Factoring Determinant
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I am trying to find the determinant of this matrix with eigenvalues in it.
$(lambda I - A)$ =
$begin{bmatrix}
lambda - 1 & 3 & 0 \
3 & lambda - 1 & 0 \
0 & 0 & lambda + 2 \
end{bmatrix}$
$det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Stuck here. This is the correct factorization, but what do the mean?
= $[(lambda - 1)^2 - 9](lambda + 2)$
= $(lambda - 4)(lambda + 2)^2$
linear-algebra quadratic-forms
asked Dec 5 at 1:44
Evan Kim
758
add a comment |
up vote
0
down vote
favorite
I am trying to find the determinant of this matrix with eigenvalues in it.
$(lambda I - A)$ =
$begin{bmatrix}
lambda - 1 & 3 & 0 \
3 & lambda - 1 & 0 \
0 & 0 & lambda + 2 \
end{bmatrix}$
$det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Stuck here. This is the correct factorization, but what do the mean?
= $[(lambda - 1)^2 - 9](lambda + 2)$
= $(lambda - 4)(lambda + 2)^2$
linear-algebra quadratic-forms
asked Dec 5 at 1:44
Evan Kim
758
has the same meaning as () does.
– xbh
Dec 5 at 1:50
Try expanding along the last row or column instead. That lets you start off with a partial factorization.
– amd
Dec 5 at 3:24
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to find the determinant of this matrix with eigenvalues in it.
$(lambda I - A)$ =
$begin{bmatrix}
lambda - 1 & 3 & 0 \
3 & lambda - 1 & 0 \
0 & 0 & lambda + 2 \
end{bmatrix}$
$det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Stuck here. This is the correct factorization, but what do the mean?
= $[(lambda - 1)^2 - 9](lambda + 2)$
= $(lambda - 4)(lambda + 2)^2$
linear-algebra quadratic-forms
asked Dec 5 at 1:44
Evan Kim
758
I am trying to find the determinant of this matrix with eigenvalues in it.
$(lambda I - A)$ =
$begin{bmatrix}
lambda - 1 & 3 & 0 \
3 & lambda - 1 & 0 \
0 & 0 & lambda + 2 \
end{bmatrix}$
$det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Stuck here. This is the correct factorization, but what do the mean?
= $[(lambda - 1)^2 - 9](lambda + 2)$
= $(lambda - 4)(lambda + 2)^2$
linear-algebra quadratic-forms
linear-algebra quadratic-forms
asked Dec 5 at 1:44
Evan Kim
758
asked Dec 5 at 1:44
Evan Kim
758
asked Dec 5 at 1:44
Evan Kim
758
asked Dec 5 at 1:44
Evan Kim
758
asked Dec 5 at 1:44
Evan Kim
758
758
has the same meaning as () does.
– xbh
Dec 5 at 1:50
Try expanding along the last row or column instead. That lets you start off with a partial factorization.
– amd
Dec 5 at 3:24
add a comment |
has the same meaning as () does.
– xbh
Dec 5 at 1:50
Try expanding along the last row or column instead. That lets you start off with a partial factorization.
– amd
Dec 5 at 3:24
has the same meaning as () does.
– xbh
Dec 5 at 1:50
has the same meaning as () does.
– xbh
Dec 5 at 1:50
Try expanding along the last row or column instead. That lets you start off with a partial factorization.
– amd
Dec 5 at 3:24
Try expanding along the last row or column instead. That lets you start off with a partial factorization.
– amd
Dec 5 at 3:24
add a comment |
1 Answer
1
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up vote
0
down vote
You got $det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Which gives result as $$(lambda-1)(lambda-1)(lambda+2)-3(3(lambda+2))+0$$
$$=(lambda-1)^2(lambda+2)-9(lambda+2)$$
Now they just factored out $(lambda+2)$ from the above equation and got $$[(lambda-1)^2-9](lambda+2)$$
answered Dec 5 at 1:51
Key Flex
7,44941232
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
0
down vote
You got $det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Which gives result as $$(lambda-1)(lambda-1)(lambda+2)-3(3(lambda+2))+0$$
$$=(lambda-1)^2(lambda+2)-9(lambda+2)$$
Now they just factored out $(lambda+2)$ from the above equation and got $$[(lambda-1)^2-9](lambda+2)$$
answered Dec 5 at 1:51
Key Flex
7,44941232
add a comment |
up vote
0
down vote
You got $det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Which gives result as $$(lambda-1)(lambda-1)(lambda+2)-3(3(lambda+2))+0$$
$$=(lambda-1)^2(lambda+2)-9(lambda+2)$$
Now they just factored out $(lambda+2)$ from the above equation and got $$[(lambda-1)^2-9](lambda+2)$$
answered Dec 5 at 1:51
Key Flex
7,44941232
add a comment |
up vote
0
down vote
up vote
0
down vote
You got $det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Which gives result as $$(lambda-1)(lambda-1)(lambda+2)-3(3(lambda+2))+0$$
$$=(lambda-1)^2(lambda+2)-9(lambda+2)$$
Now they just factored out $(lambda+2)$ from the above equation and got $$[(lambda-1)^2-9](lambda+2)$$
answered Dec 5 at 1:51
Key Flex
7,44941232
You got $det(A)= (lambda - 1) begin{vmatrix}
lambda - 1 & 0 \
0 & lambda + 2 \
end{vmatrix}$-
$3begin{vmatrix}
3 & 0 \
0 & lambda + 2 \
end{vmatrix}$
+
$0begin{vmatrix}
3 & lambda - 1 \
0 & 0 \
end{vmatrix}$
Which gives result as $$(lambda-1)(lambda-1)(lambda+2)-3(3(lambda+2))+0$$
$$=(lambda-1)^2(lambda+2)-9(lambda+2)$$
Now they just factored out $(lambda+2)$ from the above equation and got $$[(lambda-1)^2-9](lambda+2)$$
answered Dec 5 at 1:51
Key Flex
7,44941232
answered Dec 5 at 1:51
Key Flex
7,44941232
answered Dec 5 at 1:51
Key Flex
7,44941232
answered Dec 5 at 1:51
Key Flex
7,44941232
7,44941232
add a comment |
add a comment |
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has the same meaning as () does.
– xbh
Dec 5 at 1:50
Try expanding along the last row or column instead. That lets you start off with a partial factorization.
– amd
Dec 5 at 3:24