Question

PLEASE HELP !! (2/5) -50 POINTS-

Answer 1

`X=\begin{bmatrix}5\\ 14\\ -10\end{bmatrix}`

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,

`\begin{bmatrix}1&2&3\\ -3&5&5\\ \:\:\:3&-2&-1\end{bmatrix}` : Matrix 1

`\begin{bmatrix}3\\ -1\\ 8\end{bmatrix}` : Matrix 2

`\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}` ( Subtract Matrix 2 from either side )

`\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}` ( Simplify )

`\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}` ( Substitute )

`\begin{bmatrix}1&2&3\\ -3&5&5\\ 3&-2&-1\end{bmatrix}X=\begin{bmatrix}3\\ 5\\ -3\end{bmatrix}` ( Multiply either side by inverse of Matrix 1 )

`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}` - let's say that this is Matrix 3. Our solution would hence be Matrix 3.