# Thread: Inverse of 3X3 Matrix

1. ## Inverse of 3X3 Matrix

Any hints? Got everything down with the "cofactor method" for finding the inverse of a 3X3 Matrix, except this part:

2. ## Re: Inverse of 3X3 Matrix

First of all, you have calculated the matrix of minors wrong, it should be:

$\displaystyle \begin{bmatrix}2&2&2\\-2&3&3\\0&-10&0 \end{bmatrix}$.

Now to create the adjugate (or adjoint) matrix, we multiply each ij-entry by $\displaystyle (-1)^{i+j}$, that is, alternate signs, starting with postive in the upper-left corner, to get:

$\displaystyle \begin{bmatrix}2&-2&2\\2&3&-3\\0&10&0 \end{bmatrix}$.

We will want the transpose of this matrix, which is:

$\displaystyle \begin{bmatrix}2&2&0\\-2&3&10\\2&-3&0 \end{bmatrix}$.

We need to now calculate the determinant of the original matrix, to get a suitable scalar factor:

The easiest way is to expand by minors along the second column, since it has only one non-zero entry. Since this is in the 3,2 position, we take the negative of this determinant (since 3+2 = 5 is odd), the cofactor of the other 2 subdeterminants will be 0, so we don't have to calculate the matrix of the other minors (although we've already done it, so meh...).

that is, the determinant of the original matrix is:

$\displaystyle \det(A) = (-1)\ast 1 \ast \begin{vmatrix}3&2\\2&-2 \end{vmatrix} = (-1)(-10) = 10$.

So our scaling factor will be 1/10.

This gives the inverse:

$\displaystyle A^{-1} = \frac{1}{\det(A)}\text{adj}(A) = \begin{bmatrix} \frac{1}{5}&\frac{1}{5}&0\\ -\frac{1}{5}&\frac{3}{10}&1\\ \frac{1}{5}&-\frac{3}{10}&0 \end{bmatrix}$

3. ## Re: Inverse of 3X3 Matrix

Ok, corrected now.

4. ## Re: Inverse of 3X3 Matrix

Ok here are the two charts. I have one last question on the 2nd one.

As far as understanding the question in the first chart, I understand it. Wherever the two lines meet (in a cross-out pattern), that place is where the determinant goes for the new "matrix of determinant answers", so to speak.

5. ## Re: Inverse of 3X3 Matrix

Go entry-by-entry. Draw one horizontal and one vertical line that intersect at the entry you are on. You should wind up with 9 "charts" as you call them: one cross-out for each entry in the matrix.