Cramer’s Rule

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Year 12

Matrices

Cramer's Rule

Cramer’s Rule

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Question 1 of 3

Solve for

x, y

$x,y$ and

$z$

$2x+y+4z=6$

$3y-z=-13$

$x-2y+z=12$

$x=$ (3)

$y=$ (-4)

$z=$ (1)

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$x=(Dx)/D,y=(Dy)/D,z=(Dz)/D$

where

$D$ is the determinant of the Matrix of Coefficients

Determinant of a $3xx3$ Matrix

$A=$

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$ then

$|A|=\color{#D800AD}{a}$

$\begin{vmatrix} e & f \\ h & k \end{vmatrix}$

$-\color{#D800AD}{b}$

$\begin{vmatrix} d & f \\ g & k \end{vmatrix}$

$+\color{#D800AD}{c}$

$\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Determinant of a $2xx2$ Matrix

$A=[[a,b],[c,d]]$ then

$|A|=ad-bc$

To solve a system of equations using Cramer’s Rule, first form the Matrices of Coefficients, Variables, and Constants.

First, convert the system of equations into a matrix equation

$2x+y+4z$	$=$	$6$
$3y-z$	$=$	$-13$
$x-2y+z$	$=$	$12$

Coefficients

$xx$ Variables

$=$ Constants

$\begin{bmatrix} 2 & 1 & 4 \\ 0 & 3 & -1 \\ 1 & -2 & 1 \end{bmatrix}$

$\begin{bmatrix} x \\ y \\ z \end{bmatrix}$

$=$

$\begin{bmatrix} \color{#e65021}{6} \\ \color{#e65021}{-13} \\ \color{#e65021}{12} \end{bmatrix}$

Next, solve for

$D$ or the determinant of the Matrix of Coefficients

Label the values of the matrix

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$

$=$

$\begin{bmatrix} \color{#D800AD}{2} & \color{#D800AD}{1} & \color{#D800AD}{4} \\ 0 & 3 & -1 \\ 1 & -2 & 1 \end{bmatrix}$

$a=2$

$d=0$

$g=1$

$b=1$

$e=3$

$h=-2$

$c=4$

$f=-1$

$k=1$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{2} & \color{#D800AD}{1} & \color{#D800AD}{4} \\ 0 & 3 & -1 \\ 1 & -2 & 1 \end{vmatrix}$	$=$	$\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{-1} \\ \color{#9a00c7}{-2} & \color{#007DDC}{1} \end{vmatrix}$ $-\color{#D800AD}{1}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{-1} \\ \color{#9a00c7}{1} & \color{#007DDC}{1} \end{vmatrix}$ $+\color{#D800AD}{4}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{3} \\ \color{#9a00c7}{1} & \color{#007DDC}{-2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{2}(\color{#007DDC}{3\cdot1}-\color{#9a00c7}{(-1)\cdot(-2)})]-[\color{#D800AD}{1}(\color{#007DDC}{0\cdot1}-\color{#9a00c7}{(-1)\cdot1})]+[\color{#D800AD}{4}(\color{#007DDC}{0\cdot(-2)}-\color{#9a00c7}{3\cdot1})]$
	$=$	$[2(3-2)]-[1(0-(-1))]+[4(0-3)]$
	$=$	$2(1)-1(1)+4(-3)$
	$=$	$2-1-12$
$D$	$=$	$-11$

Hence, the determinant of the Matrix of Coefficients is

$D=-11$

Solve for

$Dx$ or the determinant of the Matrix of Coefficients with a replaced

$x$ column

Replace the

$x$ column with the Matrix of Constants

$\begin{bmatrix} 2 & 1 & 4 \\ 0 & 3 & -1 \\ 1 & -2 & 1 \end{bmatrix}$

$→$

$\begin{bmatrix} \color{#e65021}{6} & 1 & 4 \\ \color{#e65021}{-13} & 3 & -1 \\ \color{#e65021}{12} & -2 & 1 \end{bmatrix}$

Label the values of the matrix

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$

$=$

$\begin{bmatrix} \color{#D800AD}{6} & \color{#D800AD}{1} & \color{#D800AD}{4} \\ -13 & 3 & -1 \\ 12 & -2 & 1 \end{bmatrix}$

$a=6$

$d=-13$

$g=12$

$b=1$

$e=3$

$h=-2$

$c=4$

$f=-1$

$k=1$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{6} & \color{#D800AD}{1} & \color{#D800AD}{4} \\ -13 & 3 & -1 \\ 12 & -2 & 1 \end{vmatrix}$	$=$	$\color{#D800AD}{6}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{-1} \\ \color{#9a00c7}{-2} & \color{#007DDC}{1} \end{vmatrix}$ $-\color{#D800AD}{1}$ $\begin{vmatrix} \color{#007DDC}{-13} & \color{#9a00c7}{-1} \\ \color{#9a00c7}{12} & \color{#007DDC}{1} \end{vmatrix}$ $+\color{#D800AD}{4}$ $\begin{vmatrix} \color{#007DDC}{-13} & \color{#9a00c7}{3} \\ \color{#9a00c7}{12} & \color{#007DDC}{-2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{6}(\color{#007DDC}{3\cdot1}-\color{#9a00c7}{(-1)\cdot(-2)})]-[\color{#D800AD}{1}(\color{#007DDC}{-13\cdot1}-\color{#9a00c7}{(-1)\cdot12})]+[\color{#D800AD}{4}(\color{#007DDC}{-13\cdot(-2)}-\color{#9a00c7}{3\cdot12})]$
	$=$	$[6(3-2)]-[1(-13-(-12))]+[4(26-36)]$
	$=$	$6(1)-1(-1)+4(-10)$
	$=$	$6+1-40$
$Dx$	$=$	$-33$

Hence, the determinant of the Matrix of Coefficients with the

$x$ column replaced is

$Dx=-33$

Solve for

$Dy$ or the determinant of the Matrix of Coefficients with a replaced

$y$ column

Replace the

$y$ column with the Matrix of Constants

$\begin{bmatrix} 2 & 1 & 4 \\ 0 & 3 & -1 \\ 1 & -2 & 1 \end{bmatrix}$

$→$

$\begin{bmatrix} 2 & \color{#e65021}{6} & 4 \\ 0 & \color{#e65021}{-13} & -1 \\ 1 & \color{#e65021}{12} & 1 \end{bmatrix}$

Label the values of the matrix

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$

$=$

$\begin{bmatrix} \color{#D800AD}{2} & \color{#D800AD}{6} & \color{#D800AD}{4} \\ 0 & -13 & -1 \\ 1 & 12 & 1 \end{bmatrix}$

$a=2$

$d=0$

$g=1$

$b=6$

$e=-13$

$h=12$

$c=4$

$f=-1$

$k=1$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{2} & \color{#D800AD}{6} & \color{#D800AD}{4} \\ 0 & -13 & -1 \\ 1 & 12 & 1 \end{vmatrix}$	$=$	$\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{-13} & \color{#9a00c7}{-1} \\ \color{#9a00c7}{12} & \color{#007DDC}{1} \end{vmatrix}$ $-\color{#D800AD}{6}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{-1} \\ \color{#9a00c7}{1} & \color{#007DDC}{1} \end{vmatrix}$ $+\color{#D800AD}{4}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{-13} \\ \color{#9a00c7}{1} & \color{#007DDC}{12} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{2}(\color{#007DDC}{-13\cdot1}-\color{#9a00c7}{(-1)\cdot12})]-[\color{#D800AD}{6}(\color{#007DDC}{0\cdot1}-\color{#9a00c7}{(-1)\cdot1})]+[\color{#D800AD}{4}(\color{#007DDC}{0\cdot12}-\color{#9a00c7}{(-13)\cdot1})]$
	$=$	$[2(-13-(-12))]-[6(0-(-1))]+[4(0-(-13))]$
	$=$	$2(-1)-6(1)+4(13)$
	$=$	$-2-6+52$
$Dy$	$=$	$44$

Hence, the determinant of the Matrix of Coefficients with the

$y$ column replaced is

$Dy=44$

Solve for

$Dz$ or the determinant of the Matrix of Coefficients with a replaced

$z$ column

Replace the

$z$ column with the Matrix of Constants

$\begin{bmatrix} 2 & 1 & 4 \\ 0 & 3 & -1 \\ 1 & -2 & 1 \end{bmatrix}$

$→$

$\begin{bmatrix} 2 & 1 & \color{#e65021}{6} \\ 0 & 3 & \color{#e65021}{-13} \\ 1 & -2 & \color{#e65021}{12} \end{bmatrix}$

Label the values of the matrix

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$

$=$

$\begin{bmatrix} \color{#D800AD}{2} & \color{#D800AD}{1} & \color{#D800AD}{6} \\ 0 & 3 & -13 \\ 1 & -2 & 12 \end{bmatrix}$

$a=2$

$d=0$

$g=1$

$b=1$

$e=3$

$h=-2$

$c=6$

$f=-13$

$k=12$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{2} & \color{#D800AD}{1} & \color{#D800AD}{6} \\ 0 & 3 & -13 \\ 1 & -2 & 12 \end{vmatrix}$	$=$	$\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{-13} \\ \color{#9a00c7}{-2} & \color{#007DDC}{12} \end{vmatrix}$ $-\color{#D800AD}{1}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{-13} \\ \color{#9a00c7}{1} & \color{#007DDC}{12} \end{vmatrix}$ $+\color{#D800AD}{6}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{3} \\ \color{#9a00c7}{1} & \color{#007DDC}{-2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{2}(\color{#007DDC}{3\cdot12}-\color{#9a00c7}{(-13)\cdot(-2)})]-[\color{#D800AD}{1}(\color{#007DDC}{0\cdot12}-\color{#9a00c7}{(-13)\cdot1})]+[\color{#D800AD}{6}(\color{#007DDC}{0\cdot(-2)}-\color{#9a00c7}{3\cdot1})]$
	$=$	$[2(36-26)]-[1(0-(-13))]+[6(0-3)]$
	$=$	$2(10)-1(13)+6(-3)$
	$=$	$20-13-18$
$Dz$	$=$	$-11$

Hence, the determinant of the Matrix of Coefficients with the

$z$ column replaced is

$Dz=-11$

Finally, substitute computed values into Cramer’s Rule

$D=-11$

$Dx=-33$

$Dy=44$

$Dz=-11$

$x$	$=$	$(Dx)/D$

	$=$	$(-33)/(-11)$

	$=$	$3$

$y$	$=$	$(Dy)/D$

	$=$	$(44)/(-11)$

	$=$	$-4$

$z$	$=$	$(Dz)/D$

	$=$	$(-11)/-11$

	$=$	$1$

$x=3$

$y=-4$

$z=1$

Question 2 of 3

Solve for

$x,y$ and

$z$

$4x-y+z=-5$

$3y-2z=11$

$x-2y-z=-8$

$x=$ (-2)

$y=$ (1)

$z=$ (4)

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Cramer’s Rule

$x=(Dx)/D,y=(Dy)/D,z=(Dz)/D$

where

$D$ is the determinant of the Matrix of Coefficients

Determinant of a $3xx3$ Matrix

$A=$

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$ then

$|A|=\color{#D800AD}{a}$

$\begin{vmatrix} e & f \\ h & k \end{vmatrix}$

$-\color{#D800AD}{b}$

$\begin{vmatrix} d & f \\ g & k \end{vmatrix}$

$+\color{#D800AD}{c}$

$\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Determinant of a $2xx2$ Matrix

$A=[[a,b],[c,d]]$ then

$|A|=ad-bc$

To solve a system of equations using Cramer’s Rule, first form the Matrices of Coefficients, Variables, and Constants.

First, convert the system of equations into a matrix equation

$4x-y+z$	$=$	$-5$
$3y-2z$	$=$	$11$
$x-2y-z$	$=$	$-8$

Coefficients

$xx$ Variables

$=$ Constants

$\begin{bmatrix} 4 & -1 & 1 \\ 0 & 3 & 2 \\ 1 & -2 & -1 \end{bmatrix}$

$\begin{bmatrix} x \\ y \\ z \end{bmatrix}$

$=$

$\begin{bmatrix} \color{#e65021}{-5} \\ \color{#e65021}{-11} \\ \color{#e65021}{-8} \end{bmatrix}$

Next, solve for

$D$ or the determinant of the Matrix of Coefficients

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{4} & \color{#D800AD}{-1} & \color{#D800AD}{1} \\ 0 & 3 & 2 \\ 1 & -2 & -1 \end{vmatrix}$	$=$	$\color{#D800AD}{4}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{2} \\ \color{#9a00c7}{-2} & \color{#007DDC}{-1} \end{vmatrix}$ $-\color{#D800AD}{-1}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{2} \\ \color{#9a00c7}{1} & \color{#007DDC}{-1} \end{vmatrix}$ $+\color{#D800AD}{1}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{3} \\ \color{#9a00c7}{1} & \color{#007DDC}{-2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{4}(\color{#007DDC}{3\cdot(-1)}-\color{#9a00c7}{2\cdot(-2)})]-[\color{#D800AD}{(-1)}(\color{#007DDC}{0\cdot(-1)}-\color{#9a00c7}{2\cdot1})]+[\color{#D800AD}{1}(\color{#007DDC}{0\cdot(-2)}-\color{#9a00c7}{3\cdot1})]$
	$=$	$[4(-3-(-4))]-[-1(0-2)]+[1(0-3)]$
	$=$	$4(1)+1(-2)+1(-3)$
	$=$	$4-2-3$
$D$	$=$	$-1$

Hence, the determinant of the Matrix of Coefficients is

$D=-1$

Solve for

$Dx$ or the determinant of the Matrix of Coefficients with a replaced

$x$ column

Replace the

$x$ column with the Matrix of Constants

$\begin{bmatrix} 4 & -1 & 1 \\ 0 & 3 & 2 \\ 1 & -2 & -1 \end{bmatrix}$

$→$

$\begin{bmatrix} \color{#e65021}{-5} & -1 & 1 \\ \color{#e65021}{11} & 3 & 2 \\ \color{#e65021}{-8} & -2 & -1 \end{bmatrix}$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{-5} & \color{#D800AD}{-1} & \color{#D800AD}{1} \\ 11 & 3 & 2 \\ -8 & -2 & -1 \end{vmatrix}$	$=$	$\color{#D800AD}{-5}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{2} \\ \color{#9a00c7}{-2} & \color{#007DDC}{-1} \end{vmatrix}$ $-\color{#D800AD}{-1}$ $\begin{vmatrix} \color{#007DDC}{11} & \color{#9a00c7}{2} \\ \color{#9a00c7}{-8} & \color{#007DDC}{-1} \end{vmatrix}$ $+\color{#D800AD}{1}$ $\begin{vmatrix} \color{#007DDC}{11} & \color{#9a00c7}{3} \\ \color{#9a00c7}{-8} & \color{#007DDC}{-2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{-5}(\color{#007DDC}{3\cdot(-1)}-\color{#9a00c7}{2\cdot(-2)})]-[\color{#D800AD}{(-1)}(\color{#007DDC}{11\cdot(-1)}-\color{#9a00c7}{2\cdot(-8)})]+[\color{#D800AD}{1}(\color{#007DDC}{11\cdot(-2)}-\color{#9a00c7}{3\cdot(-8)})]$
	$=$	$[-5(-3-(-4))]-[-1(-11-(-16))]+[1(-22-(-24))]$
	$=$	$-5(1)+1(5)+1(2)$
	$=$	$-5+5+2$
$Dx$	$=$	$2$

Hence, the determinant of the Matrix of Coefficients with the

$x$ column replaced is

$Dx=2$

Solve for

$Dy$ or the determinant of the Matrix of Coefficients with a replaced

$y$ column

Replace the

$y$ column with the Matrix of Constants

$\begin{bmatrix} 4 & -1 & 1 \\ 0 & 3 & 2 \\ 1 & -2 & -1 \end{bmatrix}$

$→$

$\begin{bmatrix} 4 & \color{#e65021}{-5} & 1 \\ 0 & \color{#e65021}{11} & 2 \\ 1 & \color{#e65021}{-8} & -1 \end{bmatrix}$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{4} & \color{#D800AD}{-5} & \color{#D800AD}{1} \\ 0 & 11 & 2 \\ 1 & -8 & -1 \end{vmatrix}$	$=$	$\color{#D800AD}{4}$ $\begin{vmatrix} \color{#007DDC}{11} & \color{#9a00c7}{2} \\ \color{#9a00c7}{-8} & \color{#007DDC}{-1} \end{vmatrix}$ $-\color{#D800AD}{-5}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{2} \\ \color{#9a00c7}{1} & \color{#007DDC}{-1} \end{vmatrix}$ $+\color{#D800AD}{1}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{11} \\ \color{#9a00c7}{1} & \color{#007DDC}{-8} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{4}(\color{#007DDC}{11\cdot(-1)}-\color{#9a00c7}{2\cdot(-8)})]-[\color{#D800AD}{-5}(\color{#007DDC}{0\cdot(-1)}-\color{#9a00c7}{2\cdot1})]+[\color{#D800AD}{1}(\color{#007DDC}{0\cdot(-8)}-\color{#9a00c7}{11\cdot1})]$
	$=$	$[4(-11-(-16))]-[-5(0-2)]+[1(0-11)]$
	$=$	$4(5)+5(-2)+1(-11)$
	$=$	$20-10-11$
$Dy$	$=$	$-1$

Hence, the determinant of the Matrix of Coefficients with the

$y$ column replaced is

$Dy=-1$

Solve for

$Dz$ or the determinant of the Matrix of Coefficients with a replaced

$z$ column

Replace the

$z$ column with the Matrix of Constants

$\begin{bmatrix} 4 & -1 & 1 \\ 0 & 3 & 2 \\ 1 & -2 & -1 \end{bmatrix}$

$→$

$\begin{bmatrix} 4 & -1 & \color{#e65021}{-5} \\ 0 & 3 & \color{#e65021}{11} \\ 1 & -2 & \color{#e65021}{-8} \end{bmatrix}$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{4} & \color{#D800AD}{-1} & \color{#D800AD}{-5} \\ 0 & 3 & 11 \\ 1 & -2 & -8 \end{vmatrix}$	$=$	$\color{#D800AD}{4}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{11} \\ \color{#9a00c7}{-2} & \color{#007DDC}{-8} \end{vmatrix}$ $-\color{#D800AD}{-1}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{11} \\ \color{#9a00c7}{1} & \color{#007DDC}{-8} \end{vmatrix}$ $+\color{#D800AD}{-5}$ $\begin{vmatrix} \color{#007DDC}{0} & \color{#9a00c7}{3} \\ \color{#9a00c7}{1} & \color{#007DDC}{-2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{4}(\color{#007DDC}{3\cdot(-8)}-\color{#9a00c7}{11\cdot(-2)})]-[\color{#D800AD}{(-1)}(\color{#007DDC}{0\cdot(-8)}-\color{#9a00c7}{11\cdot1})]+[\color{#D800AD}{-5}(\color{#007DDC}{0\cdot(-2)}-\color{#9a00c7}{3\cdot1})]$
	$=$	$[4(-24-(-22))]-[-1(0-11)]+[-5(0-3)]$
	$=$	$4(-2)+1(-11)-5(-3)$
	$=$	$-8-11+15$
$Dz$	$=$	$-4$

Hence, the determinant of the Matrix of Coefficients with the

$z$ column replaced is

$Dz=-4$

Finally, substitute computed values into Cramer’s Rule

$D=-1$

$Dx=2$

$Dy=-1$

$Dz=-4$

$x$	$=$	$(Dx)/D$

	$=$	$2/(-1)$

	$=$	$-2$

$y$	$=$	$(Dy)/D$

	$=$	$(-1)/(-1)$

	$=$	$1$

$z$	$=$	$(Dz)/D$

	$=$	$(-4)/(-1)$

	$=$	$4$

$x=-2$

$y=1$

$z=4$

Question 3 of 3

Solve for

$x,y$ and

$z$

$2x-z=5$

$3x+y-2z=12$

$x+2y+z=4$

$x=$ (1)

$y=$ (3)

$z=$ (-3)

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Cramer’s Rule

$x=(Dx)/D,y=(Dy)/D,z=(Dz)/D$

where

$D$ is the determinant of the Matrix of Coefficients

Determinant of a $3xx3$ Matrix

$A=$

$\begin{bmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{bmatrix}$ then

$|A|=\color{#D800AD}{a}$

$\begin{vmatrix} e & f \\ h & k \end{vmatrix}$

$-\color{#D800AD}{b}$

$\begin{vmatrix} d & f \\ g & k \end{vmatrix}$

$+\color{#D800AD}{c}$

$\begin{vmatrix} d & e \\ g & h \end{vmatrix}$

Determinant of a $2xx2$ Matrix

$A=[[a,b],[c,d]]$ then

$|A|=ad-bc$

To solve a system of equations using Cramer’s Rule, first form the Matrices of Coefficients, Variables, and Constants.

First, convert the system of equations into a matrix equation

$2x-z=5$

$3x+y-2z=12$

$x+2y+z=4$

$2x-z$	$=$	$5$
$3x+y-2z$	$=$	$12$
$x+2y+z$	$=$	$4$

Coefficients

$xx$ Variables

$=$ Constants

$\begin{bmatrix} 2 & 0 & -1 \\ 3 & 1 & -2 \\ 1 & 2 & 1 \end{bmatrix}$

$\begin{bmatrix} x \\ y \\ z \end{bmatrix}$

$=$

$\begin{bmatrix} \color{#e65021}{5} \\ \color{#e65021}{12} \\ \color{#e65021}{4} \end{bmatrix}$

Next, solve for

$D$ or the determinant of the Matrix of Coefficients

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{2} & \color{#D800AD}{0} & \color{#D800AD}{-1} \\ 3 & 1 & -2 \\ 1 & 2 & 1 \end{vmatrix}$	$=$	$\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{1} & \color{#9a00c7}{-2} \\ \color{#9a00c7}{2} & \color{#007DDC}{1} \end{vmatrix}$ $-\color{#D800AD}{0}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{-2} \\ \color{#9a00c7}{1} & \color{#007DDC}{1} \end{vmatrix}$ $+\color{#D800AD}{-1}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{1} \\ \color{#9a00c7}{1} & \color{#007DDC}{2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{2}(\color{#007DDC}{1\cdot1}-\color{#9a00c7}{(-2)\cdot2})]-[\color{#D800AD}{0}(\color{#007DDC}{3\cdot1}-\color{#9a00c7}{(-2)\cdot1})]+[\color{#D800AD}{(-1)}(\color{#007DDC}{3\cdot2}-\color{#9a00c7}{1\cdot1})]$
	$=$	$[2(1-(-4))]-[0(3-(-2))]+[-1(6-1)]$
	$=$	$2(5)-0-1(5)$
	$=$	$10-5$
$D$	$=$	$5$

Hence, the determinant of the Matrix of Coefficients is

$D=5$

Solve for

$Dx$ or the determinant of the Matrix of Coefficients with a replaced

$x$ column

Replace the

$x$ column with the Matrix of Constants

$\begin{bmatrix} 2 & 0 & -1 \\ 3 & 1 & -2 \\ 1 & 2 & 1 \end{bmatrix}$

$→$

$\begin{bmatrix} \color{#e65021}{5} & 0 & -1 \\ \color{#e65021}{12} & 1 & -2 \\ \color{#e65021}{4} & 2 & 1 \end{bmatrix}$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{5} & \color{#D800AD}{0} & \color{#D800AD}{-1} \\ 12 & 1 & -2 \\ 4 & 2 & 1 \end{vmatrix}$	$=$	$\color{#D800AD}{5}$ $\begin{vmatrix} \color{#007DDC}{1} & \color{#9a00c7}{-2} \\ \color{#9a00c7}{2} & \color{#007DDC}{1} \end{vmatrix}$ $-\color{#D800AD}{0}$ $\begin{vmatrix} \color{#007DDC}{12} & \color{#9a00c7}{-2} \\ \color{#9a00c7}{4} & \color{#007DDC}{1} \end{vmatrix}$ $+\color{#D800AD}{-1}$ $\begin{vmatrix} \color{#007DDC}{12} & \color{#9a00c7}{1} \\ \color{#9a00c7}{4} & \color{#007DDC}{2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{5}(\color{#007DDC}{1\cdot1}-\color{#9a00c7}{-2\cdot2})]-[\color{#D800AD}{0}(\color{#007DDC}{12\cdot1}-\color{#9a00c7}{-2\cdot4})]+[\color{#D800AD}{(-1)}(\color{#007DDC}{12\cdot2}-\color{#9a00c7}{1\cdot4})]$
	$=$	$[5(1-(-4))]-[0(12-(-8))]+[-1(24-4)]$
	$=$	$5(5)-0-1(20)$
	$=$	$25-20$
$Dx$	$=$	$5$

Hence, the determinant of the Matrix of Coefficients with the

$x$ column replaced is

$Dx=5$

Solve for

$Dy$ or the determinant of the Matrix of Coefficients with a replaced

$y$ column

Replace the

$y$ column with the Matrix of Constants

$\begin{bmatrix} 2 & 0 & -1 \\ 3 & 1 & -2 \\ 1 & 2 & 1 \end{bmatrix}$

$→$

$\begin{bmatrix} 2 & \color{#e65021}{5} & -1 \\ 3 & \color{#e65021}{12} & -2 \\ 1 & \color{#e65021}{4} & 1 \end{bmatrix}$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{2} & \color{#D800AD}{5} & \color{#D800AD}{-1} \\ 3 & 12 & -2 \\ 1 & 4 & 1 \end{vmatrix}$	$=$	$\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{12} & \color{#9a00c7}{-2} \\ \color{#9a00c7}{4} & \color{#007DDC}{1} \end{vmatrix}$ $-\color{#D800AD}{5}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{-2} \\ \color{#9a00c7}{1} & \color{#007DDC}{1} \end{vmatrix}$ $+\color{#D800AD}{-1}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{12} \\ \color{#9a00c7}{1} & \color{#007DDC}{4} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{2}(\color{#007DDC}{12\cdot1}-\color{#9a00c7}{-2\cdot4})]-[\color{#D800AD}{5}(\color{#007DDC}{3\cdot1}-\color{#9a00c7}{-2\cdot1})]+[\color{#D800AD}{(-1)}(\color{#007DDC}{3\cdot4}-\color{#9a00c7}{12\cdot1})]$
	$=$	$[2(12-(-8))]-[5(3-(-2))]+[-1(12-12)]$
	$=$	$2(20)-5(5)-1(0)$
	$=$	$40-25-0$
$Dy$	$=$	$15$

Hence, the determinant of the Matrix of Coefficients with the

$y$ column replaced is

$Dy=15$

Solve for

$Dz$ or the determinant of the Matrix of Coefficients with a replaced

$z$ column

Replace the

$z$ column with the Matrix of Constants

$\begin{bmatrix} 2 & 0 & -1 \\ 3 & 1 & -2 \\ 1 & 2 & 1 \end{bmatrix}$

$→$

$\begin{bmatrix} 2 & 0 & \color{#e65021}{5} \\ 3 & 1 & \color{#e65021}{12} \\ 1 & 2 & \color{#e65021}{4} \end{bmatrix}$

Substitute the values into the Determinant Formula and solve each term

$\begin{vmatrix} \color{#D800AD}{a} & \color{#D800AD}{b} & \color{#D800AD}{c} \\ d & e & f \\ g & h & k \end{vmatrix}$	$=$	$\color{#D800AD}{a}$ $\begin{vmatrix} e & f \\ h & k \end{vmatrix}$ $-\color{#D800AD}{b}$ $\begin{vmatrix} d & f \\ g & k \end{vmatrix}$ $+\color{#D800AD}{c}$ $\begin{vmatrix} d & e \\ g & h \end{vmatrix}$	Determinant of a $3xx3$ Matrix

$\begin{vmatrix} \color{#D800AD}{2} & \color{#D800AD}{0} & \color{#D800AD}{5} \\ 3 & 1 & 12 \\ 1 & 2 & 4 \end{vmatrix}$	$=$	$\color{#D800AD}{2}$ $\begin{vmatrix} \color{#007DDC}{1} & \color{#9a00c7}{12} \\ \color{#9a00c7}{2} & \color{#007DDC}{4} \end{vmatrix}$ $-\color{#D800AD}{0}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{12} \\ \color{#9a00c7}{1} & \color{#007DDC}{4} \end{vmatrix}$ $+\color{#D800AD}{5}$ $\begin{vmatrix} \color{#007DDC}{3} & \color{#9a00c7}{1} \\ \color{#9a00c7}{1} & \color{#007DDC}{2} \end{vmatrix}$	Substitute values

	$=$	$[\color{#D800AD}{2}(\color{#007DDC}{1\cdot4}-\color{#9a00c7}{12\cdot2})]-[\color{#D800AD}{0}(\color{#007DDC}{3\cdot4}-\color{#9a00c7}{12\cdot1})]+[\color{#D800AD}{5}(\color{#007DDC}{3\cdot2}-\color{#9a00c7}{1\cdot1})]$
	$=$	$[2(4-24)]-[0(12-12)]+[5(6-1)]$
	$=$	$2(-20)-0+5(5)$
	$=$	$-40+25$
$Dz$	$=$	$-15$

Hence, the determinant of the Matrix of Coefficients with the

$z$ column replaced is

$Dz=-15$

Finally, substitute computed values into Cramer’s Rule

$D=5$

$Dx=5$

$Dy=15$

$Dz=-15$

$x$	$=$	$(Dx)/D$

	$=$	$5/5$

	$=$	$1$

$y$	$=$	$(Dy)/D$

	$=$	$(15)/5$

	$=$	$3$

$z$	$=$	$(Dz)/D$

	$=$	$(-15)/5$

	$=$	$-3$

$x=1$

$y=3$

$z=-3$

Cramer’s Rule

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Quiz summary

Information

1. Question

Hint

Keep Going!

Cramer’s Rule

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

2. Question

Hint

Fantastic!

Cramer’s Rule

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

3. Question

Hint

Excellent!

Cramer’s Rule

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

Quizzes

Cramer’s Rule

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Quiz summary

Information

1. Question

Hint

Keep Going!

Cramer’s Rule

Determinant of a 3×33xx3 Matrix

Determinant of a 2×22xx2 Matrix

2. Question

Hint

Fantastic!

Cramer’s Rule

Determinant of a 3×33xx3 Matrix

Determinant of a 2×22xx2 Matrix

3. Question

Hint

Excellent!

Cramer’s Rule

Determinant of a 3×33xx3 Matrix

Determinant of a 2×22xx2 Matrix

Quizzes

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix

Determinant of a $3xx3$ Matrix

Determinant of a $2xx2$ Matrix