Question

Find X by the use of determinant:
3x-4y+2z+8=0
x+5y-3z+2=0
5x+3y-z+6=0

Answer 1

Answer on Question#54580 – Math – Linear Algebra

1. Find $x$ by the use of determinant:

$$\left\{ \begin{array}{l} 3x - 4y + 2z + 8 = 0 \\ x + 5y - 3z + 2 = 0 \\ 5x + 3y - z + 6 = 0 \end{array} \right.$$

Solution.

At first, we will rewrite our system of equations in matrix form:

$$\left( \begin{array}{ccc} 3 & -4 & 2 \\ 1 & 5 & -3 \\ 5 & 3 & -1 \end{array} \right) \left( \begin{array}{c} x \\ y \\ z \end{array} \right) = \left( \begin{array}{c} -8 \\ -2 \\ -6 \end{array} \right).$$

Now we will find the determinant of the coefficient matrix $\left( \begin{array}{ccc} 3 & -4 & 2 \\ 1 & 5 & -3 \\ 5 & 3 & -1 \end{array} \right)$:

$$\Delta = \det \left| \begin{array}{ccc} 3 & -4 & 2 \\ 1 & 5 & -3 \\ 5 & 3 & -1 \end{array} \right| = -3 \cdot 5 + 4 \cdot 3 \cdot 5 + 2 \cdot 3 - 2 \cdot 5 \cdot 5 + 3 \cdot 3 \cdot 3 - 4 = -15 + 60 + 6 - -50 + 27 - 4 = 24.$$

To find $x$ we must know the determinant of the matrix which obtained from coefficient matrix by replacing column which correspond to $x$ variable by column from right side of system:

$$\Delta_{x} = \det \left| \begin{array}{ccc} -8 & -4 & 2 \\ -2 & 5 & -3 \\ -6 & 3 & -1 \end{array} \right| = 8 \cdot 5 - 4 \cdot 3 \cdot 6 - 2 \cdot 2 \cdot 3 + 2 \cdot 5 \cdot 6 - 8 \cdot 3 \cdot 3 + 4 \cdot 2 =$$ $$= 40 - 72 - 12 + 60 - 72 + 8 = -48.$$

And now, from Cramer's rule we can find $x$:

$$x = \frac{\Delta_{x}}{\Delta} = -\frac{48}{24} = -2.$$

Answer:

hence, $x = -2$.

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