Chapter 3: Matrices
Q1. If A = [2 3; 4 5] and B = [1 −2; 3 7], find A + B and A − B.
Solution
A + B = [2+1 3+(−2)] = [3 1]
[4+3 5+7 ] [7 12]
A − B = [2−1 3−(−2)] = [1 5]
[4−3 5−7 ] [1 −2]
Q2. Compute: [1 2 3] × [4 5 6]T
Solution
[1 2 3] × [4; 5; 6] = 1(4) + 2(5) + 3(6) = 4 + 10 + 18 = 32
Q3. If A = [3 1; −1 2], verify that A + AT is symmetric.
Solution
AT = [3 −1]
[1 2]
A + AT = [3+3 1+(−1)] = [6 0]
[−1+1 2+2 ] [0 4]
(A + AT)T = [6 0] = A + AT. ✓ Symmetric.
Q4. Find the value of x, y, z if:
[x+y 2; 5+z xy] = [6 2; 5 8]
Solution
Comparing corresponding elements:
x + y = 6 ... (i)
5 + z = 5 ⇒ z = 0
xy = 8 ... (ii)
From (i): y = 6 − x. Substituting in (ii):
x(6 − x) = 8 ⇒ x² − 6x + 8 = 0
(x − 2)(x − 4) = 0
x = 2, y = 4 OR x = 4, y = 2
So x = 2, y = 4, z = 0 (or x = 4, y = 2, z = 0)
Q5. Find A² if A = [1 2; 0 1]
Solution
A² = A × A = [1 2; 0 1] × [1 2; 0 1]
= [1(1)+2(0) 1(2)+2(1)] = [1 4]
[0(1)+1(0) 0(2)+1(1)] [0 1]
Q6. If A = [1 2; 3 4], B = [5 6; 7 8], verify that (AB)T = BTAT.
Solution
AB = [1(5)+2(7) 1(6)+2(8)] = [19 22]
[3(5)+4(7) 3(6)+4(8)] [43 50]
(AB)T = [19 43]
[22 50]
BTAT = [5 7][1 3] = [5+7 15+28] = [19 43]
[6 8][2 4] [6+8 18+32] [22 50]
(AB)T = BTAT. ✓