Question

Solve the linear equation : 2x+y-3z= 5,3 x-2y-2z= 5, and 5x-3y-z= 16.

Answer 1

ANSWER ON QUESTION #55014 – Math – Linear Algebra

Solve the linear equation $2x+y-3z=5,3x-2y-2z=5$, and $5x-3y-z=16$.

Solution

We shall solve the given system

$$\left\{ \begin{array}{l} 2x + y - 3z = 5, \\ 3x - 2y - 2z = 5, \\ 5x - 3y - z = 16 \end{array} \right.$$

with help of Cramer’s rule.

The determinants of this system are the following:

\Delta = \left| \begin{array}{ccc} 2 & 1 & -3 \\ 3 & -2 & -2 \\ 5 & -3 & -1 \end{array} \right| = 2 \left| \begin{array}{cc} -2 & -2 \\ -3 & -1 \end{array} \right| - \left| \begin{array}{ccc} 3 & -2 & -3 \\ 5 & -1 & -3 \end{array} \right| \begin{array}{cc} 3 & -2 \\ 5 & -3 \end{array} \right| = 2 $$= 2[( -2)( -1) - (-3)( -2) ] - [ 3(-1) - 5(-2) ] - 3[3(-3) - 5(-2) ] =$$ $$= 2(2 - 6) - (-3 + 10) - 3(-9 + 10) = -18 \neq 0$$

(hence, there exists a unique solution to the given system),

\Delta_x = \left| \begin{array}{ccc} 5 & 1 & -3 \\ 5 & -2 & -2 \\ 16 & -3 & -1 \end{array} \right| = 5 \left| \begin{array}{cc} -2 & -2 \\ -3 & -1 \end{array} \right| - \left| \begin{array}{ccc} 5 & -2 & -3 \\ 16 & -1 & -3 \end{array} \right| \begin{array}{cc} 5 & -2 \\ 16 & -3 \end{array} \right| = $$= 5[( -2)( -1) - (-3)( -2) ] - [5(-1) - 16(-2) ] - 3[5(-3) - 16(-2) ] = 5(2 - 6) - (-5 + 32) - 3(-15 + 32) = -98,$$ \Delta_y = \left| \begin{array}{ccc} 2 & 5 & -3 \\ 3 & 5 & -2 \\ 5 & 16 & -1 \end{array} \right| = 2 \left| \begin{array}{cc} 5 & -2 \\ 16 & -1 \end{array} \right| - 5 \left| \begin{array}{ccc} 3 & -2 & -3 \\ 5 & -1 & -3 \end{array} \right| \begin{array}{cc} 3 & 5 \\ 5 & 16 \end{array} \right| = $$= 2[5(-1) - 16(-2) ] - 5[3(-1) - 5(-2) ] - 3[3 \cdot 16 - 5 \cdot 5 ] = 2(-5 + 32) - 5(-3 + 10) - 3(48 - 25) = -50,$$ $$\Delta_z = \left| \begin{array}{ccc} 2 & 1 & 5 \\ 3 & -2 & 5 \\ 5 & -3 & 16 \end{array} \right| = 2 \left| \begin{array}{cc} -2 & 5 \\ -3 & 16 \end{array} \right| - \left| \begin{array}{cc} 3 & 5 \\ 5 & 16 \end{array} \right| + 5 \left| \begin{array}{cc} 3 & -2 \\ 5 & -3 \end{array} \right| =$$ $$= 2[( -2)16 - (-3)5 ] - [3 \cdot 16 - 5 \cdot 5 ] + 5[3 \cdot (-3) - 5(-2) ] = 2(-32 + 15) - (48 - 25) + 5(-9 + 10) = -52.$$

By Cramer’s rule, the unique solution to the given system is given by

$$x = \frac{\Delta_x}{\Delta}, \quad y = \frac{\Delta_y}{\Delta}, \quad z = \frac{\Delta_z}{\Delta}.$$

Therefore,

$$x = \frac{98}{18} = \frac{49}{9}, \quad y = \frac{50}{18} = \frac{25}{9}, \quad z = \frac{52}{18} = \frac{26}{9}$$

Answer: $x = \frac{49}{9}, \quad y = \frac{25}{9}, \quad z = \frac{26}{9}$.

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