Solve the following system of linear equations by determinant method
$x + y + 3 z = 4; 2 x + 2 y + 6 z = 7; 2 x + y + z = 10$ .

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Solution

By determinant method , the value of $D$ is $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 3 2 & 2 & 6 2 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
which implies the value of $D = 0$

Observe first $2$ equations,
$x + y + 3 z = 4$ and $x + y + 3 z = \frac{7}{2}$
They differ only in the constant term. It shows that there is no solution for given set of equations

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x + y + 2 z = 4; 2 x + 2 y + 4 z = 8; 3 x + 3 y + 6 z = 10

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x + y - 2 z = 0

3 x + 2 y - 3 z = 4

x + y + 2 z = 4; 2 x + 2 y + 4 z = 8; 3 x + 3 y + 6 z = 12

Solve the following system of linear equations, using matrix method

$2 x + 3 y + 3 z = 5, x - 2 y + z = - 4, 3 x - y - 2 z = 3$

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