Quiz 1 (Term 242)

1. Augmented matrix: $$\left[\begin{array}{ccc|c}1 & 1 & 2 & 3 \\ 1 & 2 & 1 & 1 \\ 2 & 1 & 1 & 0\end{array}\right]$$
2. $R_2 \leftarrow R_2 - R_1$ and $R_3 \leftarrow R_3 - 2R_1$: $$\left[\begin{array}{ccc|c}1 & 1 & 2 & 3 \\ 0 & 1 & -1 & -2 \\ 0 & -1 & -3 & -6\end{array}\right]$$
3. $R_3 \leftarrow R_3 + R_2$: the last row is $-4z = -8$, so $z = 2$.
4. Back-substitute: $y - z = -2 \Rightarrow y = 0$, then $x + y + 2z = 3 \Rightarrow x = -1$.
5. Three pivots and no free variables, so the solution is unique. RREF: $$\left[\begin{array}{ccc|c}1 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 2\end{array}\right]$$

Answer: $(x, y, z) = (-1,\ 0,\ 2)$

1. Augmented matrix: $$\left[\begin{array}{cccc|c}1 & 1 & 1 & -3 & 1 \\ 2 & 1 & -1 & 1 & -1\end{array}\right]$$
2. $R_2 \leftarrow R_2 - 2R_1$, then $R_2 \leftarrow -R_2$, then $R_1 \leftarrow R_1 - R_2$: $$\left[\begin{array}{cccc|c}1 & 0 & -2 & 4 & -2 \\ 0 & 1 & 3 & -7 & 3\end{array}\right]$$
3. Pivots in the $x$ and $y$ columns: $x, y$ are basic, $z, t$ are free.
4. $x = -2 + 2z - 4t$ and $y = 3 - 3z + 7t$.
5. Parametric vector form: $$\begin{bmatrix}x\\y\\z\\t\end{bmatrix} = \begin{bmatrix}-2 \\ 3 \\ 0 \\ 0\end{bmatrix} + z\begin{bmatrix}2 \\ -3 \\ 1 \\ 0\end{bmatrix} + t\begin{bmatrix}-4 \\ 7 \\ 0 \\ 1\end{bmatrix}$$

Answer: Infinitely many solutions, two free variables.

1. Each condition is a linear equation in $a, b, c$: $$\begin{cases}a - b + c = 5 \\ a + b + c = 1 \\ 4a + 2b + c = 2\end{cases}$$
2. Subtract the first from the second: $2b = -4$, so $b = -2$ and $a + c = 3$.
3. The third becomes $4a + c = 6$. Subtract $a + c = 3$: $3a = 3$, so $a = 1$ and $c = 2$.

Answer: $P(x) = x^2 - 2x + 2$, so $(a, b, c) = (1, -2, 2)$.

1. Augmented matrix: $$\left[\begin{array}{ccc|c}1 & 0 & 2 & 1 \\ 0 & -1 & 1 & 2 \\ 1 & -2 & 0 & 1\end{array}\right]$$
2. $R_3 \leftarrow R_3 - R_1$: $(0, -2, -2 \mid 0)$. Then $R_3 \leftarrow R_3 - 2R_2$: $(0, 0, -4 \mid -4)$, so $z = 1$.
3. $-y + z = 2 \Rightarrow y = -1$, and $x + 2z = 1 \Rightarrow x = -1$.

Answer: $(x, y, z) = (-1, -1, 1)$, a unique solution.

1. Swap the rows and apply $R_2 \leftarrow R_2 - 2R_1$: $$\left[\begin{array}{cccc|c}1 & 1 & 1 & -3 & 1 \\ 0 & 0 & 0 & 0 & -1\end{array}\right]$$
2. The second row reads $0 = -1$.

Answer: The system is inconsistent: no solution.

1. $\det\!\begin{bmatrix}1 & 2 & 4 \\ 5 & 6 & 2 \\ 3 & 4 & 5\end{bmatrix} = -8 \neq 0$ and $\det\!\begin{bmatrix}3 & 4 & 5 \\ 2 & 2 & -3 \\ 2 & 4 & 8\end{bmatrix} = 16 \neq 0$.
2. A square matrix with nonzero determinant has a pivot in every column, so both reduce to $I_3$.
3. Row operations are reversible: reduce the first matrix to $I_3$, then apply the inverse of each step that took the second matrix to $I_3$, in reverse order. That chain turns the first matrix into the second.

Answer: Yes: both matrices are row equivalent to $I_3$, hence to each other.

1. Coefficient determinant: $\det\begin{bmatrix}1 & a \\ a & 1\end{bmatrix} = 1 - a^2$.
2. If $a \neq \pm 1$ the determinant is nonzero: one solution, $x = y = \dfrac{b}{1 + a}$.
3. If $a = 1$ both equations are $x + y = b$: consistent for every $b$, with infinitely many solutions.
4. If $a = -1$: $x - y = b$ and $-x + y = b$. Adding them gives $0 = 2b$, so the system is inconsistent when $b \neq 0$ and has infinitely many solutions when $b = 0$.

Answer: Unique for $a \neq \pm 1$ (any $b$). Inconsistent exactly when $a = -1$ and $b \neq 0$.