Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. $$ \left|\begin{array}{rrr} 2 & 0 & 3 \ 2 & 4 & 3 \ 0 & 4 & -1 \end{array}\right| $$ 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 $$ -(0)(-2)+(4)(-2)-(4)(0)= $$ $$ \square $$ . B. 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 $$ (2)(-16)-(0)(-2)+(3)(8)=-8 $$. Compute the determinant using a cofactor expansion down the second column. 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 $$ (2)(-16)-(0)(-2)+(3)(8)= $$ $$ \square $$ . B. Using this expansion, the determinant is $$ -(0)(-2)+(4)(-2)-(4)(0)= $$ $$ \square $$ . C. Using this expansion, the determinant is $$ -(2)(-16)+(0)(-2)-(3)(8)= $$ $$ \square $$ . D. Using this expansion, the determinant is $$ (0)(-2)-(4)(-2)+(4)(0)= $$ $$ \square $$ .

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UpStudy AI · Accepted Answer

We begin with the matrix
$$
A=\begin{pmatrix}
2 & 0 & 3 \
2 & 4 & 3 \
0 & 4 & -1
\end{pmatrix}
$$

**Cofactor Expansion Across the First Row**

We use the first row entries $$a_{11}=2$$, $$a_{12}=0$$, and $$a_{13}=3$$.

1. For $$a_{11}=2$$ (position $$(1,1)$$, sign $$+$$) the minor is
   $$
   \begin{pmatrix}
   4 & 3 \
   4 & -1
   \end{pmatrix}.
   $$
   Its determinant is 
   $$
   4(-1)-3(4)=-4-12=-16.
   $$
   Contribution: $$2(-16)=-32$$.

2. For $$a_{12}=0$$ (position $$(1,2)$$, sign $$-$$) the minor is
   $$
   \begin{pmatrix}
   2 & 3 \
   0 & -1
   \end{pmatrix}.
   $$
   Its determinant is 
   $$
   2(-1)-3(0)=-2.
   $$
   Contribution: $$-0(-2)=0$$.

3. For $$a_{13}=3$$ (position $$(1,3)$$, sign $$+$$) the minor is
   $$
   \begin{pmatrix}
   2 & 4 \
   0 & 4
   \end{pmatrix}.
   $$
   Its determinant is 
   $$
   2(4)-4(0)=8.
   $$
   Contribution: $$3(8)=24$$.

Now sum the contributions:
$$
-32 + 0 + 24 = -8.
$$

Thus, by a cofactor expansion across the first row, the determinant is $$-8$$. This matches choice D:
$$
(2)(-16)-(0)(-2)+(3)(8)=-8.
$$

**Cofactor Expansion Down the Second Column**

Now, we perform the expansion down the second column. The entries in the second column are: $$a_{12}=0$$, $$a_{22}=4$$, and $$a_{32}=4$$.

1. For $$a_{12}=0$$ (position $$(1,2)$$, sign $$-$$) the minor is
   $$
   \begin{pmatrix}
   2 & 3 \
   0 & -1
   \end{pmatrix}
   $$
   and its determinant is $$-2$$. The contribution is $$-0(-2)=0$$.

2. For $$a_{22}=4$$ (position $$(2,2)$$, sign $$+$$) the minor is
   $$
   \begin{pmatrix}
   2 & 3 \
   0 & -1
   \end{pmatrix}
   $$
   and its determinant is $$-2$$. The contribution is $$4(-2)=-8$$.

3. For $$a_{32}=4$$ (position $$(3,2)$$, sign $$-$$) the minor is
   $$
   \begin{pmatrix}
   2 & 3 \
   2 & 3
   \end{pmatrix}.
   $$
   Its determinant is 
   $$
   2(3)-3(2)=6-6=0.
   $$
   The contribution is $$-4(0)=0$$.

Summing these:
$$
0-8+0 = -8.
$$

Thus, by a cofactor expansion down the second column, the determinant is $$-8$$. This matches choice B:
$$
-(0)(-2)+(4)(-2)-(4)(0) = -8.
$$

**Final Answers**

For the first row expansion, the correct choice is **D** with the answer $$\boxed{-8}$$.

For the second column expansion, the correct choice is **B** with the answer $$\boxed{-8}$$.
For the first row cofactor expansion, the determinant is $$-8$$. For the second column cofactor expansion, the determinant is $$-8$$.

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