Final Exam (Term 231)

1. Augmented matrix: $$\left[\begin{array}{cccc|c}3 & -1 & 1 & 7 & 13 \\ -2 & 1 & -1 & -3 & -9 \\ -2 & 1 & 0 & -7 & -8\end{array}\right]$$
2. $R_1 \leftarrow R_1 + R_2$ gives $(1, 0, 0, 4 \mid 4)$, a convenient first pivot.
3. Clear column 1: $R_2 \leftarrow R_2 + 2R_1$, $R_3 \leftarrow R_3 + 2R_1$, then continue to the RREF: $$\left[\begin{array}{cccc|c}1 & 0 & 0 & 4 & 4 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & -4 & 1\end{array}\right]$$
4. $w$ is free: $x = 4 - 4w$, $y = -w$, $z = 1 + 4w$.

Answer: $(x, y, z, w) = (4, 0, 1, 0) + w(-4, -1, 4, 1)$, $w \in \mathbb R$

1. $R_2 \leftarrow R_2 - 3R_1$, $R_3 \leftarrow R_3 - 4R_1$, then $R_3 \leftarrow R_3 - R_2$: $$\left[\begin{array}{ccc|c}1 & 2 & -3 & 4 \\ 0 & -7 & 14 & -10 \\ 0 & 0 & a^2-16 & a-4\end{array}\right]$$
2. If $a^2 - 16 \neq 0$ ($a \neq \pm 4$): a pivot in every column, one solution.
3. $a = 4$: the last row is $0 = 0$, so a free variable and infinitely many solutions.
4. $a = -4$: the last row is $0 = -8$, inconsistent.

Answer: None at $a = -4$; infinitely many at $a = 4$; exactly one for every other $a$.

1. $C_{ij}$ = (row $i$ of $A$) $\cdot$ (column $j$ of $B$).
2. $C_{11} = 2(2) + 4(2) + 1(7) = 19$
3. $C_{32} = 4(3) + 2(-2) + 2(5) = 18$
4. $C_{21} = 5(2) + 2(2) + (-1)(7) = 7$

Answer: $C_{11} = 19,\ C_{32} = 18,\ C_{21} = 7$

1. $C = AB = \begin{bmatrix}19 & 3 \\ 7 & 6 \\ 26 & 18\end{bmatrix}$, so $C^T = \begin{bmatrix}19 & 7 & 26 \\ 3 & 6 & 18\end{bmatrix}$.
2. $A$ is $3\times3$ and $B$ is $3\times2$, so $A + B$ is undefined and so is $(A + B)^T$.

Answer: $C^T = \begin{bmatrix}19 & 7 & 26 \\ 3 & 6 & 18\end{bmatrix}$; $(A + B)^T$ does not exist.

1. $\mathbf x = A^{-1}\mathbf b = \begin{bmatrix}2 & 5 \\ -1 & 4\end{bmatrix}\begin{bmatrix}7 \\ -3\end{bmatrix} = \begin{bmatrix}14-15 \\ -7-12\end{bmatrix}$.

Answer: $x = -1$, $y = -19$

1. Both are diagonal, so each determinant is the product of the diagonal.
2. $\det A = 3(a + b - 1) = 0 \Rightarrow a + b = 1$.
3. $\det B = 5(2a - 3b - 7) = 0 \Rightarrow 2a - 3b = 7$.
4. Solve together: $a = 1 - b$, so $2 - 2b - 3b = 7$ and $b = -1$, $a = 2$.

Answer: $a = 2$, $b = -1$

1. The matrix is diagonal, so the determinant is $(2x - 4)(x + 3)x$.
2. Set each factor to zero.

Answer: $x = 2$, $x = -3$ or $x = 0$

1. Expand along row 3, $(0, 6, 0, 4)$: only $a_{32}$ and $a_{34}$ contribute.
2. $C_{32} = (-1)^{5}\begin{vmatrix}1 & 6 & 0 \\ 3 & 5 & 4 \\ 3 & 7 & 1\end{vmatrix} = -31$
3. $C_{34} = (-1)^{7}\begin{vmatrix}1 & 2 & 6 \\ 3 & 2 & 5 \\ 3 & 0 & 7\end{vmatrix} = -(-34) = 34$
4. $\det A = 6(-31) + 4(34) = -186 + 136 = -50$.

Answer: $\det A = -50$

1. Solve $c_1 - c_2 = 3$ and $2c_1 + c_2 = 5$. Adding: $3c_1 = 8$.
2. $c_1 = \tfrac83$, $c_2 = c_1 - 3 = -\tfrac13$.

Answer: Yes: $\mathbf u = \tfrac83(1, 2) - \tfrac13(-1, 1)$.

1. Three vectors span $\mathbb R^3$ only if the matrix with them as columns is invertible.
2. $\begin{vmatrix}1 & 3 & 1 \\ 1 & 2 & 0 \\ 1 & -1 & -3\end{vmatrix} = 1(-6 - 0) - 3(-3 - 0) + 1(-1 - 2) = -6 + 9 - 3 = 0$.
3. Indeed $(1, 0, -3) = -2(1, 1, 1) + (3, 2, -1)$.

Answer: No, the vectors are dependent and span only a plane.

1. One nonzero row, already in echelon form: one pivot.
2. rank $= \dim\operatorname{Col}A = 1$; nullity $= 3 - 1 = 2$.

Answer: rank 1, dim Col 1, nullity 2

1. The pivot column of $A$ is column 1.

Answer: $\{(10, 0, 0)\}$

1. $10x_1 - 3x_2 - 2x_3 = 0$, so $x_1 = 0.3x_2 + 0.2x_3$ with $x_2, x_3$ free.
2. $$\mathbf x = x_2\begin{bmatrix}\tfrac{3}{10} \\ 1 \\ 0\end{bmatrix} + x_3\begin{bmatrix}\tfrac15 \\ 0 \\ 1\end{bmatrix}$$

Answer: Basis: $\{(3, 10, 0),\ (1, 0, 5)\}$ (any nonzero multiples work).

1. Expand $\det(A - \lambda I)$ along column 2, which has one nonzero entry, $2 - \lambda$:
2. $\det(A - \lambda I) = (2 - \lambda)\det\begin{bmatrix}-\lambda & -2 \\ 1 & 3-\lambda\end{bmatrix} = (2 - \lambda)(\lambda^2 - 3\lambda + 2)$.
3. $= (2 - \lambda)(\lambda - 1)(\lambda - 2)$.

Answer: $\lambda = 1$ and $\lambda = 2$ (multiplicity 2)

1. Triangular, so the eigenvalues are the diagonal entries 3 and 4.
2. Two distinct eigenvalues for a $2\times2$ matrix guarantee diagonalizability.

Answer: Yes.

1. $\lambda = 3$: $A - 3I = \begin{bmatrix}0 & 5 \\ 0 & 1\end{bmatrix}$ forces $x_2 = 0$: $\mathbf v_1 = (1, 0)$.
2. $\lambda = 4$: $A - 4I = \begin{bmatrix}-1 & 5 \\ 0 & 0\end{bmatrix}$ gives $x_1 = 5x_2$: $\mathbf v_2 = (5, 1)$.

Answer: $P = \begin{bmatrix}1 & 5 \\ 0 & 1\end{bmatrix},\quad D = \begin{bmatrix}3 & 0 \\ 0 & 4\end{bmatrix}$

1. $\mathbf y\cdot\mathbf u = 18 + 5 = 23$ and $\mathbf u\cdot\mathbf u = 9 + 1 = 10$.
2. $\hat{\mathbf y} = \tfrac{23}{10}\mathbf u = (6.9,\ 2.3)$.
3. $\mathbf z = \mathbf y - \hat{\mathbf y} = (-0.9,\ 2.7)$. Check: $\mathbf z\cdot\mathbf u = -2.7 + 2.7 = 0$.

Answer: $\mathbf y = (6.9, 2.3) + (-0.9, 2.7)$

Note on the original key: The paper prints "the orthogonal projection of y onto y"; the intended question is the projection onto u.