Sign rule for finding the adjugate of a 3x3 matrix?
$begingroup$
So i have this matrix
A= $$
begin{pmatrix}
1 & 3 & 0 \
-2 & -5 & 2 \
1 & 4 & 3 \
end{pmatrix}
$$
And i want to find the inverse of it. Following all the calculations, i get that the determinant is 1 and that the adjucate of the matrix by creating the matrix of co factors is
$$
begin{pmatrix}
-23 & -9 & 6 \
8 & 3 & -2 \
-3 & -1 & 1 \
end{pmatrix}
$$
I checked the answer on the answer sheet and it's right.
And i would be happy to end the excercise here BUT if i remember correctly shouldn't i apply the sign rule?
I always thought that after having calculated the matrix, i apply the sign rule which would be
$$
begin{pmatrix}
+ & - & + \
- & + & - \
+ & - & + \
end{pmatrix}
$$
If i apply the sign rule, then my matrix would look different:
$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$
and ,in order to finish the calculations, i would need to multiply the adjucate by 1 over the determinant of A which is 1.
So i would multiply 1 times each entry in the adjugate and the matrix would still be
$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$
Which, as you can clearly see, is not the correct answer.
So my question is , should i apply the sign rule? Should i not? What is it that i'm doing wrong? SHould i stop when i just find the adjugate and multiply the entries by the determinant ?
Thanks for the help!
(please do not tell me to find the inverse another way , thanks)!
matrices determinant inverse matrix-calculus
asked Dec 4 '18 at 12:03
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So i have this matrix
A= $$
begin{pmatrix}
1 & 3 & 0 \
-2 & -5 & 2 \
1 & 4 & 3 \
end{pmatrix}
$$
And i want to find the inverse of it. Following all the calculations, i get that the determinant is 1 and that the adjucate of the matrix by creating the matrix of co factors is
$$
begin{pmatrix}
-23 & -9 & 6 \
8 & 3 & -2 \
-3 & -1 & 1 \
end{pmatrix}
$$
I checked the answer on the answer sheet and it's right.
And i would be happy to end the excercise here BUT if i remember correctly shouldn't i apply the sign rule?
I always thought that after having calculated the matrix, i apply the sign rule which would be
$$
begin{pmatrix}
+ & - & + \
- & + & - \
+ & - & + \
end{pmatrix}
$$
If i apply the sign rule, then my matrix would look different:
$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$
and ,in order to finish the calculations, i would need to multiply the adjucate by 1 over the determinant of A which is 1.
So i would multiply 1 times each entry in the adjugate and the matrix would still be
$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$
Which, as you can clearly see, is not the correct answer.
So my question is , should i apply the sign rule? Should i not? What is it that i'm doing wrong? SHould i stop when i just find the adjugate and multiply the entries by the determinant ?
Thanks for the help!
(please do not tell me to find the inverse another way , thanks)!
matrices determinant inverse matrix-calculus
asked Dec 4 '18 at 12:03
BM97BM97
758
$endgroup$
add a comment |
$begingroup$
So i have this matrix
A= $$
begin{pmatrix}
1 & 3 & 0 \
-2 & -5 & 2 \
1 & 4 & 3 \
end{pmatrix}
$$
And i want to find the inverse of it. Following all the calculations, i get that the determinant is 1 and that the adjucate of the matrix by creating the matrix of co factors is
$$
begin{pmatrix}
-23 & -9 & 6 \
8 & 3 & -2 \
-3 & -1 & 1 \
end{pmatrix}
$$
I checked the answer on the answer sheet and it's right.
And i would be happy to end the excercise here BUT if i remember correctly shouldn't i apply the sign rule?
I always thought that after having calculated the matrix, i apply the sign rule which would be
$$
begin{pmatrix}
+ & - & + \
- & + & - \
+ & - & + \
end{pmatrix}
$$
If i apply the sign rule, then my matrix would look different:
$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$
and ,in order to finish the calculations, i would need to multiply the adjucate by 1 over the determinant of A which is 1.
So i would multiply 1 times each entry in the adjugate and the matrix would still be
$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$
Which, as you can clearly see, is not the correct answer.
So my question is , should i apply the sign rule? Should i not? What is it that i'm doing wrong? SHould i stop when i just find the adjugate and multiply the entries by the determinant ?
Thanks for the help!
(please do not tell me to find the inverse another way , thanks)!
matrices determinant inverse matrix-calculus
asked Dec 4 '18 at 12:03
BM97BM97
758
$endgroup$
So i have this matrix
A= $$
begin{pmatrix}
1 & 3 & 0 \
-2 & -5 & 2 \
1 & 4 & 3 \
end{pmatrix}
$$
And i want to find the inverse of it. Following all the calculations, i get that the determinant is 1 and that the adjucate of the matrix by creating the matrix of co factors is
$$
begin{pmatrix}
-23 & -9 & 6 \
8 & 3 & -2 \
-3 & -1 & 1 \
end{pmatrix}
$$
I checked the answer on the answer sheet and it's right.
And i would be happy to end the excercise here BUT if i remember correctly shouldn't i apply the sign rule?
I always thought that after having calculated the matrix, i apply the sign rule which would be
$$
begin{pmatrix}
+ & - & + \
- & + & - \
+ & - & + \
end{pmatrix}
$$
If i apply the sign rule, then my matrix would look different:
$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$
and ,in order to finish the calculations, i would need to multiply the adjucate by 1 over the determinant of A which is 1.
So i would multiply 1 times each entry in the adjugate and the matrix would still be
$$
begin{pmatrix}
-23 & 9 & 6 \
-8 & 3 & 2 \
-3 & 1 & 1 \
end{pmatrix}
$$
Which, as you can clearly see, is not the correct answer.
So my question is , should i apply the sign rule? Should i not? What is it that i'm doing wrong? SHould i stop when i just find the adjugate and multiply the entries by the determinant ?
Thanks for the help!
(please do not tell me to find the inverse another way , thanks)!
matrices determinant inverse matrix-calculus
matrices determinant inverse matrix-calculus
asked Dec 4 '18 at 12:03
BM97BM97
758
asked Dec 4 '18 at 12:03
BM97BM97
758
edited Dec 4 '18 at 12:14
BM97
asked Dec 4 '18 at 12:03
BM97BM97
758
asked Dec 4 '18 at 12:03
BM97BM97
758
asked Dec 4 '18 at 12:03
BM97BM97
758
758
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You’ve already applied the sign rule when you calculated the cofactors.
For example when you calculate the $(1,1)$ element, you do:
$$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$
And when you calculate the $(1,2)$ element, you do (note the negative sign!!):
$$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$
Then to get the adjugate you take the transpose of the cofactors.
You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.
answered Dec 4 '18 at 12:56
ip6ip6
54839
$endgroup$
$begingroup$
Thank you !! I greatly appreciate your simple, straightforward answer .
$endgroup$
– BM97
Dec 4 '18 at 13:18
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You’ve already applied the sign rule when you calculated the cofactors.
For example when you calculate the $(1,1)$ element, you do:
$$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$
And when you calculate the $(1,2)$ element, you do (note the negative sign!!):
$$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$
Then to get the adjugate you take the transpose of the cofactors.
You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.
answered Dec 4 '18 at 12:56
ip6ip6
54839
$endgroup$
$begingroup$
Thank you !! I greatly appreciate your simple, straightforward answer .
$endgroup$
– BM97
Dec 4 '18 at 13:18
add a comment |
$begingroup$
You’ve already applied the sign rule when you calculated the cofactors.
For example when you calculate the $(1,1)$ element, you do:
$$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$
And when you calculate the $(1,2)$ element, you do (note the negative sign!!):
$$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$
Then to get the adjugate you take the transpose of the cofactors.
You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.
answered Dec 4 '18 at 12:56
ip6ip6
54839
$endgroup$
$begingroup$
Thank you !! I greatly appreciate your simple, straightforward answer .
$endgroup$
– BM97
Dec 4 '18 at 13:18
add a comment |
$begingroup$
You’ve already applied the sign rule when you calculated the cofactors.
For example when you calculate the $(1,1)$ element, you do:
$$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$
And when you calculate the $(1,2)$ element, you do (note the negative sign!!):
$$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$
Then to get the adjugate you take the transpose of the cofactors.
You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.
answered Dec 4 '18 at 12:56
ip6ip6
54839
$endgroup$
You’ve already applied the sign rule when you calculated the cofactors.
For example when you calculate the $(1,1)$ element, you do:
$$M_{1,1}=detbegin{pmatrix}-5&2\ 4&3end{pmatrix}=-23$$
And when you calculate the $(1,2)$ element, you do (note the negative sign!!):
$$M_{1,2}=-detbegin{pmatrix}-2&2\ 1&3end{pmatrix}=-((-2*3)-(1*2))=8$$
Then to get the adjugate you take the transpose of the cofactors.
You’ve done everything correctly and incorporated the sign rule. No need to apply it twice.
answered Dec 4 '18 at 12:56
ip6ip6
54839
answered Dec 4 '18 at 12:56
ip6ip6
54839
answered Dec 4 '18 at 12:56
ip6ip6
54839
answered Dec 4 '18 at 12:56
ip6ip6
54839
54839
$begingroup$
Thank you !! I greatly appreciate your simple, straightforward answer .
$endgroup$
– BM97
Dec 4 '18 at 13:18
add a comment |
$begingroup$
Thank you !! I greatly appreciate your simple, straightforward answer .
$endgroup$
– BM97
Dec 4 '18 at 13:18
$begingroup$
Thank you !! I greatly appreciate your simple, straightforward answer .
$endgroup$
– BM97
Dec 4 '18 at 13:18
$begingroup$
Thank you !! I greatly appreciate your simple, straightforward answer .
$endgroup$
– BM97
Dec 4 '18 at 13:18
add a comment |
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