Question

Find z 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 # 41216 – Math – Linear Algebra

Find $z$ 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.

$$\left\{ \begin{array}{l} 3x - 4y + 2z + 8 = 0 \\ x + 5y - 3z + 2 = 0 \\ 5x + 3y - z + 6 = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{ccc} 3 & -4 & 2 \\ 1 & 5 & -3 \\ 5 & 3 & -1 \end{array} \right\} \binom{x}{y} = \binom{-8}{-2};$$

Hence:

$$z = \left| \begin{array}{ccc} 3 & -4 & -8 \\ 1 & 5 & -2 \\ 5 & 3 & -6 \\ \hline 3 & -4 & 2 \\ 1 & 5 & -3 \\ 5 & 3 & -1 \end{array} \right| = \frac{3 \cdot 5 \cdot (-6) + 1 \cdot 3 \cdot (-8) + 5 \cdot (-4) \cdot (-2) - 5 \cdot 5 \cdot (-8) - 1 \cdot (-4) \cdot (-6) - 3 \cdot 3 \cdot (-2)}{3 \cdot 5 \cdot (-1) + 1 \cdot 3 \cdot 2 + 5 \cdot (-4) \cdot (-3) - 5 \cdot 5 \cdot 2 - 1 \cdot (-4) \cdot (-1) - 3 \cdot 3 \cdot (-3)} = \frac{-90 - 24 + 40 + 200 - 24 + 18}{-15 + 6 + 60 - 50 - 4 + 27} = \frac{120}{24} = 5.$$

Answer.

$z = 5$

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