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 #41217 – Math – Linear Algebra

Question:

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

Solution:

Using the triangle rule we can find

$$\Delta = \begin{array}{ccc} 3 & -4 & 2 \\ 1 & 5 & -3 \\ 5 & 3 & -1 \end{array} = -15 + 6 + 60 - 50 - 4 + 27 = 24$$ $$\Delta_1 = \begin{array}{ccc} -8 & -4 & 2 \\ -2 & 5 & -3 \\ -6 & 3 & -1 \end{array} = -48$$ $$\Delta_2 = \begin{array}{ccc} 3 & -8 & 2 \\ 1 & -2 & -3 \\ 5 & -6 & -1 \end{array} = 72$$ $$\Delta_3 = \begin{array}{ccc} 3 & -4 & -8 \\ 1 & 5 & -2 \\ 5 & 3 & -6 \end{array} = 120$$ $$x = \frac{\Delta_1}{\Delta} = \frac{-48}{24} = -2$$ $$y = \frac{\Delta_2}{\Delta} = \frac{72}{24} = 3$$ $$z = \frac{\Delta_3}{\Delta} = \frac{120}{24} = 5$$

Answer:

$$x = -2$$

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