Final Practice Paper

1. $R_2 \leftarrow 2R_2 - R_1$: $(0, 3, -4 \mid -22)$. $R_3 \leftarrow 2R_3 - R_1$: $(0, -3, 0 \mid 6)$.
2. Row 3 gives $y = -2$; row 2: $-6 - 4z = -22 \Rightarrow z = 4$; row 1: $4x + 2 + 8 = 0 \Rightarrow x = -\tfrac52$.

Answer: Consistent, unique: $\left(-\tfrac52, -2, 4\right)$

1. Linearity only fixes $T$ on $\operatorname{Span}\{(1,0,1), (2,1,0)\}$. Try $a(1,0,1) + b(2,1,0) = (3, 2, 1)$:
2. the third entry forces $a = 1$, the second forces $b = 2$, but then the first is $1 + 4 = 5 \neq 3$.
3. $(3, 2, 1)$ is outside the span, so its image depends on how $T$ acts on the missing direction.
4. $(5, 3)$ is $T(3, 1, 1)$, the sum of the two given images, and $(9, 4)$ comes from forcing $a = 1$, $b = 2$ anyway.

Answer: Not determined: $(3, 2, 1)$ is not a combination of the two given vectors.

1. $R_2 - 2R_1$: $(0, 2, 1)$. $R_3 - R_1$: $(0, -4, 10)$. $R_3 + 2R_2$: $(0, 0, 12)$.
2. Three pivots ($\det = 24$), so the RREF is $I_3$ and $\mathbf x = \mathbf 0$ is the only solution.

Answer: Only the trivial solution; the columns are independent.

1. The columns are $T(\mathbf e_1), T(\mathbf e_2), T(\mathbf e_3)$: $(1, 2, 0)$, $(0, -1, 3)$, $(5, 0, -1)$.

Answer: $A = \begin{bmatrix}1 & 0 & 5 \\ 2 & -1 & 0 \\ 0 & 3 & -1\end{bmatrix}$

1. $\det A = 1(1 - 0) - 0 + 5(6 - 0) = 31 \neq 0$.
2. For a square standard matrix, invertible means both one-to-one and onto.

Answer: Both.

1. $T(\mathbf e_1) = (1, 0, 0, 1)$, $T(\mathbf e_2) = (-1, 1, 0, 1)$, $T(\mathbf e_3) = (1, 1, 0, 3)$, $T(\mathbf e_4) = (0, 1, 0, 2)$ are the columns.

Answer: $A = \begin{bmatrix}1 & -1 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 1 & 1 & 3 & 2\end{bmatrix}$

1. Row 3 is zero and $R_4 - R_1 - 2R_2 = 0$, so rank $A = 2 < 4$.
2. Free variables exist (not one-to-one) and not every row has a pivot (not onto).

Answer: Neither; $A$ is not invertible.

1. Area $= \left|\det\begin{bmatrix}2 & 1 \\ 3 & 4\end{bmatrix}\right| = |8 - 3| = 5$.

Answer: 5

1. $(-2I + C^{-1})C = -2C + I$.
2. $B(C - B^{-1} + 2B^{-1}C) = BC - I + 2C$.
3. Sum: $-2C + I + BC - I + 2C = BC$.

Answer: $BC$

1. $A$ is lower triangular: $3 \cdot 2 \cdot (-5) = -30$.
2. Expand $B$ down column 1: $2\cdot\begin{vmatrix}4 & -1 \\ 2 & 0\end{vmatrix} = 2(0 + 2) = 4$.
3. $\det(AB) = (-30)(4) = -120$; $\det(2A) = 2^3(-30) = -240$.
4. $\det(B^{-1}A^3B^T) = \tfrac{1}{\det B}(\det A)^3\det B = (-30)^3 = -27000$.

Answer: −30, 4, −120, −240, −27000

1. $\det A = 2(12 - 12) - 4(4 - 0) + 6(2 - 0) = -4 \neq 0$: three pivots.
2. Col $A = \mathbb R^3$ with the three columns as a basis; Nul $A = \{\mathbf 0\}$ (empty basis).

Answer: Col basis: the columns of $A$; Nul $A = \{\mathbf 0\}$

1. rank $3$, nullity $3 - 3 = 0$. $\det A^T = \det A = -4 \neq 0$, so both are invertible.

Answer: rank 3, nullity 0, both invertible

1. $A\mathbf v = (-4 + 4,\ -2 + 3,\ 0 + 2) = (0, 1, 2) \neq \mathbf 0$, so $\mathbf v \notin \operatorname{Nul}A$.
2. Any column works for Col $A$, e.g. $(2, 1, 0)$; so does $A\mathbf v = (0, 1, 2)$. (Col $A$ is all of $\mathbb R^3$.)

Answer: No; for example $(2, 1, 0) \in \operatorname{Col}A$.

1. Upper triangular: the eigenvalues are the diagonal entries.

Answer: 5, 3, 2

1. Three distinct eigenvalues for a $3\times3$ matrix guarantee three independent eigenvectors.

Answer: Yes.

1. $A - 5I = \begin{bmatrix}0 & 4 & 0 \\ 0 & -2 & 2 \\ 0 & 0 & -3\end{bmatrix}$ forces $x_3 = 0$ and $x_2 = 0$, leaving $x_1$ free.

Answer: $(1, 0, 0)$

1. $\mathbf v_1 = \mathbf x_1 = (1, 1, 0)$.
2. $\mathbf x_2\cdot\mathbf v_1 = 1 - 1 + 0 = 0$, so $\mathbf v_2 = \mathbf x_2 = (1, -1, 1)$ already.

Answer: $\{(1, 1, 0),\ (1, -1, 1)\}$

1. Normalize: $\mathbf q_1 = \tfrac{1}{\sqrt2}(1, 1, 0)$, $\mathbf q_2 = \tfrac{1}{\sqrt3}(1, -1, 1)$.
2. $R = Q^TA$: $R_{11} = \mathbf q_1\cdot\mathbf x_1 = \sqrt2$, $R_{12} = \mathbf q_1\cdot\mathbf x_2 = 0$, $R_{22} = \mathbf q_2\cdot\mathbf x_2 = \sqrt3$.

Answer: $Q = \begin{bmatrix}\tfrac{1}{\sqrt2} & \tfrac{1}{\sqrt3} \\ \tfrac{1}{\sqrt2} & -\tfrac{1}{\sqrt3} \\ 0 & \tfrac{1}{\sqrt3}\end{bmatrix},\ R = \begin{bmatrix}\sqrt2 & 0 \\ 0 & \sqrt3\end{bmatrix}$