Final Review Question Bank

1. $$\begin{bmatrix}1 & a \\ a & 9\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}4 \\ b\end{bmatrix}$$

Answer: $A\mathbf x = \mathbf b$ with $A = \begin{bmatrix}1 & a \\ a & 9\end{bmatrix}$

1. $-a\cdot\text{Eq}_1 + \text{Eq}_2$: $(9 - a^2)x_2 = b - 4a$.
2. Unique when $9 - a^2 \neq 0$.

Answer: $a \neq \pm 3$

1. Need $9 - a^2 = 0$ and $b - 4a = 0$ together: $b = 4a$.

Answer: $(3, 12)$ and $(-3, -12)$

1. $R_3 \leftarrow R_3 - 2R_1$: $(0, 3, a - 2 \mid b - 6)$. Then $R_3 \leftarrow aR_3 - 3R_2$:
2. $$\left[\begin{array}{ccc|c}1 & 2 & 1 & 3 \\ 0 & a & 5 & 10 \\ 0 & 0 & a^2-2a-15 & ab-6a-30\end{array}\right]$$
3. $a^2 - 2a - 15 = (a - 5)(a + 3)$. (At $a = 0$ the scaling step is not allowed, but then the matrix still has three pivots.)

Answer: Unique for $a \neq -3$ and $a \neq 5$

1. Need $a^2 - 2a - 15 = 0$ and $ab - 6a - 30 = 0$.
2. $a = 5$: $5b - 30 - 30 = 0 \Rightarrow b = 12$.
3. $a = -3$: $-3b + 18 - 30 = 0 \Rightarrow -3b = 12 \Rightarrow b = -4$.

Answer: $(5, 12)$ and $(-3, -4)$

Note on the original key: The key writes b = 4 for a = −3 and then lists the pair as (3, −4). Both are slips: the pair is (−3, −4).

1. $R_2 - 2R_1$: $(0, 0, 4 \mid 2)$. $R_3 - 3R_1$: $(0, 0, 5 \mid 3)$. $4R_3 - 5R_2$: $(0, 0, 0 \mid 2)$.
2. That row reads $0 = 2$.

Answer: No solution.

1. RREF: $$\left[\begin{array}{ccccc|c}1 & 0 & 1 & 0 & 24 & 21 \\ 0 & 1 & -2 & 0 & -8 & -7 \\ 0 & 0 & 0 & 1 & 2 & 3\end{array}\right]$$
2. Basic $x_1, x_2, x_4$; free $x_3, x_5$: $x_1 = 21 - x_3 - 24x_5$, $x_2 = -7 + 2x_3 + 8x_5$, $x_4 = 3 - 2x_5$.
3. $$\mathbf x = \begin{bmatrix}21 \\ -7 \\ 0 \\ 3 \\ 0\end{bmatrix} + x_3\begin{bmatrix}-1 \\ 2 \\ 1 \\ 0 \\ 0\end{bmatrix} + x_5\begin{bmatrix}-24 \\ 8 \\ 0 \\ -2 \\ 1\end{bmatrix}$$

Answer: As above, with two free variables.

Note on the original key: The key's final vector drops a row and prints −1 + 2a + 8b for x₂; from its own RREF the constant is −7.

1. $R_2 - 4R_1$, $R_3 - 7R_1$, then $R_3 - 2R_2$ gives a zero row.

Answer: 0

1. Main diagonal product minus the other diagonal product.

Answer: $ad - bc$

Note on the original key: The key prints a·b − c·d, which is a typo: the determinant is ad − bc.

1. $\det = a - 6$.

Answer: $a \neq 6$

1. $\det = a(ei - fh) - b(di - fg) + c(dh - eg)$.

Answer: $a(ei - fh) - b(di - fg) + c(dh - eg)$

1. $R_1 - 2R_2$, $R_3 + 3R_2$, $R_4 + R_2$ clear column 3; expand along it.
2. $\det A = (-1)^{5}\begin{vmatrix}1 & -2 & 5 \\ 1 & 2 & 3 \\ 3 & 1 & 2\end{vmatrix} = -(-38) = 38$.

Answer: 38

1. Swap: $-38$. Replacement: still $-38$. Scale by $-3$: $114$. Scale by $-2$: $-228$.

Answer: $\det B = -228$

Note on the original key: On the source sheet, the B printed under "Find the determinant of matrix B" has rows (0, −1, 1, 1) and (3, 6, 0, −15), which do not match its own row operations. This version uses the B those operations produce.

1. Row reduce $[A \mid I]$: $R_2 - 3R_1$, $R_1 + R_2$, $R_2 / (-2)$.
2. Or use $\tfrac{1}{\det A}\begin{bmatrix}4 & -2 \\ -3 & 1\end{bmatrix}$ with $\det A = -2$.

Answer: $\begin{bmatrix}-2 & 1 \\ \tfrac32 & -\tfrac12\end{bmatrix}$

1. Row reducing $[A \mid I]$ produces a zero row on the left, so $I$ cannot be reached.

Answer: No inverse.

1. Every row is a multiple of (2, 1, −2, −1, −3): rank 1, so the columns are dependent.

Answer: No inverse.

1. $\det = -1(2a) + 3(2 + 4) + 1(0 + 2a) = 12$, whatever $a$ is.

Answer: Every real $a$.

1. Expand along row 1: $a(-2 - 0) + 3(2 + 4) + 1(0 - 2) = -2a + 18 - 2 = 16 - 2a$.

Answer: $a \neq 8$

Note on the original key: The key gets −2a + 14 and a ≠ 7: it uses +2 for the last cofactor, but 1·0 − (−1)(−2) = −2.

1. $$\begin{bmatrix}1 & -3 & -2 \\ -5 & 9 & 1\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0 \\ 0\end{bmatrix}$$

Answer: $A\mathbf x = \mathbf 0$ with $A = \begin{bmatrix}1 & -3 & -2 \\ -5 & 9 & 1\end{bmatrix}$

1. $A\mathbf u = (5 - 9 + 4,\ -25 + 27 - 2) = (0, 0)$.

Answer: Yes.

1. $A(\mathbf u + \mathbf w) = A\mathbf u + A\mathbf w = \mathbf 0 + \mathbf 0 = \mathbf 0$.

Answer: Closed under addition.

1. RREF: $$\begin{bmatrix}1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & \tfrac12 & -\tfrac32 \\ 0 & 0 & 1 & 2 & -2\end{bmatrix}$$
2. $x_1 = 0$, $x_2 = -\tfrac12x_4 + \tfrac32x_5$, $x_3 = -2x_4 + 2x_5$ with $x_4, x_5$ free:
3. $$\operatorname{Nul}A = \operatorname{Span}\left\{\begin{bmatrix}0 \\ -\tfrac12 \\ -2 \\ 1 \\ 0\end{bmatrix},\begin{bmatrix}0 \\ \tfrac32 \\ 2 \\ 0 \\ 1\end{bmatrix}\right\}$$

Answer: Two spanning vectors, nullity 2

1. Pivots in columns 1, 2, 3: take those columns of the original $A$.
2. $$\left\{\begin{bmatrix}3 \\ 1 \\ 2\end{bmatrix},\begin{bmatrix}6 \\ -2 \\ -4\end{bmatrix},\begin{bmatrix}-1 \\ 2 \\ 5\end{bmatrix}\right\}$$
3. rank $= 5 - 2 = 3$.

Answer: rank 3

1. $R_2 - 2R_1$, $R_3 + R_1$, $R_1 - R_2$, $R_3 + 3R_2$ give $$\begin{bmatrix}1 & 0 & 2 \\ 0 & 1 & -2 \\ 0 & 0 & 0\end{bmatrix}$$

Answer: $\operatorname{Row}A = \operatorname{Span}\{(1, 0, 2), (0, 1, -2)\}$

Note on the original key: The key writes the last step as R3 = R3 + 3R3; the operation that produces its matrix is R3 = R3 + 3R2.

1. Solve $A^T\mathbf x = \mathbf 0$: $A^T = \begin{bmatrix}2 & -6 \\ -1 & 3\end{bmatrix}$ reduces to $x_1 = 3x_2$.

Answer: $\operatorname{Span}\{(3, 1)\}$

1. The left null space is $\{\mathbf x : \mathbf x^TA = \mathbf 0^T\}$.
2. Transpose both sides: $(\mathbf x^TA)^T = A^T\mathbf x = \mathbf 0$.
3. So $\mathbf x$ is in the left null space exactly when $\mathbf x \in \operatorname{Nul}A^T$.

Answer: $\operatorname{LeftNul}A = \operatorname{Nul}A^T$

1. $\det A = -32 \neq 0$, so $A^T$ has a pivot in every column and $A^T\mathbf x = \mathbf 0$ forces $\mathbf x = \mathbf 0$.

Answer: $\{\mathbf 0\}$

1. $\det(A - \lambda I) = \lambda^2 - 5\lambda - 2 = 0 \Rightarrow \lambda = \tfrac{5 \pm \sqrt{33}}{2}$.
2. For $\lambda_2 = \tfrac{5 + \sqrt{33}}{2}$: $3v_1 + (4 - \lambda_2)v_2 = 0 \Rightarrow v_1 = \tfrac{\sqrt{33} - 3}{6}v_2$.
3. For $\lambda_1$: $\mathbf v_1 = \left(\tfrac{-\sqrt{33} - 3}{6},\ 1\right)$.

Answer: $\lambda = \tfrac{5 \pm \sqrt{33}}{2}$

1. $\det(A - \lambda I) = -\lambda^3 - 2\lambda^2 + 7\lambda + 6 = 0$ has three real roots (no rational ones), found numerically.

Answer: $\lambda \approx -3.508,\ -0.756,\ 2.264$

1. Solve $(A - 2.264I)\mathbf v = \mathbf 0$ with $v_3 = 1$.

Answer: $\mathbf v_3 \approx (-5.957,\ -10.221,\ 1)$

1. $\mathbf u_1 = (1, 2, 3)$, $\|\mathbf u_1\|^2 = 14$.
2. $\mathbf u_2 = \mathbf c_2 - \tfrac{-8}{14}\mathbf u_1 = \left(-\tfrac{10}{7}, \tfrac87, -\tfrac27\right)$, $\|\mathbf u_2\|^2 = \tfrac{24}{7}$.
3. $\mathbf u_3 = \mathbf c_3 - \tfrac{12}{14}\mathbf u_1 - \tfrac{-8/7}{24/7}\mathbf u_2 = \left(-\tfrac13, -\tfrac13, \tfrac13\right)$.
4. Check: $\mathbf u_1\cdot\mathbf u_2 = \mathbf u_1\cdot\mathbf u_3 = \mathbf u_2\cdot\mathbf u_3 = 0$.

Answer: Orthogonal basis $\mathbf u_1, \mathbf u_2, \mathbf u_3$

1. $Q$ has the normalized $\mathbf u_i$ as columns; $R = Q^TA$, upper triangular.
2. $$R = \begin{bmatrix}\sqrt{14} & -\tfrac{4\sqrt{14}}{7} & \tfrac{6\sqrt{14}}{7} \\ 0 & \tfrac{2\sqrt{42}}{7} & -\tfrac{2\sqrt{42}}{21} \\ 0 & 0 & \tfrac{\sqrt3}{3}\end{bmatrix}$$

Answer: $A = QR$

1. $R_2 - 2R_1$, $R_3 - 3R_1$, then $R_3 - R_2$. The multipliers fill $L$ below the diagonal.
2. $$L = \begin{bmatrix}1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 1 & 1\end{bmatrix},\quad U = \begin{bmatrix}1 & -2 & 1 \\ 0 & 4 & -1 \\ 0 & 0 & 1\end{bmatrix}$$

Answer: Check: $LU = A$