Does the commutative law of addition permit us to state that A−B = B−A? How would you correct the statement?

Here are the solutions to the problems.

2. Commutative Law of Subtraction

The commutative law of addition states that $A+B = B+A$.
Matrix subtraction $A-B$ is defined as $A+(-1)B$.
If we assume $A-B = B-A$, then $A+(-1)B = B+(-1)A$.
This statement is generally not true for matrices. For example, if $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$:
$A-B = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}$
$B-A = \begin{bmatrix} -1 & 1 \\ 0 & 0 \end{bmatrix}$
Clearly, $A-B \neq B-A$. The commutative law of addition does not apply directly to subtraction in this way.

To correct the statement, we can say that matrix subtraction is anti-commutative, meaning:

$$A-B = -(B-A)$$

Let's verify this:

$$-(B-A) = -(B+(-1)A) = -B - (-1)A = -B + A = A-B$$

This shows that $A-B = -(B-A)$ is the correct relationship.

• Does the commutative law of addition permit us to state that $A-B = B-A$? ❌ No.
• How would you correct the statement? The correct statement is $\boxed{A-B = -(B-A)}$.

3. Test the associative law of multiplication with the following matrices:

Given matrices:

$$A = \begin{bmatrix} 5 & 3 \\ 0 & 5 \end{bmatrix}, \quad B = \begin{bmatrix} -8 & 0 & 7 \\ 1 & 3 & 2 \end{bmatrix}, \quad C = \begin{bmatrix} 1 \\ 0 \\ 7 \end{bmatrix}$$

The associative law of multiplication states that $(AB)C = A(BC)$.