Question: Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column.
StartAbsoluteValue Start 3 By 3 Matrix 1st Row 1st Column 2 2nd Column negative 3 3rd Column 4 2nd Row 1st Column 3 2nd Column 1 3rd Column 2 3rd Row 1st Column 1 2nd Column 4 3rd Column negative 1 EndMatrix EndAbsoluteValue
Part
1 of 2
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
(
negative 3
)(
negative 5
)
minus
(
1
)(
negative 6
)
plus
(
4
)(
negative 8
)
equals
enter your response here
.
B.
Using this expansion, the determinant is
minus
(
negative 3
)(
negative 5
)
plus
(
1
)(
negative 6
)
minus
(
4
)(
negative 8
)
equals
enter your response here
.
C.
Using this expansion, the determinant is
(
2
)(
negative 9
)
minus
(
negative 3
)(
negative 5
)
plus
(
4
)(
11
)
equals
enter your response here
.
D.
Using this expansion, the determinant is
minus
(
2
)(
negative 9
)
plus
(
negative 3
)(
negative 5
)
minus
(
4
)(
11
)
equals
enter your response here
.
Part
2 of 2
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
(
negative 3
)(
negative 5
)
minus
(
1
)(
negative 6
)
plus
(
4
)(
negative 8
)
equals
enter your response here
.
B.
Using this expansion, the determinant is
(
2
)(
negative 9
)
minus
(
negative 3
)(
negative 5
)
plus
(
4
)(
11
)
equals
enter your response here
.
C.
Using this expansion, the determinant is
minus
(
2
)(
negative 9
)
plus
(
negative 3
)(
negative 5
)
minus
(
4
)(
11
)
equals
enter your response here
.
D.
Using this expansion, the determinant is
minus
(
negative 3
)(
negative 5
)
plus
(
1
)(
negative 6
)
minus
(
4
)(
negative 8
)
equals
enter your response here
.
1. **State the problem:**
We want to compute the determinant of the matrix
$$\begin{bmatrix} 2 & -3 & 4 \\ 3 & 1 & 2 \\ 1 & 4 & -1 \end{bmatrix}$$
using two methods: cofactor expansion across the first row and cofactor expansion down the second column.
2. **Recall the formula for determinant by cofactor expansion:**
For a 3x3 matrix $A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$,
the determinant 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 minor determinant obtained by deleting row $i$ and column $j$.
---
### Part 1: Cofactor expansion across the first row
3. **Calculate minors for the first row:**
- $M_{11} = \begin{vmatrix} 1 & 2 \\ 4 & -1 \end{vmatrix} = (1)(-1) - (2)(4) = -1 - 8 = -9$
- $M_{12} = \begin{vmatrix} 3 & 2 \\ 1 & -1 \end{vmatrix} = (3)(-1) - (2)(1) = -3 - 2 = -5$
- $M_{13} = \begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix} = (3)(4) - (1)(1) = 12 - 1 = 11$
4. **Calculate cofactors for the first row:**
- $C_{11} = (-1)^{1+1} M_{11} = (+1)(-9) = -9$
- $C_{12} = (-1)^{1+2} M_{12} = (-1)(-5) = 5$
- $C_{13} = (-1)^{1+3} M_{13} = (+1)(11) = 11$
5. **Apply cofactor expansion across the first row:**
$$\det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = 2(-9) + (-3)(5) + 4(11)$$
6. **Calculate each term:**
- $2(-9) = -18$
- $(-3)(5) = -15$
- $4(11) = 44$
7. **Sum the terms:**
$$-18 - 15 + 44 = (-18 - 15) + 44 = -33 + 44 = 11$$
8. **Answer for part 1:**
The determinant using cofactor expansion across the first row is $\boxed{11}$.
---
### Part 2: Cofactor expansion down the second column
9. **Calculate minors for the second column:**
- $M_{12} = \begin{vmatrix} 3 & 2 \\ 1 & -1 \end{vmatrix} = -5$ (already computed)
- $M_{22} = \begin{vmatrix} 2 & 4 \\ 1 & -1 \end{vmatrix} = (2)(-1) - (4)(1) = -2 - 4 = -6$
- $M_{32} = \begin{vmatrix} 2 & 4 \\ 3 & 2 \end{vmatrix} = (2)(2) - (4)(3) = 4 - 12 = -8$
10. **Calculate cofactors for the second column:**
- $C_{12} = (-1)^{1+2} M_{12} = (-1)(-5) = 5$
- $C_{22} = (-1)^{2+2} M_{22} = (+1)(-6) = -6$
- $C_{32} = (-1)^{3+2} M_{32} = (-1)(-8) = 8$
11. **Apply cofactor expansion down the second column:**
$$\det(A) = a_{12}C_{12} + a_{22}C_{22} + a_{32}C_{32} = (-3)(5) + (1)(-6) + (4)(8)$$
12. **Calculate each term:**
- $(-3)(5) = -15$
- $(1)(-6) = -6$
- $(4)(8) = 32$
13. **Sum the terms:**
$$-15 - 6 + 32 = (-21) + 32 = 11$$
14. **Answer for part 2:**
The determinant using cofactor expansion down the second column is $\boxed{11}$.
---
### Final conclusion:
Both methods yield the same determinant value:
$$\boxed{11}$$