Solving Systems of Equations 1

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Year 12

Matrices

Solving Systems of Equations

Solving 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 Video

Correct

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 constants

Remember 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}	`=`	`12`* `-``(-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`

Question 2 of 4

2. Question

Solve for `x` and `y`

`2x-3y=9`

`x+5y=-2`

`x=` (3)

`y=` (-1)

Hint

Help Video

Correct

Excellent!

Incorrect

Inverse of a `2xx2` Matrix

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 constants

Remember 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}	`=`	`25`* `-``(-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 Video

Correct

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 Form

First, 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 Form

Notice 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 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 & | & -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`

Question 4 of 4

4. Question

Solve for `x` and `y`

`3x+5y=14`

`7x-2y=-22`

`x=` (-2)

`y=` (4)

Hint

Help Video

Correct

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 Form

First, 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 Form

Start 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 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 & | & 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

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

`x-5y`	`=`	`13`
`x-5(-2)`	`=`	`13`	Substitute `y=-2`
`x+10`	`=`	`13`
`x+10` `-10`	`=`	`13` `-10`
`x`	`=`	`3`

Solving Systems of Equations 1

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Quiz summary

Information

1. Question

Hint

Fantastic!

Inverse of a `2xx2` Matrix

Determinant of a `2xx2` Matrix

2. Question

Hint

Excellent!

Inverse of a `2xx2` Matrix

Determinant of a `2xx2` Matrix

3. Question

Hint

Nice Job!

Row-Echelon Form

4. Question

Hint

Well Done!

Row-Echelon Form

Quizzes