Cramer’s Rule
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Question 1 of 3
1. Question
Solve for x,yx,y and zz2x+y+4z=62x+y+4z=63y-z=-133y−z=−13x-2y+z=12x−2y+z=12-
x=x= (3)y=y= (-4)z=z= (1)
Hint
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Cramer’s Rule
x=DxD,y=DyD,z=DzDx=DxD,y=DyD,z=DzDwhere DD is the determinant of the Matrix of CoefficientsDeterminant of a 3×33×3 Matrix
If A=A=[abcdefghk] then |A|=a|efhk|−b|dfgk|+c|degh|Determinant of a 2×2 Matrix
If A=[abcd] then |A|=ad-bcTo 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 equation2x+y+4z = 6 3y-z = -13 x-2y+z = 12 Coefficients × Variables = Constants[21403−11−21] [xyz]=[6−1312]Next, solve for D or the determinant of the Matrix of CoefficientsLabel the values of the matrix[abcdefghk]=[21403−11−21]a=2d=0g=1b=1e=3h=-2c=4f=-1k=1Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |21403−11−21| = 2|3−1−21|−1|0−111|+4|031−2| Substitute values = [2(3⋅1−(−1)⋅(−2))]−[1(0⋅1−(−1)⋅1)]+[4(0⋅(−2)−3⋅1)] = [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=-11Solve for Dx or the determinant of the Matrix of Coefficients with a replaced x columnReplace the x column with the Matrix of Constants[21403−11−21]→[614−133−112−21]Label the values of the matrix[abcdefghk]=[614−133−112−21]a=6d=-13g=12b=1e=3h=-2c=4f=-1k=1Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |614−133−112−21| = 6|3−1−21|−1|−13−1121|+4|−13312−2| Substitute values = [6(3⋅1−(−1)⋅(−2))]−[1(−13⋅1−(−1)⋅12)]+[4(−13⋅(−2)−3⋅12)] = [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=-33Solve for Dy or the determinant of the Matrix of Coefficients with a replaced y columnReplace the y column with the Matrix of Constants[21403−11−21]→[2640−13−11121]Label the values of the matrix[abcdefghk]=[2640−13−11121]a=2d=0g=1b=6e=-13h=12c=4f=-1k=1Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |2640−13−11121| = 2|−13−1121|−6|0−111|+4|0−13112| Substitute values = [2(−13⋅1−(−1)⋅12)]−[6(0⋅1−(−1)⋅1)]+[4(0⋅12−(−13)⋅1)] = [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=44Solve for Dz or the determinant of the Matrix of Coefficients with a replaced z columnReplace the z column with the Matrix of Constants[21403−11−21]→[21603−131−212]Label the values of the matrix[abcdefghk]=[21603−131−212]a=2d=0g=1b=1e=3h=-2c=6f=-13k=12Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |21603−131−212| = 2|3−13−212|−1|0−13112|+6|031−2| Substitute values = [2(3⋅12−(−13)⋅(−2))]−[1(0⋅12−(−13)⋅1)]+[6(0⋅(−2)−3⋅1)] = [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=-11Finally, substitute computed values into Cramer’s RuleD=-11Dx=-33Dy=44Dz=-11x = DxD = -33-11 = 3 y = DyD = 44-11 = -4 z = DzD = -11-11 = 1 x=3y=-4z=1 -
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Question 2 of 3
2. Question
Solve for x,y and z4x-y+z=-53y-2z=11x-2y-z=-8-
x= (-2)y= (1)z= (4)
Hint
Help VideoCorrect
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Incorrect
Cramer’s Rule
x=DxD,y=DyD,z=DzDwhere D is the determinant of the Matrix of CoefficientsDeterminant of a 3×3 Matrix
If A=[abcdefghk] then |A|=a|efhk|−b|dfgk|+c|degh|Determinant of a 2×2 Matrix
If A=[abcd] then |A|=ad-bcTo 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 equation4x-y+z = -5 3y-2z = 11 x-2y-z = -8 Coefficients × Variables = Constants[4−110321−2−1] [xyz]=[−5−11−8]Next, solve for D or the determinant of the Matrix of CoefficientsSubstitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |4−110321−2−1| = 4|32−2−1|−−1|021−1|+1|031−2| Substitute values = [4(3⋅(−1)−2⋅(−2))]−[(−1)(0⋅(−1)−2⋅1)]+[1(0⋅(−2)−3⋅1)] = [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=-1Solve for Dx or the determinant of the Matrix of Coefficients with a replaced x columnReplace the x column with the Matrix of Constants[4−110321−2−1]→[−5−111132−8−2−1]Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |−5−111132−8−2−1| = −5|32−2−1|−−1|112−8−1|+1|113−8−2| Substitute values = [−5(3⋅(−1)−2⋅(−2))]−[(−1)(11⋅(−1)−2⋅(−8))]+[1(11⋅(−2)−3⋅(−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=2Solve for Dy or the determinant of the Matrix of Coefficients with a replaced y columnReplace the y column with the Matrix of Constants[4−110321−2−1]→[4−5101121−8−1]Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |4−5101121−8−1| = 4|112−8−1|−−5|021−1|+1|0111−8| Substitute values = [4(11⋅(−1)−2⋅(−8))]−[−5(0⋅(−1)−2⋅1)]+[1(0⋅(−8)−11⋅1)] = [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=-1Solve for Dz or the determinant of the Matrix of Coefficients with a replaced z columnReplace the z column with the Matrix of Constants[4−110321−2−1]→[4−1−503111−2−8]Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |4−1−503111−2−8| = 4|311−2−8|−−1|0111−8|+−5|031−2| Substitute values = [4(3⋅(−8)−11⋅(−2))]−[(−1)(0⋅(−8)−11⋅1)]+[−5(0⋅(−2)−3⋅1)] = [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=-4Finally, substitute computed values into Cramer’s RuleD=-1Dx=2Dy=-1Dz=-4x = DxD = 2-1 = -2 y = DyD = -1-1 = 1 z = DzD = -4-1 = 4 x=-2y=1z=4 -
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Question 3 of 3
3. Question
Solve for x,y and z2x-z=53x+y-2z=12x+2y+z=4-
x= (1)y= (3)z= (-3)
Hint
Help VideoCorrect
Excellent!
Incorrect
Cramer’s Rule
x=DxD,y=DyD,z=DzDwhere D is the determinant of the Matrix of CoefficientsDeterminant of a 3×3 Matrix
If A=[abcdefghk] then |A|=a|efhk|−b|dfgk|+c|degh|Determinant of a 2×2 Matrix
If A=[abcd] then |A|=ad-bcTo 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 equation2x-z=53x+y-2z=12x+2y+z=42x-z = 5 3x+y-2z = 12 x+2y+z = 4 Coefficients × Variables = Constants[20−131−2121] [xyz]=[5124]Next, solve for D or the determinant of the Matrix of CoefficientsSubstitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |20−131−2121| = 2|1−221|−0|3−211|+−1|3112| Substitute values = [2(1⋅1−(−2)⋅2)]−[0(3⋅1−(−2)⋅1)]+[(−1)(3⋅2−1⋅1)] = [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=5Solve for Dx or the determinant of the Matrix of Coefficients with a replaced x columnReplace the x column with the Matrix of Constants[20−131−2121]→[50−1121−2421]Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |50−1121−2421| = 5|1−221|−0|12−241|+−1|12142| Substitute values = [5(1⋅1−−2⋅2)]−[0(12⋅1−−2⋅4)]+[(−1)(12⋅2−1⋅4)] = [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=5Solve for Dy or the determinant of the Matrix of Coefficients with a replaced y columnReplace the y column with the Matrix of Constants[20−131−2121]→[25−1312−2141]Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |25−1312−2141| = 2|12−241|−5|3−211|+−1|31214| Substitute values = [2(12⋅1−−2⋅4)]−[5(3⋅1−−2⋅1)]+[(−1)(3⋅4−12⋅1)] = [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=15Solve for Dz or the determinant of the Matrix of Coefficients with a replaced z columnReplace the z column with the Matrix of Constants[20−131−2121]→[2053112124]Substitute the values into the Determinant Formula and solve each term|abcdefghk| = a|efhk|−b|dfgk|+c|degh| Determinant of a 3×3 Matrix |2053112124| = 2|11224|−0|31214|+5|3112| Substitute values = [2(1⋅4−12⋅2)]−[0(3⋅4−12⋅1)]+[5(3⋅2−1⋅1)] = [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=-15Finally, substitute computed values into Cramer’s RuleD=5Dx=5Dy=15Dz=-15x = DxD = 55 = 1 y = DyD = 155 = 3 z = DzD = -155 = -3 x=1y=3z=-3 -
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