Question

Solve the system of equations X+3y-2z=0,x-11y+14z=0,2x-y+4z=0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given system of equations are

$\rm \: x + 3y - 2z = 0$

$\rm \: x - 11y + 14z = 0$

$\rm \: 2x -y + 4z = 0$

The matrix representation of the equation is

$\rm \: \begin{gathered}\sf\left[\begin{array}{ccc} 1&3& - 2\\1& - 11&14\\2& - 1&4\end{array}\right]\end{gathered} \begin{gathered}\sf \left[\begin{array}{c}x\\y \\ z\end{array}\right] \end{gathered} = \begin{gathered}\sf \left[\begin{array}{c}0\\0\\0\end{array}\right] \end{gathered}$

where,

$\rm \: A = \begin{gathered}\sf\left[\begin{array}{ccc} 1&3& - 2\\1& - 11&14\\2& - 1&4\end{array}\right]\end{gathered}$

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

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

so that,

$\rm \: AX = B$

Now, Consider

$\rm \: |A|$

$\rm \: = \: \begin{gathered}\sf\left | \begin{array}{ccc} 1&3& - 2\\1& - 11&14\\2& - 1&4\end{array}\right | \end{gathered}$

$\rm \: = \: 1( - 44 + 14) - 3(4 - 28) - 2( - 1 + 22)$

$\rm \: = \: - 30 + 72 - 42$

$\rm \: = \: 0$

$\rm\implies \: |A| = 0$

It means, system of equations is consistent having infinitely many solutions.

Let assume that z = k, where k is some real number.

Now, we choose any two equations from given system of equations and substitute z = k.

So, the above two equations can be rewritten as

$\rm \: x + 3y = 2k$

and

$\rm \: 2x -y = - 4k$

So,

Matrix representation of the system is

$\rm \: \bigg[ \begin{matrix}1&3 \\ 2& - 1 \end{matrix} \bigg]\begin{gathered}\sf \left[\begin{array}{c}x\\y\end{array}\right] \end{gathered} = \begin{gathered}\sf \left[\begin{array}{c}2k\\ - 4k\end{array}\right] \end{gathered}$

Now, using Cramer's Rule,

Consider,

$\rm \: |A| = - 1 - 6 = - 7$

It means, system of equations is consistent having unique solution.

$\rm \: D_1 = \begin{array}{|cc|}\sf 2k &\sf 3 \\ \sf - 4k &\sf - 1 \\\end{array} = - 2k + 12k = 10k$

$\rm \: D_2 = \begin{array}{|cc|}\sf 1 &\sf 2k \\ \sf 2 &\sf - 4k \\\end{array} = - 4k - 4k = - 8k$

So,

$\rm\implies \:x = \dfrac{D_1}{ |A| } = \dfrac{10k}{ - 7} = - \dfrac{10k}{7}$

and

$\rm\implies \:y = \dfrac{D_2}{ |A| } = \dfrac{ - 8k}{ - 7} = \dfrac{8k}{7}$

So, solution of given system of equations is

$\rm \: x = - \dfrac{10k}{7} \: where \: k \: \in \: R$

$\rm \: y = \dfrac{8k}{7} \: where \: k \: \in \: R$

$\rm \: z = k \: where \: k \: \in \: R$

Answer 2

Step-by-step explanation:

Here the number of unknowns is 3 .

Transforming into echelon form (Gaussian elimination method), the augmented matrix becomes

$\left[\begin{array}{ccc|c}1 & 3 & -2 & 0 \\ 2 & -1 & 4 & 0 \\ 1 & -11 & 14 & 0\end{array}\right] \stackrel{R_{2} \rightarrow R_{2}-2 R_{1},}{R_{1} \rightarrow R_{1}-R_{1}}$

$\left[\begin{array}{ccc|c}1 & 3 & -2 & 0 \\ 0 & -7 & 8 & 0 \\ 0 & -14 & 16 & 0\end{array}\right] \stackrel{\begin{array}{l}R_{1} \rightarrow R_{1}+(-1) \\ R_{1} \rightarrow R_{1}+(-2)\end{array}}{ - - - - - - - - \longrightarrow}$