Question

PLEASE HELP!! Which equation can be used to solve 2 6 0 1 * x1 x2 = 2 -3

Answer 1

Answer: d

Step-by-step explanation: on edge

Answer 2

Equation :

`\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}\0.5}&-3\\ 0&1\end{bmatrix}}\begin{bmatrix}2\\ -3\end{bmatrix}`

Explanation:

To isolate the following matrix, we will have to divide either by matrix 1, or the co - efficient of the matrix shown below. By doing so we will have to take the inverse of the co - efficient of that same matrix on the other side. In other words,

`\begin{bmatrix}x_1\\ x_2\end{bmatrix}` - Matrix which we have to isolate,

`\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}2&6\\ \:0&1\end{bmatrix}^(-1)\begin{bmatrix}2\\ -3\end{bmatrix}` - Equation used to solve the matrix

Now as you can see this equation is not any of the given options. That is as we have to simplify it a bit further,

`\begin{bmatrix}2&6\\ 0&1\end{bmatrix}^(-1) = \frac{1}{\det \begin{bmatrix}2&6\\ 0&1\end{bmatrix}}\begin{bmatrix}1&-6\\ -0&2\end{bmatrix} = (1)/(2)\begin{bmatrix}1&-6\\ -0&2\end{bmatrix} = \begin{bmatrix}(1)/(2)&-3\\ 0&1\end{bmatrix}`

We know that 1 / 2 can be replaced with 0.5, giving us the following equation to solve for x1 and x2,

`\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}\0.5}&-3\\ 0&1\end{bmatrix}}\begin{bmatrix}2\\ -3\end{bmatrix}`

As you can see our solution is option d.