# Cramer's Rule for Inverse

I am trying to find the inverse of the following matrix using Cramer's rule:$\begin{bmatrix}2&0&0&0\\ 1&-2&1&-1\\ 0&2&2&3 \\1&0&1&1 \end{bmatrix}$. If this was a $3\times3$, I would know the process but I am completely stuck since this is a $4 \times 4$. My question is, how do you even begin to calculate $C_{1,1}, ...., C_{m,n}$? I tried searching for a tutorial with no luck and my book does not mention 4x4. My idea was to compute:$\begin{bmatrix} -2&1&-1\\ 2&2&3 \\0&1&1\end{bmatrix}$ for $C_{1,1}$ and take this long process all the way up to $C_{m,n}$. But if this is correct, do I solve the determinant of the 3x3 matrix now and that is my co factor or....? Any help would be appreciated. Also since this is a very long process, I was wondering if there is any algorithm for paper-pencil process or any easier way to compute it?

• Are you referring to the adjugate matrix method? – Git Gud Jul 20 '13 at 18:42
• I am referring to Cramer's Rule en.wikipedia.org/wiki/Cramer%27s_rule but I think that might be another name for adjugate matrix. – Jlhglkj Glkjhg Jul 20 '13 at 18:46
• My answer is an example of the method above. Beware that it's over finite field. The method is the same, but the arithmetic is a bit different. – Git Gud Jul 20 '13 at 18:47
• Cramer's Rule is to solve systems. Even though finding the inverse matrix can be done by solving a system, it's not the same thing as what I mentioned above. – Git Gud Jul 20 '13 at 18:48
• Do you how to use the same method for a $4x4$? You can determine the inverse of a matrix using cofactor by finding each $C_{m,n}$ and dividing by the determinant. I guess that was my question: how do we find the cofactor for each spot when it is a 4x4 system? – Jlhglkj Glkjhg Jul 20 '13 at 18:53

$$AB=I_4$$
$$A\begin{bmatrix}\vec{b}_1 & \vec{b}_2 & \vec{b}_3 & \vec{b}_4\end{bmatrix}=\begin{bmatrix}\hat{e}_1 & \hat{e}_2 & \hat{e}_3 & \hat{e}_4\end{bmatrix}$$
\begin{align}A\vec{b}_1&=\hat{e}_1 \\ A\vec{b}_2&=\hat{e}_2 \\ A\vec{b}_3&=\hat{e}_3 \\ A\vec{b}_4&=\hat{e}_4\end{align}
$$b_{11}=\frac{\begin{vmatrix}1&0 & 0 &0\\0 & -2 & 1 & -1\\0 & 2 & 2 & 3 \\ 0 & 1 & 1 &1 \end{vmatrix}}{\begin{vmatrix}2&0&0&0\\ 1&-2&1&-1\\ 0&2&2&3 \\1&0&1&1 \end{vmatrix}}$$