Matrices: Systems of Equations 1

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Precalculus

Matrices

Matrices: Systems of Equations

Matrices: Systems of Equations 1

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Question 1 of 4

Solve for

x

$x$ and

y

$y$

x - 3 y = 19

$x-3y=19$

5 x + 2 y = 10

$5x+2y=10$

$x =$ $x=$ (4)

$y =$ $y=$ (-5)

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Inverse of a $2 \times 2$ $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ 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

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ 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

Solve for

$x$ and

$y$

$2x-3y=9$

$x+5y=-2$

$x=$ (3)

$y=$ (-1)

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Inverse of a $2xx2$ Matrix

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ 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

$\begin{bmatrix} \color{#007DDC}{a} & \color{#9a00c7}{b} \\ \color{#9a00c7}{c} & \color{#007DDC}{d} \end{bmatrix}$ 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)
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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

$x-5y=13$

$4x-6y=24$

$\begin{bmatrix} 1 & -5 & | & 13 \\ 4 & -6 & | & 24 \end{bmatrix}$ $\begin{matrix} \color{#6F6161}{R_1} \\ \color{#6F6161}{R_2} \end{matrix}$

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

Solve for

$x$ and

$y$

$3x+5y=14$

$7x-2y=-22$

$x=$ (-2)

$y=$ (4)

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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

$3x+5y=14$

$7x-2y=-22$

$\begin{bmatrix} 3 & 5 & | & 14 \\ 7 & -2 & | & -22 \end{bmatrix}$

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

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

$x$	$=$	$-6/3$

$x$	$=$	$-2$

$x=-2$

$y=4$