Determinant of a $4\times4$ matrix without using triangular method if possible Find the unique value of $x$ for which
$$\det\begin{pmatrix}-6 &amp; -1 &amp; -3 &amp; 6\\
5 &amp; -5 &amp; x &amp; 6\\
-8 &amp; -4 &amp; -8 &amp; 2\\
-5 &amp; 1 &amp; -6 &amp; -4
\end{pmatrix}=0.$$
Using SymPy Live:

&gt;&gt;&gt; x = symbols('x')
&gt;&gt;&gt; M = Matrix([[-6, -1, -3, 6], [5, -5, x, 6], [-8, -4, -8, 2], [-5, 1, -6, -4]])
&gt;&gt;&gt; M
[-6  -1  -3  6 ]
[              ]
[5   -5  x   6 ]
[              ]
[-8  -4  -8  2 ]
[              ]
[-5  1   -6  -4]         
&gt;&gt;&gt; M.det()
210*x - 1100
I recommend this website https://matrixcalc.org/en/ to analyze matrices.


$$\det\begin{pmatrix}-6 &amp; -1 &amp; -3 &amp; 6\\
5 &amp; -5 &amp; \frac{1100}{210} &amp; 6\\
-8 &amp; -4 &amp; -8 &amp; 2\\
-5 &amp; 1 &amp; -6 &amp; -4
\end{pmatrix}=0.$$

Leibniz-Formel:

$-6 \cdot(-5) \cdot (-8) \cdot (-4)\\
-(-6) \cdot (-5) \cdot 2 \cdot (-6)\\
-(-6) \cdot x \cdot (-4) \cdot (-4)\\
+(-6) \cdot x \cdot 2 \cdot 1\\
+(-6) \cdot 6 \cdot (-4) \cdot (-6)\\
-(-6) \cdot 6 \cdot (-8) \cdot 1\\
-(-1) \cdot 5 \cdot (-8) \cdot (-4)\\
+(-1) \cdot 5 \cdot 2 \cdot (-6)\\
+(-1) \cdot x \cdot (-8) \cdot (-4)\\
-(-1) \cdot x \cdot 2 \cdot (-5)\\
-(-1) \cdot 6 \cdot (-8) \cdot (-6)\\
+(-1) \cdot 6 \cdot (-8) \cdot (-5)\\
+(-3) \cdot 5 \cdot (-4) \cdot (-4)\\
-(-3) \cdot 5 \cdot 2 \cdot 1\\
-(-3) \cdot (-5) \cdot (-8) \cdot (-4)\\
+(-3) \cdot (-5) \cdot 2 \cdot (-5)\\
+(-3) \cdot 6 \cdot (-8) \cdot 1\\
-(-3) \cdot 6 \cdot (-4) \cdot (-5)\\
-6 \cdot 5 \cdot (-4) \cdot (-6)\\
+6 \cdot 5 \cdot (-8) \cdot 1\\
+6 \cdot (-5) \cdot (-8) \cdot (-6)\\
-6 \cdot (-5) \cdot (-8) \cdot (-5)\\
-6 \cdot x \cdot (-8) \cdot 1\\
+6 \cdot x \cdot (-4) \cdot (-5)\\
=210 \cdot x-1100$

$$210 \cdot x -1100 \Rightarrow x = \frac{1100}{210}$$

The method of Gaussian elimination or the Bareiss algorithm provide the same result.

Good luck :)