How to solve a system of 3 equations with Cramer's Rule? I am given the following system of 3 simultaneous equations:
$$
\begin{align*}
4a+c &= 4\\
19a + b - 3c &= 3\\
7a + b &= 1\end{align*}
$$
How do I solve using Cramers' rule?
For one, I do know that by putting as a matrix the LHS
$$\begin{pmatrix}
4&0&1\\19&1&-3\\7&1&0
\end{pmatrix}$$
and then computing its determinant to be $24$ is of some use...but could somebody give more ideas?
 A: In the below tutorial, they mention a very easy method to understand Cramer's rule and simplification of 2 as well as 3 equation systems. See the step by step tutorial with a solved example:
Check it here: Cramer’s Rule | 2 & 3 Equation Systems. Easy & Step by Step

A: Did you read Wikipedia's article on Cramer's rule? Its worked examples are fairly clear.
In your example, to find $a$, substitute the RHS in for $a$'s column:
$$a = \frac{ \left|\begin{array}{ccc}4 & 0 & 1\\3 & 1 & -3\\1 & 1& 0\end{array}\right| } { \left|\begin{array}{ccc}4 & 0 & 1\\19 & 1 & -3 \\ 7 & 1 & 0\end{array}\right| } = \frac{12+0+2}{24} = \frac{7}{12},$$
and similarly for $b$ and $c$.
A: $$\begin{pmatrix}
4&0&1\\19&1&-3\\7&1&0
\end{pmatrix}=\frac{1}{a}\begin{pmatrix}
4a&0&1\\19a&1&-3\\7a&1&0
\end{pmatrix}$$
$$\Rightarrow \frac{1}{a}\begin{pmatrix}
(4a+0.b+1.c)&0&1\\(19a+b-3c)&1&-3\\(7a+b+0.c)&1&0
\end{pmatrix}$$
$$\Rightarrow \frac{1}{a}\begin{pmatrix}
4&0&1\\3&1&-3\\1&1&0
\end{pmatrix}$$
$$\Rightarrow \begin{pmatrix}
4&0&1\\19&1&-3\\7&1&0
\end{pmatrix}=\frac{1}{a}\begin{pmatrix}
4&0&1\\3&1&-3\\1&1&0
\end{pmatrix}$$
$$\Rightarrow \frac{det(\begin{pmatrix}
4&0&1\\3&1&-3\\1&1&0
\end{pmatrix})}{det \begin{pmatrix}
4&0&1\\19&1&-3\\7&1&0
\end{pmatrix} }=a$$
Similarly for others.
