Final Exam (Term 232)

1. $R_1 \leftarrow R_1/(-2)$, $R_2 \leftarrow R_2 - R_1$, $R_3 \leftarrow R_3 - 3R_1$, $R_2 \leftarrow -2R_2$.
2. $R_1 \leftarrow R_1 + \tfrac32R_2$, $R_3 \leftarrow R_3 - \tfrac{11}{2}R_2$, $R_3 \leftarrow R_3/(-7)$, $R_1 \leftarrow R_1 - 2R_3$.
3. RREF: $$\left[\begin{array}{ccc|c}1 & 0 & 0 & \tfrac47 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & -\tfrac97\end{array}\right]$$
4. Check in row 1: $-2(\tfrac47) + 3(-2) - 4(-\tfrac97) = \tfrac{-8 - 42 + 36}{7} = -2$.

Answer: $x_1 = \tfrac47,\ x_2 = -2,\ x_3 = -\tfrac97$

1. $\det A = 1(a^2 + 12) - 2(-a - 8) + 1(3 - 2a) = a^2 + 31$.
2. $a^2 + 31 \geq 31 > 0$ for every real $a$, so the coefficient matrix is always invertible.

Answer: Unique for every real $a$.

Note on the original key: The official key writes $a \neq \pm i\sqrt{31}$. That is the same statement, since $\pm i\sqrt{31}$ are not real.

1. Row reduction ends in $\left[\,0 \;\; 0 \;\; a^2 + 31 \mid ab - 6a + 2b - 5\,\right]$.
2. More than one solution needs $a^2 + 31 = 0$, which has no real solution.

Answer: No real pair.

Note on the original key: The key lists complex pairs: $a = \pm i\sqrt{31}$ with $b = \tfrac{11 + 6i\sqrt{31}}{2 + i\sqrt{31}}$ or $\tfrac{5 - 6i\sqrt{31}}{2 - i\sqrt{31}}$. Over the reals, no pair exists.

1. $C = B^TA^T = \begin{bmatrix}1 & -1 & -3 \\ -2 & 3 & 3\end{bmatrix}\begin{bmatrix}-1 & 2 & -2 \\ 2 & 1 & 3 \\ -3 & -1 & 1\end{bmatrix} = \begin{bmatrix}6 & 4 & -8 \\ -1 & -4 & 16\end{bmatrix}$.

Answer: $C = \begin{bmatrix}6 & 4 & -8 \\ -1 & -4 & 16\end{bmatrix}$

1. The transpose of a product reverses the order: $(AB)^T = B^TA^T$.

Answer: $C = (AB)^T$

1. Let $U\mathbf x = \mathbf z$ and solve $L\mathbf z = \mathbf b$ forward: $z_1 = 3$, $z_2 = -1 + 3 = 2$, $z_3 = 2 - 3 - 4 = -5$.
2. Solve $U\mathbf x = \mathbf z$ backward: $-6x_3 = -5 \Rightarrow x_3 = \tfrac56$; $x_2 + 3x_3 = 2 \Rightarrow x_2 = -\tfrac12$;
3. $x_1 - 2x_2 + x_3 = 3 \Rightarrow x_1 = 3 - 1 - \tfrac56 = \tfrac76$.

Answer: $\mathbf x = \left(\tfrac76,\ -\tfrac12,\ \tfrac56\right)$

1. Upper triangular: $\det = 3(a - 2)(b + 1)$.
2. The product is zero when either factor is zero.

Answer: Singular exactly when $a = 2$ or $b = -1$ (or both).

1. Undo the operations that turn the given matrix into $A$, tracking the determinant:
2. Swap $R_1, R_4$: $-1 \times (-42) = 42$.
3. Scale $R_4$ by $-\tfrac12$: $42 \times (-\tfrac12) = -21$.
4. Swap $R_2, R_3$: $21$. Scale $R_2$ by $\tfrac13$: $7$.
5. $R_3 \leftarrow R_3 - R_2$ leaves it unchanged: $7$.

Answer: $\det A = 7$

1. Row 2 has a single nonzero entry, $a_{23} = 2$.
2. $\det A = (-1)^{2+3}\cdot 2 \cdot \begin{vmatrix}-2 & 0 & -2 \\ -1 & 6 & -3 \\ -2 & 1 & -2\end{vmatrix}$.
3. The $3\times3$ minor is $-2(-12 + 3) - 0 + (-2)(-1 + 12) = 18 - 22 = -4$.
4. $\det A = -2 \cdot (-4) = 8$.

Answer: $\det A = 8$

1. $\det A = -\tfrac{15}{2} \neq 0$, so $A\mathbf x = \mathbf 0$ has only the trivial solution.

Answer: Yes.

1. Only the trivial solution, so $\operatorname{Nul}A = \{\mathbf 0\}$.

Answer: $\operatorname{Nul}A = \{\mathbf 0\}$, nullity $0$

Note on the original key: The key calls the null space "empty". A null space always contains the zero vector, so the precise answer is {0}, with dimension 0.

1. rank $= 3 - $ nullity $= 3$, so the three columns are a basis of $\operatorname{Col}A = \mathbb R^3$.

Answer: rank $3$, $\operatorname{Col}A = \operatorname{Span}\{\mathbf v_1, \mathbf v_2, \mathbf v_3\} = \mathbb R^3$

1. $A$ reduces to $I_3$, whose rows are the standard basis vectors.

Answer: $\operatorname{Row}A = \mathbb R^3$, basis $\{(1,0,0), (0,1,0), (0,0,1)\}$

1. A square matrix with independent columns also has independent rows, so $A^T$ is invertible.
2. $A^T\mathbf x = \mathbf 0$ has only $\mathbf x = \mathbf 0$.

Answer: Left null space $= \{\mathbf 0\}$

1. $\det(A - \lambda I) = -\lambda^3 + 11\lambda^2 - 39\lambda + 45 = -(\lambda - 3)^2(\lambda - 5)$.

Answer: $\lambda_1 = 3$ (multiplicity 2), $\lambda_2 = 5$

1. $A - 5I = \begin{bmatrix}-1 & 1 & -1 \\ 2 & 0 & -2 \\ 1 & 1 & -3\end{bmatrix}$ reduces to $x_1 = x_3$, $x_2 = 2x_3$.

Answer: $\mathbf v = \alpha(1, 2, 1)$, $\alpha \neq 0$

1. $A - 3I = \begin{bmatrix}1 & 1 & -1 \\ 2 & 2 & -2 \\ 1 & 1 & -1\end{bmatrix}$ has rank 1, so the eigenspace for $\lambda = 3$ is 2-dimensional: $x_1 = -x_2 + x_3$ gives $(-1, 1, 0)$ and $(1, 0, 1)$.
2. With $(1, 2, 1)$ for $\lambda = 5$ that is three independent eigenvectors: $A$ is diagonalizable.
3. $$S = \begin{bmatrix}-1 & 1 & 1 \\ 1 & 0 & 2 \\ 0 & 1 & 1\end{bmatrix},\quad D = \begin{bmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 5\end{bmatrix},\quad A^2 = SD^2S^{-1} = \begin{bmatrix}17 & 8 & -8 \\ 16 & 25 & -16 \\ 8 & 8 & 1\end{bmatrix}$$

Answer: Yes.

Note on the original key: The official key answers No, claiming the eigenvector matrix has two identical columns. That is wrong: a repeated eigenvalue only blocks diagonalization when its eigenspace is too small, and here it has dimension 2. You can check the result: squaring A directly gives the same matrix.

1. $\mathbf u_1 = \mathbf a_1 = (1, -1, 1)$.
2. $\mathbf a_2\cdot\mathbf u_1 = -1 - 1 + 2 = 0$, so $\mathbf u_2 = \mathbf a_2 = (-1, 1, 2)$.
3. $\mathbf u_3 = \mathbf a_3 - \tfrac{\mathbf a_3\cdot\mathbf u_1}{\mathbf u_1\cdot\mathbf u_1}\mathbf u_1 - \tfrac{\mathbf a_3\cdot\mathbf u_2}{\mathbf u_2\cdot\mathbf u_2}\mathbf u_2 = (2, 0, -1) - \tfrac13(1, -1, 1) + \tfrac46(-1, 1, 2) = (1, 1, 0)$.
4. Normalize: $\mathbf e_1 = \tfrac{1}{\sqrt3}(1, -1, 1)$, $\mathbf e_2 = \tfrac{1}{\sqrt6}(-1, 1, 2)$, $\mathbf e_3 = \tfrac{1}{\sqrt2}(1, 1, 0)$.

Answer: Orthonormal columns $\mathbf e_1, \mathbf e_2, \mathbf e_3$ as above

1. $Q = \begin{bmatrix}\mathbf e_1 & \mathbf e_2 & \mathbf e_3\end{bmatrix}$ and, since $Q^TQ = I$, $R = Q^TA$.
2. $R_{ij} = \mathbf e_i\cdot\mathbf a_j$: $R_{11} = \sqrt3$, $R_{12} = 0$, $R_{13} = \tfrac{1}{\sqrt3}$, $R_{22} = \sqrt6$, $R_{23} = -\tfrac{4}{\sqrt6}$, $R_{33} = \sqrt2$.
3. $$Q = \begin{bmatrix}\tfrac{\sqrt3}{3} & -\tfrac{\sqrt6}{6} & \tfrac{\sqrt2}{2} \\ -\tfrac{\sqrt3}{3} & \tfrac{\sqrt6}{6} & \tfrac{\sqrt2}{2} \\ \tfrac{\sqrt3}{3} & \tfrac{\sqrt6}{3} & 0\end{bmatrix},\quad R = \begin{bmatrix}\sqrt3 & 0 & \tfrac{\sqrt3}{3} \\ 0 & \sqrt6 & -\tfrac{2\sqrt6}{3} \\ 0 & 0 & \sqrt2\end{bmatrix}$$

Answer: $A = QR$ with $Q$, $R$ as above