Matrices — Practice

Question 1

A matrix has 15 elements. What are all the possible orders it can have?

Accepted Answer

15 = 1x15, 15x1, 3x5, 5x3. Answer: Possible orders: 1x15, 15x1, 3x5, 5x3.

Question 2

If A=[[1,2],[3,4]] and B=[[0,1],[1,0]], find AB and BA.

Accepted Answer

AB = [[1*0+2*1, 1*1+2*0],[3*0+4*1, 3*1+4*0]] = [[2,1],[4,3]]. BA = [[0*1+1*3, 0*2+1*4],[1*1+0*3, 1*2+0*4]] = [[3,4],[1,2]]. Answer: AB=[[2,1],[4,3]], BA=[[3,4],[1,2]] — different, confirming AB is not equal to BA here.

Question 3

If A=[[1,2],[3,4]] and B=[[0,1],[-1,2]], verify that (AB)'=B'A'.

Accepted Answer

AB=[[1*0+2*-1,1*1+2*2],[3*0+4*-1,3*1+4*2]]=[[-2,5],[-4,11]], so (AB)'=[[-2,-4],[5,11]]. B'=[[0,-1],[1,2]], A'=[[1,3],[2,4]]. B'A'=[[0*1+-1*2,0*3+-1*4],[1*1+2*2,1*3+2*4]]=[[-2,-4],[5,11]]. Answer: (AB)'=B'A'=[[-2,-4],[5,11]], confirmed equal.

Question 4

Express A=[[4,2],[6,8]] as the sum of a symmetric and a skew-symmetric matrix.

Accepted Answer

A'=[[4,6],[2,8]]. (1/2)(A+A')=[[4,4],[4,8]] (symmetric part). (1/2)(A-A')=[[0,-2],[2,0]] (skew-symmetric part). Answer: A = [[4,4],[4,8]] + [[0,-2],[2,0]].

Question 5

Why can't a matrix have two different inverses?

Accepted Answer

If B and C were both inverses of A, then B=BI=B(AC)=(BA)C=IC=C, forcing B=C. Answer: The inverse, when it exists, must be unique.

Question 6

If A=[[3,1],[-1,2]], show that A^2-5A+7I=0 (the zero matrix).

Accepted Answer

A^2 = [[3,1],[-1,2]]@[[3,1],[-1,2]] = [[9-1,3+2],[-3-2,-1+4]] = [[8,5],[-5,3]]. 5A=[[15,5],[-5,10]]. 7I=[[7,0],[0,7]]. A^2-5A+7I = [[8-15+7, 5-5+0],[-5+5+0, 3-10+7]] = [[0,0],[0,0]]. Answer: A^2-5A+7I is the zero matrix, as required.

Question 7

If A is a skew-symmetric matrix of odd order, what can you say about its determinant? (Beyond this chapter's scope, but a natural extension — just reason about the diagonal here.)

Accepted Answer

Every skew-symmetric matrix has an all-zero diagonal, since a_ii=-a_ii forces a_ii=0. Answer: For any skew-symmetric matrix, the trace (sum of diagonal entries) is always 0, regardless of order.

Question 8

If A is of order 2x3 and B is of order 3x2, is AB defined? Is BA defined? What are their orders?

Accepted Answer

AB: inner dimensions 3 and 3 match, so AB is defined, of order 2x2. BA: inner dimensions 2 and 2 match, so BA is also defined, of order 3x3. Answer: Both are defined, but AB (2x2) and BA (3x3) have different orders, so they can't even be compared, let alone be equal.

Practice — Matrices

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