Information
You have already completed the quiz before. Hence you can not start it again.
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
-
Question 1 of 4
Solve for xx and yy
x-3y=19x−3y=19
5x+2y=105x+2y=10
Incorrect
Inverse of a 2×22×2 Matrix
If [abcd] then A−1=1|A|[d−b−ca]
Determinant of a 2×2 Matrix
If [abcd] then |A|=ad−bc
First, convert the system of equations into a matrix equation
Note that A is the matrix of coefficients, X is the matrix of variables, and C is the matrix of constants
Remember that we are looking for the values of x and y, which is contained in the matrix X
Start finding A-1 by solving for the determinant
|A| |
= |
ad -bc |
Determinant Formula |
|
|1−352| |
= |
1⋅2 -(-3)⋅5 |
Substitute values |
|
|
= |
2-(-15) |
|
= |
17 |
Substitute respective values into the Inverse Formula
A-1 |
= |
1|A|[d−b−ca] |
Inverse Formula |
|
|
= |
117[2−(−3)−51] |
Substitute values |
|
A-1 |
= |
117[23−51] |
Finally, substitute the matrices into the X formula
X |
= |
A-1C |
|
|
= |
117[23−51][1910] |
Substitute values |
|
|
= |
117[(2⋅19)+(3⋅10)(−5⋅19)+(1⋅10)] |
|
|
= |
117[38+30−95+10] |
|
|
= |
117[68−85] |
|
X |
= |
[4−5] |
|
[xy] |
= |
[4−5] |
-
Question 2 of 4
Incorrect
Inverse of a 2×2 Matrix
If [abcd] then A−1=1|A|[d−b−ca]
Determinant of a 2×2 Matrix
If [abcd] then |A|=ad−bc
First, convert the system of equations into a matrix equation
Note that A is the matrix of coefficients, X is the matrix of variables, and C is the matrix of constants
Remember that we are looking for the values of x and y, which is contained in the matrix X
Start finding A-1 by solving for the determinant
|A| |
= |
ad -bc |
Determinant Formula |
|
|2−315| |
= |
2⋅5 -(-3)⋅1 |
Substitute values |
|
|
= |
10-(-3) |
|
= |
13 |
Substitute respective values into the Inverse Formula
A-1 |
= |
1|A|[d−b−ca] |
Inverse Formula |
|
|
= |
113[5−(−3)−12] |
Substitute values |
|
A-1 |
= |
113[53−12] |
Finally, substitute the matrices into the X formula
X |
= |
A-1C |
|
|
= |
113[53−12][9−2] |
Substitute values |
|
|
= |
113[(5⋅9)+(3⋅−2)(−1⋅9)+(2⋅−2)] |
|
|
= |
113[45+(−6)−9+(−4)] |
|
|
= |
113[39−13] |
|
X |
= |
[3−1] |
|
[xy] |
= |
[3−1] |
-
Question 3 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
Notice that R1 is already in Row-Echelon Form. Hence, we only have to transform R2
Transforming the First Element of R2:
[1−5|134−6|24] R1R2
−4R1+R2→R2
−4R1:−420|−52+R2:4−6|24R2→014|−28
Transforming the Second Element of R2:
[1−5|13014|−28] R1R2
R2÷14→R2
New matrix:
[1−5|1301|−2] 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-5y |
= |
13 |
x-5(-2) |
= |
13 |
Substitute y=-2 |
x+10 |
= |
13 |
x+10 -10 |
= |
13 -10 |
x |
= |
3 |
-
Question 4 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 transforming R2
Transforming the First Element of R2:
[35|147−2|−22] R1R2
7R1−3R2→R2
7R1:2135|98−3R2:−216|66R2→041|164
Transforming the Second Element of R2:
[35|14041|164] R1R2
R2÷41→R2
New matrix:
[35|1401|4] R1R2
Proceed with transforming R1
Transforming the First Element of R1:
New matrix:
[153|14301|−2] 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+53y |
= |
143 |
|
x+53(4) |
= |
143 |
Substitute y=4 |
|
x+203 |
= |
143 |
|
x+203 -203 |
= |
143 -203 |
Subtract 203 from both sides |
|
x |
= |
-63 |
|
x |
= |
-2 |