1. Systems of linear equations(a) Consider the following system of linear equationsz − 4w = −4y − 3z + 14 = 133x + z + w = 1x + w = 2.Determine whether or not the system is consistent, and if it is, give a particular solution.[10 Marks](b) Compute the reduced row echelon form of the matrix0 0 1 1 11 3 −2 0 −12 6 −3 1 0 .[5 Marks]Suppose a system of linear equations is represented as an augmented matrix with reduced row echelon form1 0 4 00 1 1 00 0 0 10 0 0 0 .(c) Is the system consistent?[5 Marks](d) Give a basis for the solution set of the corresponding homogeneous system.[5 Marks][End of Question 1: Total 25 marks]COMP1043-MOCK Turn over

(a) To determine whether the system of linear equations is consistent and to find a particular solution, we can use the method of substitution or elimination. Let's write down the system of equations:

1. $z - 4w = -4$
2. $y - 3z + 14 = 13$
3. $3x + z + w = 1$
4. $x + w = 2$

First, solve equation (4) for $x$: $x = 2 - w$

Next, substitute $x = 2 - w$ into equation (3): $3(2 - w) + z + w = 1$ $6 - 3w + z + w = 1$ $6 - 2w + z = 1$ $z - 2w = -5$

Now we have two equations involving $z$ and $w$:

1. $z - 4w = -4$
2. $z - 2w = -5$

Subtract the second equation from the first: $(z - 4w) - (z - 2w) = -4 - (-5)$ $-2w = 1$ $w = -\frac{1}{2}$

Substitute $w = -\frac{1}{2}$ back into the equation $z - 2w = -5$: $z - 2(-\frac{1}{2}) = -5$ $z + 1 = -5$ $z = -6$

Now substitute $z = -6$ and $w = -\frac{1}{2}$ into equation (2) to find $y$: $y - 3(-6) + 14 = 13$ $y + 18 + 14 = 13$ $y + 32 = 13$ $y = 13 - 32$ $y = -19$

Finally, substitute $w = -\frac{1}{2}$ into $x = 2 - w$: $x = 2 - (-\frac{1}{2})$ $x = 2 + \frac{1}{2}$ $x = \frac{5}{2}$

Thus, the particular solution is: $x = \frac{5}{2}, y = -19, z = -6, w = -\frac{1}{2}$

The system is consistent.

(b) To compute the reduced row echelon form (RREF) of the matrix: $\begin{pmatrix} 0 & 0 & 1 & 1 & 1 \\ 1 & 3 & -2 & 0 & -1 \\ 2 & 6 & -3 & 1 & 0 \end{pmatrix}$

First, perform row operations to get leading 1s and zeros below and above them.

1. Swap $R1$ and $R2$: $\begin{pmatrix} 1 & 3 & -2 & 0 & -1 \\ 0 & 0 & 1 & 1 & 1 \\ 2 & 6 & -3 & 1 & 0 \end{pmatrix}$

2. Replace $R3$ with $R3 - 2R1$: $\begin{pmatrix} 1 & 3 & -2 & 0 & -1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 2 \end{pmatrix}$

3. Replace $R3$ with $R3 - R2$: $\begin{pmatrix} 1 & 3 & -2 & 0 & -1 \\ 0 & 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$

4. Replace $R2$ with $R2 - R3$: $\begin{pmatrix} 1 & 3 & -2 & 0 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$

5. Replace $R1$ with $R1 + 2R2$: $\begin{pmatrix} 1 & 3 & 0 & 2 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$

6. Replace $R1$ with $R1 - 3R2$: $\begin{pmatrix} 1 & 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$

The reduced row echelon form is: $\begin{pmatrix} 1 & 0 & 0 & -1 & -1 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}$

(c) To determine if the system is consistent, examine the augmented matrix in reduced row echelon form: $\begin{pmatrix} 1 & 0 & 4 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

The last row indicates $0 = 1$, which is a contradiction. Therefore, the system is inconsistent.

(d) For the corresponding homogeneous system, the augmented matrix would be: $\begin{pmatrix} 1 & 0 & 4 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

The solution set can be written in terms of free variables. Let $x_3 = t$, where $t$ is a free variable. Then: $x_1 = -4t$ $x_2 = -t$ $x_3 = t$

The basis for the solution set is: $\{ (-4, -1, 1) \}$