Activity 2.2.5

Consider the matrices $$A = \left[\begin{array}{rrr} 1 & 3 & 2 \\ -3 & 4 & -1 \\ \end{array}\right],~~~ B = \left[\begin{array}{rr} 3 & 0 \\ 1 & 2 \\ -2 & -1 \\ \end{array}\right]$$.

1. Before computing, first explain why the shapes of $A$ and $B$ enable us to form the product $AB$. Then describe the shape of $AB$.

2. Compute the product $AB$.

3. Sage can multiply matrices using the * operator. Define the matrices $A$ and $B$ in the Sage cell below and check your work by computing $AB$.

4. Are we able to form the matrix product $BA$? If so, use the Sage cell above to find $BA$. Is it generally true that $AB = BA$?

5. Suppose we form the three matrices. $$A = \left[\begin{array}{rr} 1 & 2 \\ 3 & -2 \\ \end{array}\right], B = \left[\begin{array}{rr} 0 & 4 \\ 2 & -1 \\ \end{array}\right], C = \left[\begin{array}{rr} -1 & 3 \\ 4 & 3 \\ \end{array}\right]$$. Compare what happens when you compute $A(B+C)$ and $AB + AC$. State your finding as a general principle.

6. Compare the results of evaluating $A(BC)$ and $(AB)C$ and state your finding as a general principle.

7. When we are dealing with real numbers, we know if $a\neq 0$ and $ab = ac$, then $b=c$. Define matrices $$A = \left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \\ \end{array}\right], B = \left[\begin{array}{rr} 3 & 0 \\ 1 & 3 \\ \end{array}\right], C = \left[\begin{array}{rr} 1 & 2 \\ 2 & 2 \\ \end{array}\right]$$ and compute $AB$ and $AC$. If $AB = AC$, is it necessarily true that $B = C$?

8. Again, with real numbers, we know that if $ab = 0$, then either $a = 0$ or $b=0$. Define $$A = \left[\begin{array}{rr} 1 & 2 \\ -2 & -4 \\ \end{array}\right], B = \left[\begin{array}{rr} 2 & -4 \\ -1 & 2 \\ \end{array}\right]$$ and compute $AB$. If $AB = 0$, is it necessarily true that either $A=0$ or $B=0$?