Cramer’s Rule

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Precalculus

Matrices

Cramer's Rule

Cramer’s Rule

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Question 1 of 3

1. Question

Solve for `x,y` and `z`

`2x+y+4z=6`

`3y-z=-13`

`x-2y+z=12`

`x=` (3)

`y=` (-4)

`z=` (1)

Hint

Help Video

Correct

Keep Going!

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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

If $$A=$$\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

If `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

`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

2. Question

Solve for `x,y` and `z`

`4x-y+z=-5`

`3y-2z=11`

`x-2y-z=-8`

`x=` (-2)

`y=` (1)

`z=` (4)

Hint

Help Video

Correct

Fantastic!

Incorrect

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

Determinant of a `2xx2` Matrix

If `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

3. Question

Solve for `x,y` and `z`

`2x-z=5`

`3x+y-2z=12`

`x+2y+z=4`

`x=` (1)

`y=` (3)

`z=` (-3)

Hint

Help Video

Correct

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Incorrect

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

Determinant of a `2xx2` Matrix

If `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`

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