Sample Final (Term 223)

1. $R_2 \leftarrow R_2 - R_1$: $(0, 1, 2 \mid 8)$. $R_3 \leftarrow R_3 - R_1$: $(0, 3, 6 \mid 24)$. $R_3 \leftarrow R_3 - 3R_2$: a zero row.
2. RREF: $$\left[\begin{array}{ccc|c}1 & 0 & -1 & -2 \\ 0 & 1 & 2 & 8 \\ 0 & 0 & 0 & 0\end{array}\right]$$
3. $z$ is free: $x = -2 + z$, $y = 8 - 2z$.

Answer: $(x, y, z) = (-2, 8, 0) + z(1, -2, 1)$

Note on the original key: The answer table on the paper has a single box for each of x, y, z, but the system is consistent with a free variable, so it has infinitely many solutions.

1. $R_2 \leftarrow R_2 - R_1$: $(0, 5, 6, 6 \mid 1)$. $R_3 \leftarrow R_3 - R_1$: $(0, -10, -12, -12 \mid 4)$.
2. $R_3 \leftarrow R_3 + 2R_2$: $(0, 0, 0, 0 \mid 6)$, which reads $0 = 6$.

Answer: Inconsistent: no solution.

1. $R_2 \leftarrow R_2 - 2R_1$: $(0, 0, 5, -3)$. $R_3 \leftarrow R_3 + R_1$: $(0, 0, 5, -3)$. $R_3 \leftarrow R_3 - R_2$: zero row.
2. Pivots in columns 1 and 3: rank 2, and nullity $= 4 - 2 = 2$.

Answer: rank 2, nullity 2

1. For a complex matrix, unitary means $A^{H}A = I$ with $A^H = \overline{A}^{\,T}$ (the conjugate transpose).
2. Column 1 has squared length $\tfrac{1}{25}(|{-1+2i}|^2 + |2-4i|^2) = \tfrac{5 + 20}{25} = 1$, and so does column 2.
3. Their Hermitian inner product is $\tfrac{1}{25}\left[(-1-2i)(-4-2i) + (2+4i)(-2-i)\right] = \tfrac{1}{25}\left[(0 + 10i) + (0 - 10i)\right] = 0$.

Answer: Yes: $A^HA = I$.

Note on the original key: The hint on the paper says $U^TU = I$. With complex entries you need the conjugate transpose: the plain $A^TA$ here has diagonal $\tfrac{(1-2i)^2}{5}$ and $\tfrac{(2+i)^2}{5}$, not 1.

1. $AB = \begin{bmatrix}1 & 3 & 0 \\ 3 & 5 & 2 \\ 2 & 2 & 2\end{bmatrix} \neq I$.
2. In fact $\det A = 0$ (column 3 = column 1 + column 2), so $A$ has no inverse at all.

Answer: No.

1. Row 3 is all zeros, so $\det A = 0$ (expand along that row).
2. A matrix with zero determinant is not invertible.

Answer: No inverse.

1. $R_2 \leftarrow R_2 - \tfrac52R_1$: $(0, -\tfrac32, -3)$. $R_3 \leftarrow R_3 - 4R_1$: $(0, -3, -15)$.
2. $R_3 \leftarrow R_3 - 2R_2$: $(0, 0, -9)$. Replacements keep the determinant.
3. $\det A = 2 \cdot (-\tfrac32)(-9) = 27$.

Answer: $\det A = 27$

1. Clear column 3 with $R_1 \leftarrow R_1 - 2R_2$, $R_3 \leftarrow R_3 + 3R_2$, $R_4 \leftarrow R_4 + R_2$ (determinant unchanged):$$\begin{bmatrix}1 & -2 & 0 & 5 \\ 2 & 3 & 1 & -2 \\ 1 & 2 & 0 & 3 \\ 3 & 1 & 0 & 2\end{bmatrix}$$
2. Expand along column 3: $\det B = (-1)^{2+3}\cdot 1\cdot\begin{vmatrix}1 & -2 & 5 \\ 1 & 2 & 3 \\ 3 & 1 & 2\end{vmatrix}$.
3. The minor is $1(4 - 3) + 2(2 - 9) + 5(1 - 6) = 1 - 14 - 25 = -38$, so $\det B = 38$.

Answer: $\det B = 38$

1. The rows of $C$ are rows 3, 4, 5, 2, 1 of $D$, so $C = PD$ for a permutation matrix $P$, and $\det C = \det P\cdot\det D$.
2. The permutation $(3, 4, 5, 2, 1)$ splits into the cycles $(1\,3\,5)(2\,4)$: an even 3-cycle times one swap, so it is odd and $\det P = -1$.
3. $\det C = -\det D = -(-54) = 54$.

Answer: $\det C = -\det D = 54$ (using the given value)

Note on the original key: Computing $D$ directly gives $\det D = +54$, so the printed value has a sign slip. With the true value, $\det C = -54$. Both answers are accepted here; the row-permutation argument is what the question tests.

1. $U$ is block upper triangular: a $2\times2$ block on top-left and a $3\times3$ block on bottom-right, zeros below.
2. $\begin{vmatrix}2 & 3 \\ -1 & 5\end{vmatrix} = 10 + 3 = 13$.
3. $\begin{vmatrix}2 & 1 & 5 \\ 3 & -1 & 4 \\ 5 & 2 & 6\end{vmatrix} = 2(-6 - 8) - 1(18 - 20) + 5(6 + 5) = -28 + 2 + 55 = 29$.
4. $\det U = 13 \cdot 29 = 377$.

Answer: $\det U = 377$

1. Solve $c_1 + 3c_2 = -2$, $4c_1 - 6c_2 = 10$: $c_1 = 1$, $c_2 = -1$.

Answer: Yes: $\mathbf v = \mathbf a_1 - \mathbf a_2$

1. Column 3 = column 1 + column 2, so Col $A$ is a plane. Reduce $[A \mid \mathbf v]$:
2. $R_2 - R_1$: $(0, -1, -1 \mid 1)$; $R_3 - 2R_1$: $(0, -1, -1 \mid 4)$; subtract: $(0, 0, 0 \mid 3)$, inconsistent.

Answer: No.

1. $\det A = 16 \neq 0$, so every $\mathbf v$ is in Col $A$; solving gives $c = (1, -3, 1)$.
2. Check: $(1 + 3 + 1,\ 9 - 9 + 1,\ 1 - 3 + 1) = (5, 1, -1)$.

Answer: Yes: $\mathbf v = \mathbf a_1 - 3\mathbf a_2 + \mathbf a_3$

1. RREF: $$\begin{bmatrix}1 & 0 & -16 \\ 0 & 1 & -19 \\ 0 & 0 & 0\end{bmatrix}$$
2. $x_1 = 16x_3$, $x_2 = 19x_3$: $\operatorname{Nul}A = \operatorname{Span}\{(16, 19, 1)\}$, nullity 1.
3. Pivot columns 1 and 2 of $A$: $\operatorname{Col}A$ basis $\{(1, 5, 7), (-1, -4, -6)\}$, rank 2.

Answer: Nul basis (16, 19, 1); nullity 1; Col basis = columns 1 and 2; rank 2

1. rank $= \dim\operatorname{Col}A = 2$ and nullity $= 2$.
2. Rank theorem: $n = \text{rank} + \text{nullity} = 4$.

Answer: 4 columns

1. rank $= n - $ nullity $= 5 - 3 = 2$.

Answer: rank 2

1. $A = I - J$ where $J$ is the all-ones matrix, whose eigenvalues are $3, 0, 0$.
2. So $A$ has eigenvalues $1 - 3 = -2$ and $1 - 0 = 1$ (twice). Expanding $\det(A - \lambda I) = -(\lambda + 2)(\lambda - 1)^2$ agrees.

Answer: $\lambda = -2,\ 1,\ 1$

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

Answer: $\lambda = 1,\ 2,\ 3$

1. $A\mathbf v_1 = (-2, -3, -5)$, $A\mathbf v_2 = (10, 2, 1)$, $A\mathbf v_3 = (-27, -19, -15)$.
2. None of these is a scalar multiple of the vector it came from. (The eigenvalues of $A$ are $1$ and $2 \pm \sqrt{13}$.)

Answer: None of them.

1. $\lambda^2 - 4\lambda - 5 = (\lambda - 5)(\lambda + 1)$: distinct eigenvalues $-1, 5$.
2. $\lambda = -1$: $(-2, 1)$. $\lambda = 5$: $(1, 1)$.

Answer: Yes: $P = \begin{bmatrix}-2 & 1 \\ 1 & 1\end{bmatrix},\ D = \begin{bmatrix}-1 & 0 \\ 0 & 5\end{bmatrix}$

1. $A^{10} = SD^{10}S^{-1}$ with $S = \begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}$, $S^{-1} = \tfrac12\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}$, $D^{10} = \operatorname{diag}(1, 9^{10})$.
2. $A^{10} = \tfrac12\begin{bmatrix}1+9^{10} & 1-9^{10} \\ 1-9^{10} & 1+9^{10}\end{bmatrix}$ and $9^{10} = 3\,486\,784\,401$.

Answer: $A^{10} = \begin{bmatrix}1743392201 & -1743392200 \\ -1743392200 & 1743392201\end{bmatrix}$

1. $\dfrac{\mathbf v\cdot\mathbf u}{\mathbf u\cdot\mathbf u}\mathbf u = \tfrac43(1, 1, 1)$.

Answer: $\left(\tfrac43, \tfrac43, \tfrac43\right)$

1. $\mathbf z = (\tfrac23, -\tfrac13, -\tfrac13)$, orthogonal to $\mathbf u$.
2. $\dfrac{\mathbf w\cdot\mathbf u}{\mathbf u\cdot\mathbf u} = -\tfrac13$ and $\dfrac{\mathbf w\cdot\mathbf z}{\mathbf z\cdot\mathbf z} = \dfrac{-2/3}{2/3} = -1$.
3. $\hat{\mathbf w} = -\tfrac13(1, 1, 1) - (\tfrac23, -\tfrac13, -\tfrac13) = (-1, 0, 0)$.

Answer: $(-1, 0, 0)$