Answer:
[tex]X=\begin{bmatrix}5\\ 14\\ -10\end{bmatrix}[/tex]
Step-by-step explanation:
Our approach here is to isolate X, and simplify this solution. We want to begin by subtracting matrix 2, as shown below, from either side - the first step in isolating X. Afterwards we can multiply either side by the inverse of matrix 1, the co - efficient of X, such that X is now isolated. We can then simplify this value.
Given,
[tex]\begin{bmatrix}1&2&3\\ -3&5&5\\ \:\:\:3&-2&-1\end{bmatrix}[/tex] : Matrix 1
[tex]\begin{bmatrix}3\\ -1\\ 8\end{bmatrix}[/tex] : Matrix 2
[tex]\begin{bmatrix}1&2&3\\ -3&5&5\\ 3&-2&-1\end{bmatrix}X+\begin{bmatrix}3\\ -1\\ 8\end{bmatrix}=\begin{bmatrix}6\\ 4\\ 5\end{bmatrix}[/tex] ( Subtract Matrix 2 from either side )
[tex]\begin{bmatrix}1&2&3\\ -3&5&5\\ 3&-2&-1\end{bmatrix}X=\begin{bmatrix}6\\ 4\\ 5\end{bmatrix}-\begin{bmatrix}3\\ -1\\ 8\end{bmatrix}[/tex] ( Simplify )
[tex]\begin{bmatrix}6\\ 4\\ 5\end{bmatrix}-\begin{bmatrix}3\\ -1\\ 8\end{bmatrix} = \begin{bmatrix}6-3\\ 4-\left(-1\right)\\ 5-8\end{bmatrix}=\begin{bmatrix}3\\ 5\\ -3\end{bmatrix}[/tex] ( Substitute )
[tex]\begin{bmatrix}1&2&3\\ -3&5&5\\ 3&-2&-1\end{bmatrix}X=\begin{bmatrix}3\\ 5\\ -3\end{bmatrix}[/tex] ( Multiply either side by inverse of Matrix 1 )
[tex]X=\begin{bmatrix}1&2&3\\ -3&5&5\\ 3&-2&-1\end{bmatrix}^{-1}\begin{bmatrix}3\\ 5\\ -3\end{bmatrix}=\begin{bmatrix}5\\ 14\\ -10\end{bmatrix}[/tex] - let's say that this is Matrix 3. Our solution would hence be Matrix 3.
