Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. $$ \left|\begin{array}{rrr}2 & 0 & 3 \ 2 & 4 & 3 \ 0 & 4 & -1\end{array} ight| $$ Compute the determinant using a cofactor expansion across the first row. Select the correct choice below and fill in the answer box to complete your choice. (Simplify your answer.) A. Using this expansion, the determinant is $$ -(0)(-2)+(4)(-2)-(4)(0)=\square $$. Using this expansion, the determinant is $$ -(2)(-16)+(0)(-2)-(3)(8)=\square $$. C. Using this expansion, the determinant is $$ (0)(-2)-(4)(-2)+(4)(0)=\square $$. D. Using this expansion, the determinant is $$ (2)(-16)-(0)(-2)+(3)(8)=\square $$.

Question

Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. $$ \left|\begin{array}{rrr}2 & 0 & 3 \ 2 & 4 & 3 \ 0 & 4 & -1\end{array}ight| $$ Compute the determinant using a cofactor expansion across the first row. Select the correct choice below and fill in the answer box to complete your choice. (Simplify your answer.) A. Using this expansion, the determinant is $$ -(0)(-2)+(4)(-2)-(4)(0)=\square $$. Using this expansion, the determinant is $$ -(2)(-16)+(0)(-2)-(3)(8)=\square $$. C. Using this expansion, the determinant is $$ (0)(-2)-(4)(-2)+(4)(0)=\square $$. D. Using this expansion, the determinant is $$ (2)(-16)-(0)(-2)+(3)(8)=\square $$.

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We begin with the matrix

$$
\begin{pmatrix}
2 & 0 & 3 \
2 & 4 & 3 \
0 & 4 & -1
\end{pmatrix}.
$$

**Cofactor expansion across the first row:**

The elements in the first row are $$ a_{11}=2 $$, $$ a_{12}=0 $$, and $$ a_{13}=3 $$.

1. For $$ a_{11}=2 $$ (position $$(1,1)$$):
   - The cofactor sign is $$ (-1)^{1+1}=1 $$.
   - The submatrix (remove row 1 and column 1) is
     $$
     \begin{pmatrix}
     4 & 3 \
     4 & -1
     \end{pmatrix},
     $$
     so its determinant is
     $$
     (4)(-1)-(3)(4)=-4-12=-16.
     $$
   - The contribution is $$ 2(-16) $$.

2. For $$ a_{12}=0 $$ (position $$(1,2)$$):
   - The cofactor sign is $$ (-1)^{1+2}=-1 $$.
   - The submatrix (remove row 1 and column 2) is
     $$
     \begin{pmatrix}
     2 & 3 \
     0 & -1
     \end{pmatrix},
     $$
     and its determinant is
     $$
     (2)(-1)-(3)(0)=-2.
     $$
   - The cofactor is $$ -(-2)=2 $$ but since the element is $$0$$, the contribution is $$ 0 $$.

3. For $$ a_{13}=3 $$ (position $$(1,3)$$):
   - The cofactor sign is $$ (-1)^{1+3}=1 $$.
   - The submatrix (remove row 1 and column 3) is
     $$
     \begin{pmatrix}
     2 & 4 \
     0 & 4
     \end{pmatrix},
     $$
     with determinant
     $$
     (2)(4)-(4)(0)=8.
     $$
   - The contribution is $$ 3(8) $$.

Thus, the determinant computed across the first row is

$$
2(-16) + 0 + 3(8) = -32 + 24 = -8.
$$

**Cofactor expansion down the second column:**

The entries in the second column are $$ 0 $$, $$ 4 $$, and $$ 4 $$.

1. For the entry $$ 0 $$ (position $$(1,2)$$):
   - The cofactor sign is $$ (-1)^{1+2}=-1 $$.
   - The submatrix (removing row 1 and column 2) is
     $$
     \begin{pmatrix}
     2 & 3 \
     0 & -1
     \end{pmatrix},
     $$
     with determinant $$ -2 $$, but the contribution is $$ 0 $$ because the entry is $$ 0 $$.

2. For the entry $$ 4 $$ (position $$(2,2)$$):
   - The cofactor sign is $$ (-1)^{2+2}=1 $$.
   - The submatrix (removing row 2 and column 2) is
     $$
     \begin{pmatrix}
     2 & 3 \
     0 & -1
     \end{pmatrix},
     $$
     with determinant
     $$
     (2)(-1)-(3)(0)=-2.
     $$
   - The contribution is $$ 4(-2)=-8 $$.

3. For the entry $$ 4 $$ (position $$(3,2)$$):
   - The cofactor sign is $$ (-1)^{3+2}=-1 $$.
   - The submatrix (removing row 3 and column 2) is
     $$
     \begin{pmatrix}
     2 & 3 \
     2 & 3
     \end{pmatrix},
     $$
     with determinant
     $$
     (2)(3)-(3)(2)=6-6=0.
     $$
   - The contribution is $$ 4(-1)(0)=0 $$.

So, the determinant computed down the second column is

$$
0 + (-8) + 0 = -8.
$$

**Matching the Answer Choices:**

Reviewing the choices, option D states:

$$
\text{Using this expansion, the determinant is } (2)(-16)-(0)(-2)+(3)(8)=\square.
$$

Calculating this gives:

$$
(2)(-16)= -32,\quad -(0)(-2)= 0,\quad (3)(8)= 24,\quad\text{and } -32+0+24=-8.
$$

Thus, the correct answer is option D with the determinant equal to $$-8$$.
The determinant is $$-8$$.

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