# Find det(A) for matrix, find C and adj(A)

I am asked to find the determinant of matrix A by using cofactors method. I understand how to do that portion, but then it asks to also find matrix C, and adj(A). How can I go about that?

for det(A) I got -20.

$A = \begin{bmatrix} 1 & 3 & -3 \\ -3 & -3 & 2 \\ -4 & 4 & -6 \\ \end{bmatrix}$

• Do you know how to find the adjugate matrix? en.wikipedia.org/wiki/Adjugate_matrix – gbox Mar 25 '17 at 21:12
• You can choose to calculate the determinant by determining the minors one by one. Then you get the cofactor matrix. You can see my answer to this question: math.stackexchange.com/a/2203000/213607 which explains the relation between ${\bf M, C}, \textrm{adj}({\bf A})$ and $\det({\bf A})$ – mathreadler Mar 25 '17 at 21:18
• Okay, so C is just the cofactor matrix then correct? @mathreadler – cisco Mar 25 '17 at 21:20
• I have not read from the source of your problem but I suppose so. In this context $\bf C$ is probably cofactor matrix. – mathreadler Mar 25 '17 at 21:21

## 1 Answer

$$A = \begin{bmatrix} 1 & 3 & -3 \\ -3 & -3 & 2 \\ -4 & 4 & -6 \\ \end{bmatrix}$$ $$\det A = 4$$.

We know $$A^{-1}=\frac{1}{\det A} adj (A)$$, where $$adj$$ is adjoint of matrix $$A$$.

$$adj(A)= \left( \begin{array}{ccc} -26 & -30 & -3 \\ 10 & -6 & 7 \\ -24 & -16 & 6 \end{array} \right)$$.

$$\therefore A^{-1}= \frac{1}{4}\left( \begin{array}{ccc} -26 & -30 & -3 \\ 10 & -6 & 7 \\ -24 & -16 & 6 \end{array} \right)$$

$$A^{-1}= \left( \begin{array}{ccc} -26/4 & -30/4 & -3/4 \\ 10/4 & -6/4 & 7/4 \\ -24/4 & -16/4 & 6/4 \end{array} \right)$$