Question: "Solve the simultaneous Linear equations using matrix inverse..."

Tyra

Answered question

2021-03-02

Solve the simultaneous Linear equations using matrix inverse Method.
8x+3y=2
6x+2y=4

Answer & Explanation

davonliefI

Skilled2021-03-03Added 79 answers

Step 1
To solve simultaneous linear equations, we have to find to form the equations into three matrices.
8x+3y=2
6x+2y=4
These can be shown in matrix form as,
$[\begin{matrix} 8 & 3 \\ 6 & 2 \end{matrix}] [\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} 2 \\ 4 \end{matrix}]$
Step 2
So if Ax=B , where A,x and B are matrices.
Then we can multiply it with $A^{- 1}$ to find x
$A^{- 1} \cdot A x = A^{- 1} \cdot B \Rightarrow x = A^{- 1} B$
So find $A^{- 1}, A = [\begin{matrix} 8 & 3 \\ 6 & 2 \end{matrix}], x = [\begin{matrix} x \\ y \end{matrix}], B = [\begin{matrix} 2 \\ 4 \end{matrix}]$
$[\begin{matrix} 8 & 3 & | & 1 & 0 \\ 6 & 2 & | & 0 & 1 \end{matrix}] R_{1} \Rightarrow \frac{R_{1}}{8}, R_{2} \Rightarrow R_{2} - 6 R_{1}$ ,

Applying these row operations we get,
$[\begin{matrix} 1 & \frac{3}{8} & | & \frac{1}{8} & 0 \\ 0 & \frac{- 1}{4} & | & \frac{- 3}{4} & 1 \end{matrix}] R_{2} \Rightarrow - 4 R_{2}, R_{1} \Rightarrow R_{1} - \frac{3}{8} R_{2}$
$[\begin{matrix} 1 & 0 & | & - 1 & \frac{3}{2} \\ 0 & 1 & | & 3 & - 4 \end{matrix}], So A^{- 1} = [\begin{matrix} - 1 & \frac{3}{2} \\ 3 & - 4 \end{matrix}]$
Step 3
$A^{- 1} A X = A^{- 1} B \Rightarrow x = A^{- 1} B$
$[\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} - 1 & \frac{3}{2} \\ 3 & - 4 \end{matrix}] [\begin{matrix} 2 \\ 4 \end{matrix}]$
$[\begin{matrix} x \\ y \end{matrix}] = [\begin{matrix} - 2 + \frac{3}{2} \times 4 \\ 6 + - 4 \times 4 \end{matrix}] = [\begin{matrix} - 2 + 6 \\ 6 - 16 \end{matrix}] = [\begin{matrix} 4 \\ - 10 \end{matrix}]$
x=5 , y=-10
Step 4
So the answers are x = 4 and y = -10

Jeffrey Jordon

Expert2022-01-27Added 2605 answers

Answer is given below (on video)

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