Solve three linear systems simultaneously, using an augmented matrix

Question

I have no idea how to start. Do I just get x1 x2 x3 x4 with b1, b2, and b3, then we use substitution method to get x1 x2 x3 x4? I don't really know ..

Answer 1

The first part of the question asks you to augment the matrix $A$ with the vectors $b_1$, $b_2$, and $b_3$. This is the matrix $$ X= \begin{bmatrix} A & \mid & b_1 & b_2 & b_3 \end{bmatrix} = \left[\begin{array}{rrrr|rrr} 4 & 2 & 1 & -3 & 4 & 0 & 4 \\ 1 & 2 & -1 & 0 & 2 & 1 & 2 \\ 3 & -1 & 2 & 4 & 8 & 8 & -7 \\ 0 & 2 & 4 & 3 & 9 & 4 & 5 \end{array}\right] $$ Performing row reductions gives the reduced row-echelon form $$ \DeclareMathOperator{rref}{rref}\rref X = \left[\begin{array}{rrrr|rrr} 1 & 0 & 0 & 0 & 1 & \frac{251}{265} & -1 \\ 0 & 1 & 0 & 0 & 1 & \frac{9}{265} & 2 \\ 0 & 0 & 1 & 0 & 1 & \frac{4}{265} & 1 \\ 0 & 0 & 0 & 1 & 1 & \frac{342}{265} & -1 \end{array}\right] $$ This reduced matrix tells us about the solutions to the equations \begin{align*} Ax &= b_1 & Ax &= b_2 & Ax &= b_3 \end{align*} The solution to $Ax=b_1$ is $$ \begin{bmatrix}x_1\\ x_2\\ x_3\\ x_4\end{bmatrix}= \begin{bmatrix}1\\1\\1\\1\end{bmatrix} $$ Can you see the solutions to the other two equations?

Answer 2

I think they mean to apply the standard procedure simultaneously to $b_1, b_2, b_3$ to get the three solution(s). Thus instead of a matrix augmented with one column, you'll have one augmented with three.