Exercise 3.4 · Q1 (MCQ): When are A and B inverses?

Basics you need

Solution

1. Recall the definition

   By Definition 6, $A$ and $B$ are inverses of each other exactly when they are square matrices of the same order and

   $$AB = I \quad\text{and}\quad BA = I, \qquad\text{that is}\qquad AB = BA = I.$$

   The definition is symmetric in $A$ and $B$⁠: if $B$ is the inverse of $A$⁠, then $A$ is the inverse of $B$ (Note 2 after Definition 6). So "inverse of each other" means precisely $AB = BA = I$⁠, which is option (D).

   A multiple choice question about a definition is answered by the definition itself. The phrase "only if" asks for a condition that must hold whenever the matrices are inverses; (D) must hold, because it is what being inverses means.
   Definition 6 and the Note after it, Section 3.7. Lesson: invertible matrices

2. Why (A) fails: commuting is not enough

   (A) asks only that $AB = BA$⁠. Take the $2\times2$ matrices $A = I$ and $B = 2I$⁠, whose entries are shown below.

   $$A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}.$$

   Entry by entry, row of the first factor times column of the second:

   $$\begin{aligned} (AB)_{11} &= 1\cdot2 + 0\cdot0 = 2,\\ (AB)_{12} &= 1\cdot0 + 0\cdot2 = 0,\\ (AB)_{21} &= 0\cdot2 + 1\cdot0 = 0,\\ (AB)_{22} &= 0\cdot0 + 1\cdot2 = 2,\\ (BA)_{11} &= 2\cdot1 + 0\cdot0 = 2,\\ (BA)_{12} &= 2\cdot0 + 0\cdot1 = 0,\\ (BA)_{21} &= 0\cdot1 + 2\cdot0 = 0,\\ (BA)_{22} &= 0\cdot0 + 2\cdot1 = 2. \end{aligned}$$

   So

   $$AB = BA = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \ne I.$$

   The condition in (A) holds, but $A$ and $B$ are not inverses. So (A) is not the condition for being inverses: it is necessary (since $AB = BA = I$ forces $AB = BA$⁠) but far from enough.

   To show a condition does not characterise inverses, one explicit pair that satisfies it without being inverses is enough. Matrices that commute are common (every matrix commutes with itself), but most pairs of commuting matrices are not inverses.

3. Why (B) fails: a zero product rules inverses out

   If $AB = O$⁠, then $AB \ne I$⁠, because the diagonal entries of $O$ are 0 while those of $I$ are 1. So no pair satisfying (B) can be a pair of inverses. For instance $A = B = O$ satisfies $AB = BA = O$⁠, and the zero matrix has no inverse at all, since $OX = O \ne I$ for every $X$⁠.

   Option (B) looks like (D) with $I$ replaced by 0. That replacement turns "undo" into "destroy": multiplying by the inverse should bring you back to $I$⁠, not to $O$⁠.
   For numbers, $ab = 0$ can never mean $b = \frac1a$⁠; the matrix case is the same, with $I$ in the role of 1.

4. Why (C) fails: one of the two products is wrong

   (C) demands $AB = O$⁠. As in Step 3, that alone already means $AB \ne I$⁠, so $A$ and $B$ cannot be inverses, whatever $BA$ is. (In fact, for square matrices of the same order, $BA = I$ together with $AB = O$ can never happen at all, but we do not need that to reject (C).)

   An inverse must work from both sides. Getting $I$ on one side and $O$ on the other is the opposite of what the definition asks.

5. Why the definition needs both products, and square matrices

   For rectangular matrices one product can be an identity while the other is not. Take $A$ of order $2\times3$ and $B$ of order $3\times2$⁠:

   $$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix}.$$

   $AB$ is defined (3 columns of $A$⁠, 3 rows of $B$⁠) and is $2\times2$⁠:

   $$\begin{aligned} (AB)_{11} &= 1\cdot1 + 0\cdot0 + 0\cdot0 = 1,\\ (AB)_{12} &= 1\cdot0 + 0\cdot1 + 0\cdot0 = 0,\\ (AB)_{21} &= 0\cdot1 + 1\cdot0 + 0\cdot0 = 0,\\ (AB)_{22} &= 0\cdot0 + 1\cdot1 + 0\cdot0 = 1. \end{aligned}$$

   So $AB = I_2$⁠, the $2\times2$ identity.

   $BA$ is defined (2 columns of $B$⁠, 2 rows of $A$⁠) and is $3\times3$⁠. Its $(i,j)$ entry is (row $i$ of $B$⁠) times (column $j$ of $A$⁠):

   $$\begin{aligned} (BA)_{11} &= 1\cdot1 + 0\cdot0 = 1,\\ (BA)_{12} &= 1\cdot0 + 0\cdot1 = 0,\\ (BA)_{13} &= 1\cdot0 + 0\cdot0 = 0,\\ (BA)_{21} &= 0\cdot1 + 1\cdot0 = 0,\\ (BA)_{22} &= 0\cdot0 + 1\cdot1 = 1,\\ (BA)_{23} &= 0\cdot0 + 1\cdot0 = 0,\\ (BA)_{31} &= 0\cdot1 + 0\cdot0 = 0,\\ (BA)_{32} &= 0\cdot0 + 0\cdot1 = 0,\\ (BA)_{33} &= 0\cdot0 + 0\cdot0 = 0. \end{aligned}$$

   Row 3 of $B$ is all zeros, which is why row 3 of $BA$ is all zeros:

   $$BA = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix} \ne I_3.$$

   So $AB = I$ on its own does not make $A$ and $B$ inverses. This is why Definition 6 asks for square matrices of the same order and for both $AB = I$ and $BA = I$⁠, and why a rectangular matrix has no inverse (Note 1 after Definition 6).

   This example shows that the "$= BA$" part of option (D) is doing real work. (For square matrices of the same order it turns out that either product being $I$ forces the other to be $I$ too, but that needs determinants, from Chapter 4. The definition asks for both.)
   Note 1, Section 3.7: for $AB$ and $BA$ both to be defined and equal, $A$ and $B$ must be square of the same order.

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