§ 3.5 Determinants and Cramer’s Rule. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.5...
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§ 3.5 Determinants and Cramer’s Rule
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Transcript of § 3.5 Determinants and Cramer’s Rule. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.5...
§ 3.5
Determinants and Cramer’s Rule
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 3.5
Determinants
EXAMPLEEXAMPLE
Evaluate the determinant of the matrix.
SOLUTIONSOLUTION
52
43
7815815425352
43
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 3.5
Determinants
Definition of the Determinant of a 2 by 2 Definition of the Determinant of a 2 by 2 matrixmatrix
The determinant of the matrix
dc
ba bcdadc
bais defined by
The determinant of a matrix may be positive or negative. The determinant can also have 0 as its value.
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 3.5
Determinants – Cramer’s Rule
Solving a Linear System in Two Variables Using Determinants
Cramer’s Rule
If
then
and
where
222
111
cybxa
cybxa
22
11
22
11
ba
ba
bc
bc
x
22
11
22
11
ba
ba
ca
ca
y
022
11 ba
ba
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 3.5
Determinants – Cramer’s Rule
EXAMPLEEXAMPLE
Use Cramer’s rule to solve the system:
SOLUTIONSOLUTION
132
173
yx
yx
Because we’re using Cramer’s rule to solve this system, we must first write the system in standard form.
132
173
yx
yx
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 3.5
Determinants – Cramer’s Rule
Now we need to determine the different values as defined in Cramer’s rule:
5149723332
73
22
11
ba
ba
132
173
222
111
cba
cba
523121312
13
22
11
ca
ca
CONTINUECONTINUEDD
Therefore
1073713131
71
22
11
bc
bc
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 3.5
Determinants – Cramer’s Rule
Now we can determine x and y.CONTINUECONTINUE
DD
25
10
22
11
22
11
ba
ba
bc
bc
x 15
5
22
11
22
11
ba
ba
ca
ca
y
Therefore, the solution to the system is (-2,-1). As always, the ordered pair should be checked by substituting these values into the original equations.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 3.5
Determinants
Evaluating the Determinant of a 3x3 Matrix1) Each of the three terms in the definition contains two factors – a numerical factor and a second-order determinant.
2) The numerical factor in each term is an element from the first column of the third-order determinant.
3) The minus sign precedes the second term.
4) The second-order determinant that appears in each term is obtained by crossing out the row and the column containing the numerical factor.
The minor of an element is the determinant that remains after deleting the row and column of that element. For this reason, we call this
method expansion by minors.
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 3.5
Determinants
EXAMPLEEXAMPLE
Evaluate the determinant of the matrix.
SOLUTIONSOLUTION
864
532
012
864
532
012
864
532
012
864
532
012
The minor for 2 is .
The minor for -2 is .
The minor for 4 is .53
01
86
01
86
53
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 3.5
Determinants
Therefore
864
532
012
53
014
86
012
86
532
CONTINUECONTINUEDD
= 2[(3)(8) – (-6)(5)] + 2[(1)(8) – (-6)(0)] + 4[(1)(5) – (3)(0)]
= 2[24 – (-30)] + 2[8 – 0] + 4[5 – 0]
= 2[24 + 30] + 2[8 – 0] + 4[5 – 0]
= 2[54] + 2[8] + 4[5] = 108 + 16 + 20 = 144
Therefore, the determinant of the matrix is 144.
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 3.5
Determinants – Cramer’s Rule
EXAMPLEEXAMPLE
Use Cramer’s rule to solve the system:
SOLUTIONSOLUTION 432
1
1373
zyx
zyx
zyx
We first need to determine the different values as defined in Cramer’s rule:
4321
1111
13731
3333
2222
1111
dcba
dcba
dcba
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 3.5
Determinants – Cramer’s Rule
10
321
111
731
333
222
111
cba
cba
cba
20
324
111
7313
333
222
111
cbd
cbd
cbd
6
341
111
7131
333
222
111
cda
cda
cda
Therefore
CONTINUECONTINUEDD
24
421
111
1331
333
222
111
dba
dba
dba
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 3.5
Determinants – Cramer’s Rule
210
20
333
222
111
333
222
111
cba
cba
cba
cbd
cbd
cbd
x
5
3
10
6
333
222
111
333
222
111
cba
cba
cba
cda
cda
cda
y
Now I can determine x, y and z.
CONTINUECONTINUEDD
5
12
10
24
333
222
111
333
222
111
cba
cba
cba
dba
dba
dba
z
Therefore, the solution to the system is (-2, 3/5, 12/5). As always, the ordered pair should be checked by substituting these values into the original equations.
