Module 3 Practice: Determinants
10 problems on determinants, cofactors, properties, Cramer's Rule, and geometric interpretation.
Problem 1
Compute det([7 2; 3 5]).
Solution
7(5) - 2(3) = 35 - 6 = 29.
Problem 2
Compute det([1 2 0; 3 1 -1; 2 0 4]) using cofactor expansion along row 1.
Solution
1*det([1 -1; 0 4]) - 2*det([3 -1; 2 4]) + 0 = 1(4) - 2(12+2) = 4 - 28 = -24.
Problem 3
If det(A) = 3, what is det(A^T)?
Solution
det(A^T) = det(A) = 3.
Problem 4
If det(A) = 2 and det(B) = -5, what is det(AB)?
Solution
det(AB) = det(A)*det(B) = 2*(-5) = -10.
Problem 5
A 4x4 matrix A has det(A) = 3. What is det(2A)?
Solution
det(2A) = 2^4 * det(A) = 16 * 3 = 48.
Problem 6
Use Cramer's Rule to solve: x + 2y = 5, 3x - y = 1.
Solution
det(A) = -1-6 = -7. det(A_1) = det([5 2;1 -1]) = -5-2 = -7. det(A_2) = det([1 5;3 1]) = 1-15 = -14. x = -7/(-7)=1, y = -14/(-7)=2.
Problem 7
Find the area of the parallelogram spanned by (1, 3) and (4, 2).
Solution
det([1 4; 3 2]) = 2 - 12 = -10. Area = |-10| = 10.
Problem 8
det([1 0 0; 0 5 0; 0 0 -3]) = ?
Solution
Diagonal matrix: det = product of diagonals = 1*5*(-3) = -15.
Problem 9
You row reduce a matrix with 2 row swaps and no scaling, getting diagonal entries 1, 2, 3. What is the original det?
Solution
Triangular det = 6. Two swaps negate twice: (-1)^2 = 1. Original det = 6.
Problem 10
If A has two identical rows, what is det(A)?
Solution
0. Swapping the identical rows gives the same matrix but negates det, so det = -det, implying det = 0.