Factoring Determinant












0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    has the same meaning as () does.
    $endgroup$
    – xbh
    Dec 5 '18 at 1:50










  • $begingroup$
    Try expanding along the last row or column instead. That lets you start off with a partial factorization.
    $endgroup$
    – amd
    Dec 5 '18 at 3:24
















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    has the same meaning as () does.
    $endgroup$
    – xbh
    Dec 5 '18 at 1:50










  • $begingroup$
    Try expanding along the last row or column instead. That lets you start off with a partial factorization.
    $endgroup$
    – amd
    Dec 5 '18 at 3:24














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 5 '18 at 1:44









Evan KimEvan Kim

1068




1068












  • $begingroup$
    has the same meaning as () does.
    $endgroup$
    – xbh
    Dec 5 '18 at 1:50










  • $begingroup$
    Try expanding along the last row or column instead. That lets you start off with a partial factorization.
    $endgroup$
    – amd
    Dec 5 '18 at 3:24


















  • $begingroup$
    has the same meaning as () does.
    $endgroup$
    – xbh
    Dec 5 '18 at 1:50










  • $begingroup$
    Try expanding along the last row or column instead. That lets you start off with a partial factorization.
    $endgroup$
    – amd
    Dec 5 '18 at 3:24
















$begingroup$
has the same meaning as () does.
$endgroup$
– xbh
Dec 5 '18 at 1:50




$begingroup$
has the same meaning as () does.
$endgroup$
– xbh
Dec 5 '18 at 1:50












$begingroup$
Try expanding along the last row or column instead. That lets you start off with a partial factorization.
$endgroup$
– amd
Dec 5 '18 at 3:24




$begingroup$
Try expanding along the last row or column instead. That lets you start off with a partial factorization.
$endgroup$
– amd
Dec 5 '18 at 3:24










1 Answer
1






active

oldest

votes


















0












$begingroup$

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 $$[(lambda-1)^2-9](lambda+2)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Evan Kim If you find my answer helpful then accept the answer $✓$ :)
    $endgroup$
    – Key Flex
    Jan 26 at 19:56











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

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 $$[(lambda-1)^2-9](lambda+2)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Evan Kim If you find my answer helpful then accept the answer $✓$ :)
    $endgroup$
    – Key Flex
    Jan 26 at 19:56
















0












$begingroup$

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 $$[(lambda-1)^2-9](lambda+2)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Evan Kim If you find my answer helpful then accept the answer $✓$ :)
    $endgroup$
    – Key Flex
    Jan 26 at 19:56














0












0








0





$begingroup$

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 $$[(lambda-1)^2-9](lambda+2)$$






share|cite|improve this answer









$endgroup$



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 $$[(lambda-1)^2-9](lambda+2)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 5 '18 at 1:51









Key FlexKey Flex

7,94461233




7,94461233












  • $begingroup$
    @Evan Kim If you find my answer helpful then accept the answer $✓$ :)
    $endgroup$
    – Key Flex
    Jan 26 at 19:56


















  • $begingroup$
    @Evan Kim If you find my answer helpful then accept the answer $✓$ :)
    $endgroup$
    – Key Flex
    Jan 26 at 19:56
















$begingroup$
@Evan Kim If you find my answer helpful then accept the answer $✓$ :)
$endgroup$
– Key Flex
Jan 26 at 19:56




$begingroup$
@Evan Kim If you find my answer helpful then accept the answer $✓$ :)
$endgroup$
– Key Flex
Jan 26 at 19:56


















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