I have a 3-by-3 matrix,
A=$\left [ \begin{matrix}
1 & 2 & 3 \\
1 & 0 & 1 \\
1 & 1 & -1\\
\end{matrix} \right]$
the known terms are (-6, 2, -5), at the right of "=" symbol.
(1) I've calculated the determinant,
(2) I've used Cramer's rule to find x, y, and z.
but the result isn't correct (the right solutions is (x, y,z) = (1, -5, 1)).
(1) determinant of A,
I used row operations: row 3 <-> row 2. I swapped the rows. (therefore the determinant is $-\det A$).
(2) row 2 <– $row 2 – row 1$ and I got:
A=$\left [ \begin{matrix}
1 & 2 & 3 \\
0 & -1 & -4 \\
1 & 1 & -1\\
\end{matrix} \right]$
then, I used Laplace in the first column, and I got:
$-\det A$=$\left [ \begin{matrix}
-1 & -4 \\
0 & 1 \\
\end{matrix} \right]$ + $\left [ \begin{matrix}
2 & 3 \\
-1 & -4 \\
\end{matrix} \right]$,
doing algebra here, I get: -1-(-8+3) = -6 but it was -detA, and therefore detA = 6.
I used Cramer's rule, therefore I put the known terms in first column, then in second, and so on.
x = A=$\left [ \begin{matrix}
-6 & 2 & 3 \\
2 & 0 & 1 \\
-5 & 1 & -1\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
x is equal to 3, it should've been equal to 1.
y = A=$\left [ \begin{matrix}
1 & -6 & 3 \\
0 & 2 & 1 \\
1 & -5 & -1\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
-2-6-6-5 = -14-5 = -19 it should have been equal to -5. (Sarrus' rule)
z = A=$\left [ \begin{matrix}
1 & 2 & -6 \\
0 & 0 & 2 \\
1 & 1 & -5\\
\end{matrix} \right] $ (this matrix divided by the determinant of the original matrix)
4-2 = 2. it should've been equal to 1. (Sarrus' Rule).
Best Answer
The determinant of the matrix representing $x$ is, expanding down the middle column (which you definitely can do, no row operations needed!):
$$\det\begin{pmatrix}-6&2&3\\2&0&1\\-5&1&-1\end{pmatrix}=(-2)\cdot\det\begin{pmatrix}2&1\\-5&-1\end{pmatrix}+0+(-1)\cdot\det\begin{pmatrix}-6&3\\2&1\end{pmatrix}$$
Which is: $$(-2)(-2-(-5))+(-1)(-6-6)=-2\cdot3+12=6$$Cramer tells that $x$ is then this determinant divided by the original one, namely $6/6=1$. When you’re saying: “this matrix divided by the determinant of the original matrix” this is wrong. That would imply $x$ is a matrix divided by a scalar, i.e. another matrix. You just take the determinants of both, and find their ratios.