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Question 1 of 4
Incorrect
Row-Echelon Form
[1xx01x]
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
Use row operations for the matrix to achieve Row-Echelon Form
Start with simplifying the two rows
[6−3|−9−4−2|26] R1R2
R1÷3→R1
[2−1|−3−4−2|26] R1R2
R2÷2→R2
New matrix:
[2−1|−3−2−1|13] R1R2
Proceed with transforming R2
Transforming the First Element of R2:
[2−1|−3−2−1|13] R1R2
R1+R2→R2
R1:2−1|−3R2:−2−1|13R2→0−2|10
Transforming the Second Element of R2:
[2−1|−30−2|10] R1R2
R2÷−2→R2
New matrix:
[2−1|−301|−5] R1R2
Transforming the First Element of R1:
[2−1|−301|−5] R1R2
R1÷2→R1
New matrix:
[1−12|−3201|−5] R1R2
This new matrix is in Row-Echelon Form
Remember that each row in this matrix corresponds to a term in an equation
Convert R2 into an equation
This time, convert R1 into an equation then substitute the value of y to find x
x-12y |
= |
-32 |
|
x-12(-5) |
= |
-32 |
Substitute y=-5 |
|
x+52 |
= |
-32 |
|
x+52 -52 |
= |
-32 -52 |
Subtract 52 from both sides |
|
x |
= |
-82 |
|
x |
= |
-4 |
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Question 2 of 4
Solve for x,y and z
5x+3z=-3
y+4z=15
2x-4y-z=-6
Incorrect
Row-Echelon Form
[1xxx01xx001x]
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
5x+3z=-3
y+4z=15
2x-4y-z=-6
[503|−3014|152−4−1|−6] R1R2R3
Use row operations for the matrix to achieve Row-Echelon Form
Start with transforming R3
Transforming the First Element of R3:
[503|−3014|152−4−1|−6] R1R2R3
2R1−5R3→R3
2R1:1006|−6−5R3:−10205|30R3→02011|24
Transforming the Second Element of R3:
[503|−3014|1502011|24] R1R2R3
20R2−R3→R3
20R2:02080|300−R3:0−20−11|−24R3→0069|276
Transforming the Third Element of R3:
[503|−3014|150069|276] R1R2R3
R3÷69→R3
New matrix:
[503|−3014|15001|4] R1R2R3
Notice that R2 is already in Row-Echelon Form. Hence, continue with transforming R1
Transforming the First Element of R1:
[503|−3014|15001|4] R1R2R3
R1÷5→R1
New matrix:
[1035|−35014|15001|4] R1R2R3
This new matrix is in Row-Echelon Form
Remember that each row in this matrix corresponds to a term in an equation
Convert R3 into an equation
Next, convert R2 into an equation then substitute the value of z to find y
y+4z |
= |
15 |
y+4(4) |
= |
15 |
Substitute z=4 |
y+16 |
= |
15 |
y+16 -16 |
= |
15 -16 |
y |
= |
-1 |
Lastly, convert R1 into an equation then substitute the value of z to find x
x+35z |
= |
-35 |
|
x+35(4) |
= |
-35 |
Substitute z=4 |
|
x+125 |
= |
-35 |
|
x+125 -125 |
= |
-35 -125 |
|
x |
= |
-155 |
|
x |
= |
-3 |
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Question 3 of 4
Solve for x,y and z
4x+2y+z=-3
x-3y=-11
2x+y-2z=1
Incorrect
Row-Echelon Form
[1xxx01xx001x]
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
4x+2y+z=-3
x-3y=-11
2x+y-2z=1
[421|−31−30|−1121−2|1] R1R2R3
Use row operations for the matrix to achieve Row-Echelon Form
Start with swapping R1 and R2, since the first element of R2 is already 1
[1−30|−11421|−321−2|1] R1R2R3
Proceed with transforming the first column
Transforming the First Element of R3:
[1−30|−11421|−321−2|1] R1R2R3
−2R1+R3→R3
−2R1:−260|22R3:21−2|1R3→07−2|23
Transforming the First Element of R2:
[1−30|−11421|−307−2|23] R1R2R3
−4R1+R2→R2
−4R1:−4120|44R2:421|−3R3→0141|41
New matrix:
[1−30|−110141|4107−2|23] R1R2R3
Continue with transforming R3
Transforming the Second Element of R3:
[1−30|−110141|4107−2|23] R1R2R3
R2−2R3→R3
R2:0141|41−2R3:0−144|−46R3→005|−5
Transforming the Third Element of R3:
[1−30|−110141|41005|−5] R1R2R3
R3÷5→R3
New matrix:
[1−30|−110141|41001|−1] R1R2R3
Transforming the Second Element of R2:
[1−30|−110141|41001|−1] R1R2R3
R2÷14→R2
New matrix:
[1−30|−1101114|4114001|−1] R1R2R3
This new matrix is in Row-Echelon Form
Remember that each row in this matrix corresponds to a term in an equation
Convert R3 into an equation
Next, convert R2 into an equation then substitute the value of z to find y
y+114z |
= |
4114 |
|
y+114(-1) |
= |
4114 |
Substitute z=-1 |
|
y-114 |
= |
4114 |
|
y-114 +114 |
= |
4114 +114 |
Add 114 to both sides |
|
y |
= |
4214 |
|
y |
= |
3 |
Lastly, convert R1 into an equation then substitute the value of y to find x
x-3y |
= |
-11 |
x-3(3) |
= |
-11 |
Substitute y=3 |
x-9 |
= |
-11 |
x-9 +9 |
= |
-11 +9 |
Add 9 to both sides |
x |
= |
-2 |
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Question 4 of 4
Solve for x,y and z
3x-2y=16
2x+y+5z=1
x+4y-z=-3
Incorrect
Row-Echelon Form
[1xxx01xx001x]
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-2y=16
2x+y+5z=1
x+4y-z=-3
[3−20|16215|114−1|−3] R1R2R3
Use row operations for the matrix to achieve Row-Echelon Form
Start with transforming R2
Transforming the First Element of R2:
[3−20|16215|114−1|−3] R1R2R3
−2R3+R2→R2
−2R3:−2−82|6R2:215|1R2→0−77|7
Transforming the Second Element of R2:
[3−20|160−77|714−1|−3] R1R2R3
R2÷−7→R2
New matrix:
[3−20|1601−1|−114−1|−3] R1R2R3
Next, swap R1 and R3, since the first element of R3 is already 1
[14−1|−301−1|−13−20|16] R1R2R3
Continue with transforming R3
Transforming the First Element of R3:
[14−1|−301−1|−13−20|16] R1R2R3
−3R1+R3→R3
−3R1:−3−123|9R3:3−20|16R3→0−143|25
Transforming the Second Element of R3:
[14−1|−301−1|−10−143|25] R1R2R3
14R2+R3→R3
14R2:014−14|−14R3:0−143|25R3→00−11|11
Transforming the Third Element of R3:
[14−1|−301−1|−100−11|11] R1R2R3
R3÷−11→R3
New matrix:
[14−1|−301−1|−1001|−1] R1R2R3
This new matrix is in Row-Echelon Form
Remember that each row in this matrix corresponds to a term in an equation
Convert R3 into an equation
Next, convert R2 into an equation then substitute the value of z to find y
y-z |
= |
-1 |
y-(-1) |
= |
-1 |
Substitute z=-1 |
y+1 |
= |
-1 |
y+1 -1 |
= |
-1 -1 |
Subtract 1 from both sides |
y |
= |
-2 |
Lastly, convert R1 into an equation then substitute the values of y and z to find x
x+4y-z |
= |
-3 |
x+4(-2)-(-1) |
= |
-3 |
Substitute y=-2 and z=-1 |
x-8+1 |
= |
-3 |
x-7 |
= |
-3 |
x-7 +7 |
= |
-3 +7 |
Add 7 to both sides |
x |
= |
4 |