Math Snap

STEP 1

What is this asking?
Which statement about matrix multiplication and the identity matrix is always true?
Watch out!
Matrix multiplication isn't like regular multiplication, so don't assume all the same rules apply!

STEP 2

1. Analyze statement A
2. Analyze statement B
3. Analyze statement C
4. Analyze statement D

STEP 3

Statement A says: If $AB = AC$ and $A \neq 0$, then $B = C$.
Is this always true?
Nope! Think about it: if $A$ was a regular number and had an inverse, we could multiply both sides by $A^{-1}$ to get $B=C$.
But with matrices, $A$ might not have an inverse!

STEP 4

Let's create a counterexample.
Let's use $2 \times 2$ matrices for simplicity.
Let $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$, and $C = \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix}$.
Notice that $A \neq 0$.

STEP 5

Now, let's calculate $AB$:
$$AB = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}.$$

STEP 6

And calculate $AC$:
$$AC = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 3 & 3 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}.$$

STEP 7

So, $AB = AC$, but $B \neq C$!
Statement A is false!

STEP 8

Statement B says: If $A^2 = A$, then $A = 0$ or $A = I$.
This is also false!
The identity matrix $I$ definitely works, since $I^2 = I$.
And the zero matrix works, since $0^2 = 0$.
But there are other matrices that work too!

STEP 9

Consider the matrix $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.
Then
$$A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = A.$$ So $A^2 = A$, but $A$ is not the zero matrix or the identity matrix.
Statement B is false!

STEP 10

Statement C says: If $A^2 = 0$, then $A = 0$.
Let's see if we can find a counterexample.

STEP 11

Consider $A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$.
Then
$$A^2 = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0.$$ So, $A^2 = 0$, but $A \neq 0$.
Statement C is false!

STEP 12

Statement D says: If $AB = BA$, then $(A+B)^2 = A^2 + 2AB + B^2$.
Let's expand $(A+B)^2$:
$$(A+B)^2 = (A+B)(A+B) = A(A+B) + B(A+B) = A^2 + AB + BA + B^2.$$

STEP 13

We are given that $AB = BA$.
So, we can substitute $AB$ for $BA$ in our expansion:
$$(A+B)^2 = A^2 + AB + AB + B^2 = A^2 + 2AB + B^2.$$ So, if $AB = BA$, then $(A+B)^2 = A^2 + 2AB + B^2$.
Statement D is true!

The correct statement is D.