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 VideoCorrect
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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 CoefficientsDeterminant 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 CoefficientsLabel 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` columnReplace 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` columnReplace 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` columnReplace 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` -
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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 VideoCorrect
Fantastic!
Incorrect
Cramer’s Rule
`x=(Dx)/D,y=(Dy)/D,z=(Dz)/D`where `D` is the determinant of the Matrix of CoefficientsDeterminant 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`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 CoefficientsSubstitute 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` columnReplace 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` columnReplace 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` columnReplace 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` -
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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
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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 CoefficientsDeterminant 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-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 CoefficientsSubstitute 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` columnReplace 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` columnReplace 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` columnReplace 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
- 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