Name: Solving System of Linear Equations using Matrices
Uploaded: 2019-07-19
Duration: 11 min 30 s
Description: A system of linear equations can be represented in matrix form using its coefficient, variable and constant matrices. Learn about the inverse matrix method!

Question

If $A = (\begin{matrix} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ 0 & 3 & - 5 \end{matrix})$ , find $A^{- 1}$ . Using $A^{- 1}$ , solve the following system of linear equations: $A^{T} = (\begin{matrix} 2 & 1 & 0 \\ 1 & - 2 & 3 \\ 1 & - 1 & - 5 \end{matrix})$ ; $x - 2 y - z = \frac{3}{2}$ ; $3 y - 5 z = 9$ .

Accepted Answer

Given that: $A = (\begin{matrix} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ 0 & 3 & - 5 \end{matrix})$

The determinant of matrix $A$ is $|A| = |\begin{matrix} 2 & - 1 & 1 \\ 1 & - 2 & - 1 \\ 0 & 3 & - 5 \end{matrix}|$

$|A| = 2 (- 2 \times - 5 - (- 1) \times 3) - (1 \times - 5 - (- 1) \times 0) + (1 \times 3 - (- 2) \times 0)$

$|A| = 2 (10 + 3) - (- 5) + (3)$ $= 26 + 5 + 3$ $= 34$

$| A | \neq 0$

$∴$ $A^{- 1}$ is possible.

$A^{T} = (\begin{matrix} 2 & 1 & 0 \\ 1 & - 2 & 3 \\ 1 & - 1 & - 5 \end{matrix})$

$Adj (A) = (\begin{matrix} 13 & 5 & 3 \\ 8 & - 10 & - 6 \\ 1 & 3 & - 5 \end{matrix})$

We know that $A^{- 1} = \frac{1}{| α |} a d j (A)$

$∴ A^{- 1} = \frac{1}{34} (\begin{matrix} 13 & 5 & 3 \\ 8 & - 10 & - 6 \\ 1 & 3 & - 5 \end{matrix})$

Given set of lines are:

$2 x + y + z = 1$ ,

$x - 2 y - z = \frac{3}{2}$ ,

$3 y - 5 z = 9$

Converting the following equations in matrix form: $A X = B$

Where $A = (\begin{matrix} 2 & 1 & 1 \\ 1 & - 2 & - 1 \\ 0 & 3 & - 5 \end{matrix}), X = (\begin{matrix} x \\ y \\ z \end{matrix}), B = (\begin{matrix} 1 \\ \frac{3}{2} \\ 9 \end{matrix})$

Pre - multiplying by $A^{- 1}$

$A^{- 1} A X = A^{- 1} B$

$I X = A^{- 1} B$

$X = A^{- 1} B$

$[\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{34} [\begin{matrix} 13 & 8 & 1 \\ 5 & - 10 & 3 \\ 3 & - 6 & - 5 \end{matrix}] [\begin{array}{l} 1 \\ 3 \\ \frac{2}{2} \\ 9 \end{array}]$

$[\begin{array}{l} x \\ y \\ z \end{array}] \frac{1}{34} [\begin{matrix} 1 \times 13 + \frac{3}{2} \times 8 + 9 \times 1 \\ 1 \times 5 + \frac{3}{2} \times - 10 + 9 \times 3 \\ 1 \times 3 + \frac{3}{2} \times - 6 + 9 \times - 5 \end{matrix}]$

$[\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{34} [\begin{matrix} 13 + 12 + 9 \\ 5 - 15 + 27 \\ 3 - 9 - 45 \end{matrix}]$

$[\begin{array}{l} x \\ y \\ z \end{array}] = \frac{1}{34} [\begin{matrix} 34 \\ 17 \\ - 51 \end{matrix}] = [\begin{matrix} \frac{1}{2} \\ \frac{1}{2} \\ - \frac{2}{2} \end{matrix}]$

$∴$ $x = 1, y = \frac{1}{2}, z = \frac{- 3}{2}$ .

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