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Find the unique value of $x$ for which $$\det\begin{pmatrix}-6 & -1 & -3 & 6\\ 5 & -5 & x & 6\\ -8 & -4 & -8 & 2\\ -5 & 1 & -6 & -4 \end{pmatrix}=0.$$

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closed as off-topic by JMoravitz, Jack's wasted life, iadvd, Parcly Taxel, user91500 Oct 30 '16 at 5:49

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  • 3
    $\begingroup$ Where are you blocked ? $\endgroup$ – Jean Marie Oct 29 '16 at 20:47
  • $\begingroup$ Do you know how to find the determinant of a $2\times 2$ matrix? Knowing this, do you know how to find the determinant of a $3\times 3$ matrix? Knowing this, do you then know how to find the determinant of a $4\times 4$ matrix? Knowledge of Laplacian Expansion will be useful here. $\endgroup$ – JMoravitz Oct 29 '16 at 20:56
  • $\begingroup$ i know how to find it with triangular method and if there are zeros in the matrix thats why i got lost $\endgroup$ – ISuckAtMathPleaseHELPME Oct 29 '16 at 21:14
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I recommend this website https://matrixcalc.org/en/ to analyze matrices.

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

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 :)

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  • $\begingroup$ would you mind giiving me a direction of how you got there? it'd be really helpful. Thanks $\endgroup$ – ISuckAtMathPleaseHELPME Oct 29 '16 at 21:15
  • $\begingroup$ I hope it helps you .... $\endgroup$ – Darío A. Gutiérrez Oct 29 '16 at 21:51
  • $\begingroup$ is there any way this can be solved with reduced row echelon form or the triangular form $\endgroup$ – ISuckAtMathPleaseHELPME Oct 29 '16 at 22:10
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Using SymPy Live:

>>> x = symbols('x')
>>> M = Matrix([[-6, -1, -3, 6], [5, -5, x, 6], [-8, -4, -8, 2], [-5, 1, -6, -4]])
>>> M
[-6  -1  -3  6 ]
[              ]
[5   -5  x   6 ]
[              ]
[-8  -4  -8  2 ]
[              ]
[-5  1   -6  -4]         
>>> M.det()
210*x - 1100
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