Linear Algebra

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January 31, 2019 Name:__________________ MATH 225: Linear Algebra Recap of Section 2.1 – Determinants by Cofactor Expansion ______________________________________________________________________________ ______________________________________________________________________________ Instructions: Please complete the following exercises and be prepared to discuss in class on February 5, 2019. You may work in groups and refer to your lecture notes or textbook. Show your work for full credit. Exercises: 1) Evaluate the determinant of the 3x3 matrix below by cofactor expansion along: a. The first row b. The first column c. The second row d. The second column e. The third row f. The third column g. The pneumonic shortcut (the arrow technique for evaluating 3x3 determinants) 3 2 1 0 –1 9 0 5 –4 2) Evaluate the determinant of the 4x4 matrix below by cofactor expansion along: a. The first row b. The first column c. The second row d. The second column e. The third row f. The third column g. The fourth row h. The fourth column 3 2 4 2 3 2 1 10 0 0 –3 3 5 –2 0 2 ...
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Bilal_Mursaleen
School: University of Virginia

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Determinants
Q1: - Evaluate the determinant of the 3x3 matrix below by cofactor expansion
along:
a. The first row
b. The first column
c. The second row
d. The second column
e. The third row
f. The third column
g. The pneumonic shortcut (the arrow technique for evaluating 3x3
determinants)
3
[2
1

0
0
−1 5 ]
9 −4

Q2: - Evaluate the determinant of the 4x4 matrix below by cofactor expansion
along:
a. The first row
b. The first column
c. The second row
d. The second column
e. The third row
f. The third column
g. The fourth row
h. The fourth column
3
2
[
4
2

3
0
5
2
0 −2
]
1 −3 0
10 3
2

Solution
Q1: Sol: - Given matrix is
3
[2
1

0
0
−1 5 ]
9 −4

a. Determinant using “The first row”
Sol: 3 0
0
|2 −1 5 |
1 9 −4
−1
= 3(−1)1+1 |
9
= 3(−1)2 |

2 −1
5
2 5
| + (0)(−1)1+2 |
| + (0)(−1)1+3 |
|
1 9
−4
1 −4

−1 5
|+0+0
9 −4

−1 5
= 3|
|
9 −4
= 3[(−1)(−4) − (9)(5)]
= 3[4 − 45]
= 3[−41]
= −123
b. Determinant using “The first column”
Sol: 3 0
0
|2 −1 5 |
1 9 −4
−1
= 3(−1)1+1 |
9
= 3(−1)2 |

0 0
0
5
| + 2(−1)2+1 |
| + 1(−1)3+1 |
9 −4
−1
−4

0
−1 5
| + 2(−1)3 |
9
9 −4

0
0 0
| + 1(−1)4 |
|
−4
−1 5

0
|
5

0 0
0 0
−1 5
= 3|
| − 2|
| + 1|
|
9 −4
−1 5
9 −4
= 3[(−1)(−4) − (9)(5)] − 2[(0)(−4) − (9)(0)] + 1[(0)(5) − (−1)(0)]
= 3[4 − 45] − 2[0 − 0] + 1[0 − 0]
= 3[−41] − 2[0] + 1[0]
= −123 − 0 + 0
= −123
c. Determinant using “The second row”
Sol: 3 0
0
|2 −1 5 |
1 9 −4
0 0
3
3 0
= 2(−1)2+1 |
| + (−1)(−1)2+2 |
| + 5(−1)2+3 |
9 −4
1 −4
1
0 0
3 0
3 0
= 2(−1)3 |
| + (−1)(−1)4 |
| + 5(−1)5 |
|
9 −4
1 −4
1 9
0 0
3 0
3 0
= 2(−1) |
| + (−1)(1) |
| + 5(−1) |
|
9 −4
1 −4
1 9
0 0
3 0
3 0
= −2 |
|−|
|− 5|
|
9 −4
1 −4
1 9

0
|
9

= −2[(0)(−4) − (9)(0)] − [(3)(−4) − (1)(0)] − 5[(3)(9) − (1)(0)]
= −2[0 − 0] − [−12 − 0] − 5[27 − 0]
= −2[0] − [−12] − 5[27]
= 0 + 12 − 135
= −123
d. Determinant using “The second column”
Sol: 3 0
0
|2 −1 5 |
1 9 −4

3 0
3 0
2 5
= (0)(−1)1+2 |
| + (−1)(−1)2+2 |
| + 9(−1)3+2 |
|
1 −4
2 5
1 −4
3 0
3 0
2 5
= (0)(−1)3 |
| + (−1)(−1)4 |
| + 9(−1)5 |
|
1 −4
2 5
1 −4
3
3 0
5
| + (−1)(1) |
| + 9(−1) |
1 −4
2
−4
3 0
3 0
5
|−|
| − 9|
|
1 −4
2 5
−4
3 0
0
| − 9|
|
−4
2 5

2
= (0)(−1) |
1
2
1
3
=0−|
1
= (0) |

0
|
5

= 0 − [(3)(−4) − (1)(0)] − 9[(3)(5) − (2)(0)]
= 0 − [−12 − 0] − 9[15 − 0]
= 0 − [−12] − 9[15]
= 0 + 12 − 135
= −123
e. Determinant using “The third row”
Sol: 3 0
0
|2 −1 5 |
1 9 −4
0 0
3 0
3
| + 9(−1)3+2 |
| + (−4)(−1)3+3 |
−1 5
2 5
2
0 0
3 0
3 0
= 1(−1)4 |
| + 9(−1)5 |
| + (−4)(−1)6 |
|
−1 5
2 5
2 −1
0 0
3 0
3 0
= 1(1) |
| + 9(−1) |
| + (−4)(1) |
|
−1 5
2 5
2 −1
0 0
3 0
3 0
= 1|
|−9|
| − 4|
|
−1 5
2 5
2 −1
= 1(−1)3+1 |

0
|
−1

= 1[(0)(5) − (−1)(0)] − 9[(3)(5) − (2)(0)] − 4[(3)(−1) − (2)(0)]
= 1[0 − 0] − 9[15 − 0] − 4[−3 − 0]
= 1[0] − 9[15] − 4[−3]

= 0 − 135 + 12
= −123
f. Determinant using “The third column”
Sol: 3 0
0
|2 −1 5 |
1 9 −4
2 −1
3 0
3 0
= (0)(−1)1+3 |
| + 5(−1)2+3 |
| + (−4)(−1)3+3 |
|
1 9
1 9
2 −1
2 −1
3 0
3 0
= (0)(−1)4 |
| + 5(−1)5 |
| + (−4)(−1)6 |
|
1 9
1 9
2 −1
3 0
3 0
= 0 + 5(−1) |
| + (−4)(1) |
|
1 9
2 −1
3 0
3 0
= 0 − 5|
| − 4|
|
1 9
2 −1
= 0 − 5[(3)(9) − (1)(0)] − 4[(3)(−1) − (2)(0)]
= 0 − 5[27 − 0] − 4[−3 − 0]
= 0 − 5[27] − 4[−3]
= 0 − 135 + 12
= −123
g. The pneumonic shortcut (the arrow technique for evaluating 3x3
determinants)
Sol: 3 0
0
|2 −1 5 |
1 9 −4
Rewriting the entries of the matrix and first two columns, we get
3
2
1

0
0
−1 5
9 −4

3 0
2 −1
1 9

3
2
1

0
0
−1 5
9 −4

3 0
2 −1
1 9

Now we find the product of the entries of each diagonal
(3)(−1)(−4) = 12
(0)(5)(1) = 0
(0)(2)(9) = 0
So, sum of the product of the entries of each diagonal is
𝑆1 = 12 + 0 + 0 = 12
Now again, we find the product of the entries of each diagonal (top right to
bottom left)
3
2
1

0
0
−1 5
9 −4

3 0
2 −1
1 9

(1)(−1)(0) = 0
(9)(5)(3) = 135
(−4)(2)(0) = 0
So, sum of the product of the entries of each diagonal is
𝑆2 = 0 + 135 + 0 = 135
Now,
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 = 𝑆1 − 𝑆2
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 = 12 − 135
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡 = −123

Q2: Sol: - Given matrix is
3
2
[
4
2

3
0
5
2
0 −2
]
1 −3 0
10 3
2

a. Determinant using “The first row”
Sol: 3
2
|
4
2

3
0
5
2
0 −2
|
1 −3 0
10 3
2

2
0 −2
2
= 3(−1)1+1 | 1 −3 0 | + 3(−1)1+2 |4
10 3
2
2
2
0 −2
2
= 3(−1)2 | 1 −3 0 | + 3(−1)3 |4
10 3
2
2

2
= 3(1) | 1
10

0
−3
3

2
0
= 3 | 1 −3
10 3

0 −2
2 2
−3 0 | + (0)(−1)1+3 |4 1
3
2
2 10

−2
2 2
0
0 | + 5(−1)1+4 |4 1 −3|
2
2 10 3

0 −2
2 2 −2
2
−3 0 | + (0)(−1)4 |4 1
0 | + 5(−1)5 |4
3
2
2 10 2
2

−2
2
0 | + 3(−1) |4
2
2

0
−3
3

−2
2
0 | + 0 + 5(−1) |4
2
2

2
1
10

2
0
1 −3|
10 3

0
−3|
3

−2
2 0 −2
2 2
0
0 | − 3 |4 −3 0 | + 0 − 5 |4 1 −3| … (1)
2
2 3
2
2 10 3

Now, we find the determinants of 3x3 matrices using 1st row, we get
2
|1
10

0
−3
3

−2
−3
0 | = 2(−1)1+1 |
3
2

2
|4
2

0
−3
3

−2
−3
0 | = 2(−1)1+1 |
3
2

2
|4
2

2
1
10

0
1
−3| = 2(−1)1+1 |
10
3

0
1
| + (0)(−1)1+2 |
2
10
0
4
| + (0)(−1)1+2 |
2
2
−3
4
| + 2(−1)1+2 |
3
2

0
1
| + (−2)(−1)1+3 |
2
10

0
4
| + (−2)(−1)1+3 |
2
2
−3
4
| + (0)(−1)1+3 |
3
2

Putting the values of determinants into (1), we get

−3
| = −78
3

−3
| = −48
3
1
| = 30
10

= 3(...

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Review

Anonymous
Good stuff. Would use again.

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