Answer:

Explanation:
To find x₁ and x₂ :
![\left[\begin{array}{ccc}-4&1\\5&4\\\end{array}\right] * \left[\begin{array}{ccc}x_1\\x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\\\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/29grxggke49jejd32uccy9druaxgmgsh54.png)
Step 1
Multiply first 2 x 2 matrix with 2 x 1 vector, we get
![\left[\begin{array}{ccc}-4x_1&+ x_2\\5x_1&+ 4x_2\\\end{array}\right] + \left[\begin{array}{ccc}11\\-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/4d7shcn6g6r734ogjh3g2sjg84ocq4f9jz.png)
Step 2
Add the 2 x 1 matrices on LHS, we get
![\left[\begin{array}{ccc}-4x_1&+x_2&+11\\5x_1&+4x_2&-19\\\end{array}\right] = \left[\begin{array}{ccc}-11\\40\end{array}\right]](https://img.qammunity.org/2021/formulas/mathematics/college/sxx9fxut0qgb5z0iuime6sdn1g84cze7qu.png)
Step 3,
we get

and

Step 4,
Simplify, we get

Step 5,
multiply eqn(1) by 4
we get

Step 6,
eqn (2) - eqn(3)
we get

substituting in eqn (1), we get

so, we get

Therefore
