Compute the determinant of the following matrix using a cofactor expansion across the first row. $$ \left|\begin{array}{rrr}5 & 4 & -4 \ 3 & 0 & 5 \ 6 & 5 & 2\end{array} ight| $$ Compute the determinant using a cofactor expansion across the first row, Select the correct choice below and the answer box to complete your choice. (Simplify your answer.) A. Using this expansion, the determinant is $$ (5)(-25)+(4)(-24)+(-4)(15)= $$ B. Using this expansion, the determinant is $$ (5)(-25)-(3)(-24)+(6)(15)=\square $$ C. Using this expansion, the determinant is $$ (5)(-25)+(3)(-24)+(6)(15)=\square $$ D. Using this expansion, the determinant is $$ (5)(-25)-(4)(-24)+(-4)(15)=\square $$.

Question

Compute the determinant of the following matrix using a cofactor expansion across the first row. $$ \left|\begin{array}{rrr}5 & 4 & -4 \ 3 & 0 & 5 \ 6 & 5 & 2\end{array}ight| $$ Compute the determinant using a cofactor expansion across the first row, Select the correct choice below and the answer box to complete your choice. (Simplify your answer.) A. Using this expansion, the determinant is $$ (5)(-25)+(4)(-24)+(-4)(15)= $$ B. Using this expansion, the determinant is $$ (5)(-25)-(3)(-24)+(6)(15)=\square $$ C. Using this expansion, the determinant is $$ (5)(-25)+(3)(-24)+(6)(15)=\square $$ D. Using this expansion, the determinant is $$ (5)(-25)-(4)(-24)+(-4)(15)=\square $$.

UpStudy AI · Accepted Answer

We are given the matrix

$$
\begin{pmatrix}
5 & 4 & -4 \
3 & 0 & 5 \
6 & 5 & 2
\end{pmatrix}.
$$

We will compute its determinant by expanding across the first row.

1. The formula for the determinant using the first row is

$$
\det(A) = a_{11} \, C_{11} + a_{12} \, C_{12} + a_{13} \, C_{13},
$$

where $$ C_{ij} = (-1)^{i+j} M_{ij} $$ and $$ M_{ij} $$ is the determinant of the submatrix obtained by removing row $$ i $$ and column $$ j $$.

2. For the first row, we have:
   - $$ a_{11} = 5 $$
   - $$ a_{12} = 4 $$
   - $$ a_{13} = -4 $$

3. Compute each minor:

- For $$ M_{11} $$ (remove first row and first column):

$$
     M_{11} = \det \begin{pmatrix} 0 & 5 \ 5 & 2 \end{pmatrix} = 0\cdot2 - 5\cdot5 = -25.
     $$
     
     The cofactor is
     
     $$
     C_{11} = (-1)^{1+1} \cdot (-25) = 1\cdot (-25) = -25.
     $$

- For $$ M_{12} $$ (remove first row and second column):

$$
     M_{12} = \det \begin{pmatrix} 3 & 5 \ 6 & 2 \end{pmatrix} = 3\cdot2 - 5\cdot6 = 6 - 30 = -24.
     $$
     
     The cofactor is
     
     $$
     C_{12} = (-1)^{1+2} \cdot (-24) = (-1)\cdot(-24) = 24.
     $$

- For $$ M_{13} $$ (remove first row and third column):

$$
     M_{13} = \det \begin{pmatrix} 3 & 0 \ 6 & 5 \end{pmatrix} = 3\cdot5 - 0\cdot6 = 15.
     $$
     
     The cofactor is
     
     $$
     C_{13} = (-1)^{1+3} \cdot 15 = 1\cdot15 = 15.
     $$

4. Now, substitute these into the expansion:

$$
\det(A) = (5)(-25) + (4)(24) + (-4)(15).
$$

5. Compute each term:

- $$ 5(-25) = -125 $$
   - $$ 4(24) = 96 $$
   - $$ -4(15) = -60 $$

6. Sum the terms:

$$
-125 + 96 - 60 = -89.
$$

Thus, the determinant is $$-89$$.

The correct choice is:

D. Using this expansion, the determinant is $$ (5)(-25)-(4)(-24)+(-4)(15)= -89 $$.
The determinant is $$-89$$.

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