Major 2 (Term 251)

1. $\det A = 0\cdot C_{11} + 2\cdot C_{12} + 1\cdot C_{13}$.
2. $C_{12} = -\begin{vmatrix}3 & 4 \\ 2 & 5\end{vmatrix} = -(15 - 8) = -7$ and $C_{13} = \begin{vmatrix}3 & 1 \\ 2 & 0\end{vmatrix} = 0 - 2 = -2$.
3. $\det A = 2(-7) + 1(-2) = -16$.

Answer: $\det A = -16$

1. Swap $R_1 \leftrightarrow R_2$ (sign flips): $$\begin{bmatrix}3 & 1 & 4 \\ 0 & 2 & 1 \\ 2 & 0 & 5\end{bmatrix}$$
2. $R_3 \leftarrow R_3 - \tfrac23 R_1$: $(0, -\tfrac23, \tfrac73)$. Then $R_3 \leftarrow R_3 + \tfrac13 R_2$: $(0, 0, \tfrac83)$.
3. Triangular: product of the diagonal $= 3 \cdot 2 \cdot \tfrac83 = 16$. One swap, so $\det A = -16$.

Answer: $\det A = -16$

1. $\det A = -16 \neq 0$, and a square matrix is invertible exactly when its determinant is nonzero.

Answer: Yes, $A$ is invertible.

1. The coefficient matrix is $A$ and $\mathbf b = (0, 0, 1)$. Replace one column of $A$ by $\mathbf b$ each time.
2. $\det A_1(\mathbf b) = \begin{vmatrix}0 & 2 & 1 \\ 0 & 1 & 4 \\ 1 & 0 & 5\end{vmatrix} = 1\cdot(8 - 1) = 7$
3. $\det A_2(\mathbf b) = \begin{vmatrix}0 & 0 & 1 \\ 3 & 0 & 4 \\ 2 & 1 & 5\end{vmatrix} = 1\cdot(3 - 0) = 3$
4. $\det A_3(\mathbf b) = \begin{vmatrix}0 & 2 & 0 \\ 3 & 1 & 0 \\ 2 & 0 & 1\end{vmatrix} = 1\cdot(0 - 6) = -6$
5. $x = \tfrac{7}{-16}$, $y = \tfrac{3}{-16}$, $z = \tfrac{-6}{-16}$.

Answer: $x = -\tfrac{7}{16},\ y = -\tfrac{3}{16},\ z = \tfrac38$

1. Row 1 is $2\times$ row 1 of $A$, row 2 is $2\times$ row 3 of $A$, row 3 is $2\times$ row 2 of $A$.
2. Factor 2 out of each row: $2^3 = 8$. The rows are $A$'s with rows 2 and 3 swapped: factor $-1$.
3. $8 \cdot (-1) \cdot (-16) = 128$.

Answer: $128$

1. Row 2 is $(3, 1, 4) - (2, 0, 5)$: $A$ with $R_2 \leftarrow R_2 - R_3$.
2. Adding a multiple of one row to another does not change the determinant.

Answer: $-16$

1. $\det(-A^2) = (-1)^3\det(A^2) = -(\det A)^2$ for a $3\times 3$ matrix.
2. $-(-16)^2 = -256$.

Answer: $-256$

1. $A\mathbf w = (-12 + 12,\ -6 + 6) = (0, 0)$, so $\mathbf w \in \operatorname{Nul} A$.
2. $\operatorname{Col} A = \operatorname{Span}\{(-6, -3)\} = \operatorname{Span}\{(2, 1)\}$ because column 2 is $-2\times$ column 1.
3. $\mathbf w = -\tfrac13(-6, -3)$, so $\mathbf w \in \operatorname{Col} A$ as well.

Answer: $\mathbf w$ is in both spaces.

1. The relation gives $\mathbf v_3 = \tfrac15(3\mathbf v_2 - \mathbf v_1)$, so $\mathbf v_3$ adds nothing to the span.
2. $\mathbf v_1$ and $\mathbf v_2$ are not multiples of each other, so they are independent.
3. By the Spanning Set Theorem, the two remaining vectors form a basis (any two of the three would work).

Answer: Basis: $\{\mathbf v_1, \mathbf v_2\}$, so $\dim H = 2$.

1. $B$ has pivots in columns 1 and 2, so rank $A = 2$ and $\dim\operatorname{Nul}A = 4 - 2 = 2$.
2. Nul $A$: from $B$, $x_1 = -6x_3 - 5x_4$ and $2x_2 = -5x_3 - 3x_4$. Basis: $$\left\{\begin{bmatrix}-6 \\ -\tfrac52 \\ 1 \\ 0\end{bmatrix},\begin{bmatrix}-5 \\ -\tfrac32 \\ 0 \\ 1\end{bmatrix}\right\}$$
3. Col $A$: the pivot columns of the original $A$: $$\left\{\begin{bmatrix}-2 \\ 2 \\ -3\end{bmatrix},\begin{bmatrix}4 \\ -6 \\ 8\end{bmatrix}\right\}$$
4. Row $A$: the nonzero rows of the echelon form $B$: $\{(1, 0, 6, 5),\ (0, 2, 5, 3)\}$.
5. $\dim\operatorname{Col}A = \dim\operatorname{Row}A = \operatorname{rank}A = 2$.

Answer: rank = 2, dim Nul = 2, dim Col = 2, dim Row = 2

1. Rank theorem: rank $A$ + nullity $A = n = 7$, so nullity $= 3$.
2. rank $A^T$ = rank $A = 4$ (row rank equals column rank).

Answer: nullity $A = 3$, rank $A^T = 4$