Module 2 Practice Problems: Matrix Algebra
10 problems on matrix operations, properties, inverses, and LU factorization.
Problem 1
Compute A + B where A = [ 1 -2 ; 3 0 ] and B = [ 4 1 ; -1 5 ].
Solution
[ 1+4, -2+1 ; 3+(-1), 0+5 ] = [ 5 -1 ; 2 5 ].
Problem 2
Compute 3A - 2B using A and B from Problem 1.
Solution
3A = [ 3 -6 ; 9 0 ]. 2B = [ 8 2 ; -2 10 ]. 3A - 2B = [ -5 -8 ; 11 -10 ].
Problem 3
Compute AB where A = [ 1 2 ; 3 4 ] and B = [ 0 1 ; 1 0 ].
Solution
AB: (1,1)=0+2=2, (1,2)=1+0=1, (2,1)=0+4=4, (2,2)=3+0=3. AB = [ 2 1 ; 4 3 ].
Problem 4
Using A and B from Problem 3, compute BA. Is AB = BA?
Solution
BA: (1,1)=3, (1,2)=4, (2,1)=1, (2,2)=2. BA = [ 3 4 ; 1 2 ]. AB is not equal to BA.
Problem 5
Find A^T if A = [ 1 0 3 ; 2 -1 4 ].
Solution
A is 2x3, so A^T is 3x2: A^T = [ 1 2 ; 0 -1 ; 3 4 ].
Problem 6
Find the inverse of A = [ 4 3 ; 3 2 ] using the 2x2 formula.
Solution
det = 4(2) - 3(3) = 8 - 9 = -1. A^{-1} = (1/(-1))[2 -3; -3 4] = [-2 3; 3 -4].
Problem 7
Is the matrix [ 1 2 ; 2 4 ] invertible? Explain.
Solution
det = 1(4) - 2(2) = 0. Since the determinant is zero, the matrix is NOT invertible (singular). Note that row 2 is 2 times row 1.
Problem 8
Solve Ax = b where A = [ 1 1 ; 1 2 ] and b = [ 3 ; 5 ], using the inverse.
Solution
det = 2-1=1. A^{-1} = [ 2 -1 ; -1 1 ]. x = A^{-1}b = [ 2(3)+(-1)(5) ; (-1)(3)+1(5) ] = [ 1 ; 2 ].
Problem 9
Find the LU factorization of A = [ 1 3 ; 2 8 ].
Solution
R_2 - 2R_1 → R_2: U = [ 1 3 ; 0 2 ]. Multiplier = 2. L = [ 1 0 ; 2 1 ]. Verify: LU = [ 1 3 ; 2+0, 6+2 ] = [ 1 3 ; 2 8 ] = A.
Problem 10
What elementary matrix E performs R_2 - 4R_1 on a 3x3 matrix? What is E^{-1}?
Solution
E = [ 1 0 0 ; -4 1 0 ; 0 0 1 ]. E^{-1} = [ 1 0 0 ; 4 1 0 ; 0 0 1 ] (change -4 to +4).