# [SOLVED] Inverse matrix with adj(A)

• February 9th 2010, 03:17 PM
thekrown
So I'm trying to find A^-1 (inverse) matrix. The theorem I'm supposed to use for this is of a matrix using its Adjoint.

A^-1 = [1 / det(A)] * [adj(A)]

The matrix used is A =

2 0 0
8 1 0
-5 3 6

I expanded the matrix using cofactors and obtained

c11=6, c12=48, c13=29
c21=0, c22=12, c23=6
c31=0, c32=0, c33=2

so the matrix cofactor is

6 48 29
0 12 6
0 0 2

and the adjoint of A is

6 0 0
48 12 0
29 6 2

now what do I do? I'm stuck here.
• February 9th 2010, 04:51 PM
dedust
Quote:

Originally Posted by thekrown
So I'm trying to find A^-1 (inverse) matrix. The theorem I'm supposed to use for this is of a matrix using its Adjoint.

A^-1 = [1 / det(A)] * [adj(A)]

The matrix used is A =

2 0 0
8 1 0
-5 3 6

I expanded the matrix using cofactors and obtained

c11=6, c12=48, c13=29
c21=0, c22=12, c23=6
c31=0, c32=0, c33=2

so the matrix cofactor is

6 48 29
0 12 6
0 0 2

and the adjoint of A is

6 0 0
48 12 0
29 6 2

now what do I do? I'm stuck here.

now you only need to compute the deteminant of A.

$\det A = 6 \left|\begin{matrix} 2 & 0 \\ 8 & 1 \end{matrix}\right| - 0 \cdot \left|\begin{matrix} 2 & 0 \\ -5 & 3 \end{matrix}\right| + 0 \cdot \left|\begin{matrix} 8 & 1 \\ -5 & 3 \end{matrix}\right| = 12$

and finally,

$A^{-1} = \frac{1}{12} \left(\begin{matrix} 6 & 0 & 0 \\ 48 & 12 & 0 \\ 29 & 6 & 2\end{matrix}\right)$
• February 9th 2010, 06:31 PM
thekrown
Alright so the end result should then be

A^-1 = 1/12 * 6 (det of the adj(A))

and so for my final answer I have A^-1 = 1/2, yes?
• February 9th 2010, 10:54 PM
nahduma
A(-1) is a matrix
No, the inverse of a matrix is not a value, like it's determinant.
The inverse of a matrix A = adjoint(A)/det(A).

Anyway, I don't see how the determinant of adj(A)=6. It's 144 . (12²)
The answer dedust posted is the inverse of A.
• February 10th 2010, 11:56 AM
thekrown
Okay so I understand my mistake. The answer is 1/det(A) * ajd(A) (where ads(A) is the matrix starting with line 6 0 0).

However, nahduma really confused me. Just so I can correct my mistake of trying to find the det of such a matrix:

6 0 0
48 12 0
29 6 2

I originally ended up with

(-1)^1+1 6det

12 0
6 2

+ (-1)^1+2 0 det (matrix doesn't matter here is mult by zero).

+ (-1)^1+3 0 det (matrix doesn't matter here it's mult by zero).

So with this I ended up with 6 multiplied by ad-bc which is 12x2 - 6x0. We get 6x24 = 144.

Originally I forgot to count the ad-bc part and ended up with just 6.

Now back to nahduma's post, did you just get 12^2 by simplifying 144 or did you use a method which gave you 12^2?
• February 10th 2010, 07:42 PM
nahduma