# how to solve this linear system of three equations using Cramer's rule?

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).

• Your interpretation of Cramer's formulæ is wrong: each of $x, y,z$ is the determinant of the matrices you mention, divided by the determinant of $A$. Nov 27 '21 at 9:19
• Isn't clear the sequence of operations on the determinant! Nov 27 '21 at 9:19

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.
The determinant of$$A_x=\begin{bmatrix}-6&2&3\\2&0&1\\-5&1&-1\end{bmatrix}$$is $$6$$, which is equal to $$\det(A)$$. Therefore, $$\frac{\det(A_x)}{\det(A)}=1$$, which is what you should have got.