Question

Solve the following equations by cramer's method
3x+y-4z=0, -x+2y+z=4 and 2x-y+3z=8

Answer 1

Given Question :-

Solve the following equations by cramer's method

• 3x+y-4z=0
• -x+2y+z=4
• 2x-y+3z=8

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Step by Step explanation :-

The system of equations are

3x + y - 4z = 0

- x + 2y + z = 4

2x - y + 3z = 8

☆ The matrix form of the above equation are as follow : -

$\begin{gathered}\sf\left[\begin{array}{ccc}3&1& - 4\\ - 1&2&1\\2&-1&3\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered} = \begin{gathered}\sf \left[\begin{array}{c}0\\4\\8\end{array}\right]\end{gathered}$

where,

$\begin{gathered}\sf A=\left[\begin{array}{ccc}3&1& - 4\\ - 1&2&1\\2&-1&3\end{array}\right]\end{gathered}$

$\begin{gathered}\sf B=\left[\begin{array}{c}0\\4\\8\end{array}\right]\end{gathered}$

$\begin{gathered}\sf X=\left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered}$

Now,

$\tt \:  \longrightarrow \: |A| = \begin{vmatrix} 3 & 1 & - 4 \\ - 1 & 2 & 1 \\ 2 & - 1 & 3 \end{vmatrix}$

$\tt\implies \: |A| = 3(6 + 1) - 1( - 3 - 2) - 4(1 - 4)$

$\tt\implies \: |A| \: = 21 + 5 + 12 = 38$

$\tt\implies \:system \: is \: consistent \: having \: unique \: solution.$

$\tt \:  \longrightarrow \:Consider \: D_1 \: = \begin{vmatrix}0 & 1 & - 4\\ 4 & 2 & 1 \\ 8 & - 1 & 3 \end{vmatrix}$

$\tt\implies \: D_1 = - 1(12 - 8) - 4( - 4 - 16)$

$\tt\implies \:D_1 = 76$

$\tt \:  \longrightarrow \:\:Consider \: D_2 \: = \begin{vmatrix} 3 & 0 & - 4 \\ - 1 & 4 & 1 \\ 2 & 8 & 3 \end{vmatrix}$

$\tt\implies \:D_2 = 3(12 - 8) - 4( - 8 - 8)$

$\tt\implies \:D_2 \: = 76$

$\tt \:  \longrightarrow \:Consider \: D_3 \: = \begin{vmatrix} 3 & 1 & 0 \\ - 1& 2 & 4 \\ 2 & - 1 & 8 \end{vmatrix}$

$\tt\implies \:D_3 \: = 3(16 + 4) - 1( - 8 - 8)$

$\tt\implies \:D_3 \: = 76$

$\tt \:  \longrightarrow \: \therefore \: x \: = \dfrac{ D_1}{ |A| } = \dfrac{76}{38} = 2$

$\tt \:  \longrightarrow \: \therefore \: y \: = \dfrac{ D_2}{ |A| } = \dfrac{76}{38} = 2$

$\tt \:  \longrightarrow \: \therefore \: z \: = \dfrac{ D_3}{ |A| } = \dfrac{76}{38} = 2$