#9a00c7}{c} & \color{#007DDC}{d}
\end{bmatrix} then $$A^{-1}=\frac{1}{|A|}$$\begin{bmatrix}
\color{#007DDC}{d} & -\color{#9a00c7}{b} \\
-\color{#9a00c7}{c} & \color{#007DDC}{a}
\end{bmatrix}Determinant of a `2xx2` Matrix
If \begin{bmatrix}
\color{#007DDC}{a} & \color{#9a00c7}{b} \\
\color{#9a00c7}{c} & \color{#007DDC}{d}
\end{bmatrix} then $$|A|=\color{#007DDC}{ad}-\color{#9a00c7}{bc}$$First, convert the system of equations into a matrix equation`2x-3y` `=` `9` `x+5y` `=` `-2` $$\:\:\:\:\:\:\:A\:\:\:\:\:\:\:\:\:\:\:\,X\:=\:\:\:C$$\begin{bmatrix}
2 & -3 \\
1 & 5
\end{bmatrix} \begin{bmatrix}
x \\
y
\end{bmatrix}$$=$$\begin{bmatrix}
\color{#e65021}{9} \\
\color{#e65021}{-2}
\end{bmatrix}Note that `A` is the matrix of coefficients, `X` is the matrix of variables, and `C` is the matrix of constantsRemember that we are looking for the values of `x` and `y`, which is contained in the matrix `X``AX` `=` `C` `X` `=` `A^(-1)C` Start finding `A^(-1)` by solving for the determinant`|A|` `=` `ad` `-``bc` Determinant Formula \begin{vmatrix}
\color{#007DDC}{2} & \color{#9a00c7}{-3} \\
\color{#9a00c7}{1} & \color{#007DDC}{5}
\end{vmatrix}`=` `2*5` `-``(-3)*1` Substitute values `=` `10-(-3)` `=` `13` Substitute respective values into the Inverse Formula`A^(-1)` `=` $$\frac{1}{|A|}$$\begin{bmatrix}
\color{#007DDC}{d} & -\color{#9a00c7}{b} \\
-\color{#9a00c7}{c} & \color{#007DDC}{a}
\end{bmatrix}Inverse Formula `=` $$\frac{1}{13}$$\begin{bmatrix}
\color{#007DDC}{5} & -\color{#9a00c7}{(-3)} \\
-\color{#9a00c7}{1} & \color{#007DDC}{2}
\end{bmatrix}Substitute values `A^(-1)` `=` $$\frac{1}{13}$$\begin{bmatrix}
\color{#007DDC}{5} & \color{#9a00c7}{3} \\
\color{#9a00c7}{-1} & \color{#007DDC}{2}
\end{bmatrix}Finally, substitute the matrices into the `X` formula`X` `=` `A^(-1)C` `=` $$\frac{1}{13}$$\begin{bmatrix}
\color{#007DDC}{5} & \color{#9a00c7}{3} \\
\color{#9a00c7}{-1} & \color{#007DDC}{2}
\end{bmatrix}\begin{bmatrix}
\color{#e65021}{9} \\
\color{#e65021}{-2}
\end{bmatrix}Substitute values `=` $$\frac{1}{13}$$\begin{bmatrix}
(\color{#007DDC}{5}\cdot\color{#e65021}{9})+(\color{#9a00c7}{3}\cdot\color{#e65021}{-2}) \\
(\color{#9a00c7}{-1}\cdot\color{#e65021}{9})+(\color{#007DDC}{2}\cdot\color{#e65021}{-2})
\end{bmatrix}`=` $$\frac{1}{13}$$\begin{bmatrix}
45+(-6) \\
-9+(-4)
\end{bmatrix}`=` $$\frac{1}{13}$$\begin{bmatrix}
39 \\
-13
\end{bmatrix}`X` `=` \begin{bmatrix}
3 \\
-1
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}`=` \begin{bmatrix}
3 \\
-1
\end{bmatrix}`x=3``y=-1` -
-
Question 3 of 4
3. Question
Solve for `x` and `y``x-5y=13``4x-6y=24`-
`x=` (3)`y=` (-2)
Hint
Help VideoCorrect
Nice Job!
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`x-5y=13``4x-6y=24`\begin{bmatrix}
1 & -5 & | & 13 \\
4 & -6 & | & 24
\end{bmatrix} \begin{matrix}
\color{#6F6161}{R_1} \\
\color{#6F6161}{R_2}
\end{matrix}Use row operations for the matrix to achieve Row-Echelon FormNotice that `R_1` is already in Row-Echelon Form. Hence, we only have to transform `R_2`Transforming the First Element of `R_2`:\begin{bmatrix}
1 & -5 & | & 13 \\
4 & -6 & | & 24
\end{bmatrix} \begin{matrix}
\color{#6F6161}{R_1} \\
\color{#6F6161}{R_2}
\end{matrix}$$-4R_1+R_2→R_2$$\begin{matrix}
-4R_1: & -4 & 20 & | & -52 \\
+\:R_2: & 4 & -6 & | & 24 \\
\hline
R_2→ & 0 & 14 & | & -28
\end{matrix}Transforming the Second Element of `R_2`:\begin{bmatrix}
1 & -5 & | & 13 \\
0 & 14 & | & -28
\end{bmatrix} \begin{matrix}
\color{#6F6161}{R_1} \\
\color{#6F6161}{R_2}
\end{matrix}$$R_2\div14→R_2$$\begin{matrix}
R_2\div14→ 0 & 1 & | & -2
\end{matrix}New matrix:\begin{bmatrix}
1 & -5 & | & 13 \\
0 & 1 & | & -2
\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 & | & -2
\end{bmatrix}\begin{matrix}
\:\: & \color{#00880A}{y} & = & \color{#00880A}{-2}
\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 & -5 & | & 13
\end{bmatrix}`x-5y` `=` `13` `x-5(-2)` `=` `13` Substitute `y=-2` `x+10` `=` `13` `x+10` `-10` `=` `13` `-10` `x` `=` `3` `x=3``y=-2` -
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Question 4 of 4
4. Question
Solve for `x` and `y``3x+5y=14``7x-2y=-22`-
`x=` (-2)`y=` (4)
Hint
Help VideoCorrect
Well Done!
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`3x+5y=14``7x-2y=-22`\begin{bmatrix}
3 & 5 & | & 14 \\
7 & -2 & | & -22
\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 transforming `R_2`Transforming the First Element of `R_2`:\begin{bmatrix}
3 & 5 & | & 14 \\
7 & -2 & | & -22
\end{bmatrix} \begin{matrix}
\color{#6F6161}{R_1} \\
\color{#6F6161}{R_2}
\end{matrix}$$7R_1-3R_2→R_2$$\begin{matrix}
\;\;7R_1: & 21 & 35 & | & 98 \\
-3R_2: & -21 & 6 & | & 66 \\
\hline
R_2→ & 0 & 41 & | & 164
\end{matrix}Transforming the Second Element of `R_2`:\begin{bmatrix}
3 & 5 & | & 14 \\
0 & 41 & | & 164
\end{bmatrix} \begin{matrix}
\color{#6F6161}{R_1} \\
\color{#6F6161}{R_2}
\end{matrix}$$R_2\div41→R_2$$\begin{matrix}
R_2\div41→ 0 & 1 & | & 4
\end{matrix}New matrix:\begin{bmatrix}
3 & 5 & | & 14 \\
0 & 1 & | & 4
\end{bmatrix} \begin{matrix}
\color{#6F6161}{R_1} \\
\color{#6F6161}{R_2}
\end{matrix}Proceed with transforming `R_1`Transforming the First Element of `R_1`:\begin{bmatrix}
3 & 5 & | & 14 \\
0 & 1 & | & 4
\end{bmatrix} \begin{matrix}
\color{#6F6161}{R_1} \\
\color{#6F6161}{R_2}
\end{matrix}$$R_1\div3→R_1$$\begin{matrix}
R_1\div3→ 1 & \frac{5}{3} & | & \frac{14}{3}
\end{matrix}New matrix:\begin{bmatrix}
1 & \frac{5}{3} & | & \frac{14}{3} \\[0.3em]
0 & 1 & | & -2
\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 & | & 4
\end{bmatrix}\begin{matrix}
\:\: & \color{#00880A}{y} & = & \color{#00880A}{4}
\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{5}{3} & | & \frac{14}{3}
\end{bmatrix}`x+5/3y` `=` `14/3` `x+5/3(4)` `=` `14/3` Substitute `y=4` `x+20/3` `=` `14/3` `x+20/3` `-20/3` `=` `14/3` `-20/3` Subtract `20/3` from both sides `x` `=` `-6/3` `x` `=` `-2` `x=-2``y=4` -
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