There are 2 matrices and 1 vector: $$ A = \begin{bmatrix} 5 & -7 & -4 & 9 \\ 6 & -8 & -7 & 5 \\ 4 & -4 & -9 & -9 \\ -9 & 11 & 16 & 7 \end{bmatrix}, \quad B = \begin{bmatrix} 9 & 11 & -6 & -7 & 13 \\ -7 & -8 & 5 & 6 & -9 \\ 11 & 7 & -7 & -9 & -6 \\ -3 & 4 & 1 & 8 & 7 \end{bmatrix}, \quad v = \begin{bmatrix} 4 \\ 3 \\ 2 \\ 1 \end{bmatrix} $$
- For each $b \in \mathbb{R}^4$, does the equation $Bx = b$ have a solution? Row Reduced $B$ =\begin{bmatrix} 1 & 0 & -7/13 & 0 & 0 \\ 0 & 1 & -2/13 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}
- Solve the equation $Ax = v$ by reducing the augmented matrix corresponding to this linear system. Row Reduced $Ax = v$ =\begin{bmatrix} 1 & 0 & 0 & 4/3 & 0 \\ 0 & 1 & 0 & -5/3 & 0\\ 0 & 0 & 1 & 7/3 & 0\\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}
- Solve the equation $Bx = v$ by reducing the augmented matrix corresponding to this linear system. Row Reduced $Bx = v$ =$$ \left[\begin{array}{ccccc|c} 1 & 0 & -7/13 & 0 & 0 & -1525/156\\ 0 & 1 & -2/13 & 0 & 0 &1115/156\\ 0 & 0 & 0 & 1 & 0 &-5\\ 0 & 0 & 0 & 0 & 1 & -29/12 \end{array} \right] $$ This content is very new to me. I'm good at getting it to the row-reduced echelon form but solving and interpreting it from that point, not so much. I would appreciate any help at all.
Solve.
(1) For reduced matrix B, we know there are four rows and five columns, so either there is no solution due to the contradiction, or there will be infinitely many solutions due to the free variable. It's easy to determine that there is a free variable, which would be $x_3$. So yes, there are infinitely many solutions.
(2) For the second question, I would like to remind you that $v$ should not be a $4\times 5$ matrix, which will be indicating that $x$ is a $4\times5$ matrix. So I suppose you want to show that the reduced row echelon form of the augmented matrix is the matrix you put in the question. However, the answer is obvious.
There is no such a $x$ matrix satisfies this equation because you can never satisies the forth row, which is $0=1$.
(3) Since we can't have a row which has a pivot located at the third colomn and the elements in the third colomn are not always 0, so we will have the free variable. The solution will be:
\begin{align} x_1 &= \frac{7}{13}\cdot x_3 - \frac{1525}{156}\\ x_2 &= \frac{2}{13}\cdot x_3 + \frac{1115}{156}\\ x_4 &= -5\\ x_5 &= -\frac{29}{12} \end{align}
Hope this will help you. Thank you!