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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?

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2 Answers 2

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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 $c$.

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$$\begin{pmatrix} 4&0&1\\19&1&-3\\7&1&0 \end{pmatrix}=\frac{1}{a}\begin{pmatrix} 4a&0&1\\19a&1&-3\\7a&1&0 \end{pmatrix}$$

$$\Rightarrow \frac{1}{a}\begin{pmatrix} (4a+0.b+1.c)&0&1\\(19a+b-3c)&1&-3\\(7a+b+0.c)&1&0 \end{pmatrix}$$

$$\Rightarrow \frac{1}{a}\begin{pmatrix} 4&0&1\\3&1&-3\\1&1&0 \end{pmatrix}$$

$$\Rightarrow \begin{pmatrix} 4&0&1\\19&1&-3\\7&1&0 \end{pmatrix}=\frac{1}{a}\begin{pmatrix} 4&0&1\\3&1&-3\\1&1&0 \end{pmatrix}$$

$$\Rightarrow \frac{det(\begin{pmatrix} 4&0&1\\3&1&-3\\1&1&0 \end{pmatrix})}{det \begin{pmatrix} 4&0&1\\19&1&-3\\7&1&0 \end{pmatrix} }=a$$

Similarly for others.

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