Name: Properties of Determinants (Part 3)
Uploaded: 2019-07-19
Duration: 3 min 53 s
Description: Suppose we multiply a row of a determinant with a particular value, then what would be the change in the value of the determinant? Watch this video to learn...

Question

Using matrix method solve the following system of linear equations :

$x - y + z = 4, 2 x + y - 3 z = 0 & x + y + z = 2$

Accepted Answer

Given that, $x - y + z = 4, 2 x + y - 3 z = 0 & x + y + z = 2$ .

This can be written as $A X = B$ where $A = [\begin{matrix} 1 & - 1 & 1 \\ 2 & 1 & - 3 \\ 1 & 1 & 1 \end{matrix}], X = [\begin{array}{l} x \\ y \\ z \end{array}], B = [\begin{array}{l} 4 \\ 0 \\ 2 \end{array}]$

Here, $| A | = 1 (1 + 3) + 1 (2 + 3) + 1 (2 - 1)$ $= 10$

Since, $| A | \neq 0$

Hence, the system of equations is consistent and has a unique solution given by $X = A^{- 1} B$

$A^{- 1} = \frac{adj A}{| A |}$ and $adj A = C^{T}$

$C_{11} = (- 1)^{1 + 1} |\begin{matrix} 1 & - 3 \\ 1 & 1 \end{matrix}| = 1 + 3 = 4$

$C_{12} = (- 1)^{1 + 2} |\begin{matrix} 2 & - 3 \\ 1 & 1 \end{matrix}| = - (2 + 3) = - 5$

$C_{13} = (- 1)^{1 + 3} |\begin{array}{l} 2 & 1 \\ 1 & 1 \end{array}| = 2 - 1 = 1$

$C_{21} = (- 1)^{2 + 1} |\begin{matrix} - 1 & 1 \\ 1 & 1 \end{matrix}| = 2$

$C_{22} = (- 1)^{2 + 2} |\begin{array}{l} 1 & 1 \\ 1 & 1 \end{array}| = 0$

$C_{23} = (- 1)^{2 + 3} |\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}| = - 2$

$C_{31} = (- 1)^{3 + 1} |\begin{matrix} - 1 & 1 \\ 1 & - 3 \end{matrix}| = 2$

$C_{32} = (- 1)^{3 + 2} |\begin{matrix} 1 & 1 \\ 2 & - 3 \end{matrix}| = 5$

$C_{33} = (- 1)^{3 + 3} |\begin{matrix} 1 & - 1 \\ 2 & 1 \end{matrix}| = 3$

Therefore, cofactor matrix is $C = [\begin{matrix} 4 & - 5 & 1 \\ 2 & 0 & - 2 \\ 2 & 5 & 3 \end{matrix}]$

$\Rightarrow adj A = C^{T} = [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]$

$\Rightarrow A^{- 1} = \frac{adj A}{| A |} = \frac{1}{10} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}]$

Solution is given by $X = A^{- 1} B$

$[\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{10} [\begin{matrix} 4 & 2 & 2 \\ - 5 & 0 & 5 \\ 1 & - 2 & 3 \end{matrix}] [\begin{array}{l} 4 \\ 0 \\ 2 \end{array}]$

$= \frac{1}{10} [\begin{matrix} 16 + 4 \\ - 20 + 10 \\ 4 + 6 \end{matrix}]$

$= \frac{1}{10} [\begin{matrix} 20 \\ - 10 \\ 10 \end{matrix}]$

$= [\begin{matrix} 2 \\ - 1 \\ 1 \end{matrix}]$

Therefore, $x = 2, y = - 1, z = 1$

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