Q.If $A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$, then verify $(BA)^2 \neq B^2 A^2$.

Idea: $(BA)^2$ is defined, but $B^2$ isn't (since $B$ is not square), so the two sides can't be equal.

Step 1 — compute $BA$. $B$ is $2\times3$, $A$ is $3\times2$, so $BA$ is $2\times2$:

Step 2 — square it.

$(BA)^2 = \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix}\begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix} = \begin{bmatrix} 121-91 & -77+14 \\ 143-26 & -91+4 \end{bmatrix} = \begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix}.$ …

The first thing to fix is the shapes. $A$ is $3\times2$ and $B$ is $2\times3$. The product $BA$ is $(2\times3)(3\times2) = 2\times2$ — square, so it can be squared. But $B^2 = B\cdot B$ would be $(2\times3)(2\times3)$, and the inner dimensions ($3$ and $2$) don't match, so $B^2$ does not exist. That single observation already tells us the two sides cannot be equal, but let us compute $(BA)^2$ so the left side is explicit.

Step 1 — compute $BA$

• $(1,1):\ 2\cdot3+1\cdot1+2\cdot2 = 6+1+4 = 11$
• $(1,2):\ 2\cdot(-4)+1\cdot1+2\cdot0 = -8+1+0 = -7$
• $(2,1):\ 1\cdot3+2\cdot1+4\cdot2 = 3+2+8 = 13$
• $(2,2):\ 1\cdot(-4)+2\cdot1+4\cdot0 = -4+2+0 = -2$

So $BA = \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix}$.

Step 2 — square $BA$

• $(1,1):\ 11\cdot11+(-7)\cdot13 = 121-91 = 30$
• $(1,2):\ 11\cdot(-7)+(-7)\cdot(-2) = -77+14 = -63$
• $(2,1):\ 13\cdot11+(-2)\cdot13 = 143-26 = 117$
• $(2,2):\ 13\cdot(-7)+(-2)\cdot(-2) = -91+4 = -87$

So $(BA)^2 = \begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix}$.

Step 3 — why $B^2A^2$ fails …

Method: Checking a product identity for existence before value

Use this when asked to test an identity such as $(BA)^2$ vs $B^2A^2$ for non-square matrices.

Steps

Step 1: Test conformability of every piece first.

A power $B^2=B\cdot B$ exists only if $B$ is square. If $B$ is $2\times3$, then $B^2$ (and hence $B^2A^2$) is undefined.

Step 2: Compute the side that is defined.

E.g. $BA$ is square, so $(BA)^2$ exists — compute it. …

Common Mistakes

Mistake 1: Assuming $B^2$ exists for a rectangular $B$.

Why it's wrong: $B$ is $2\times3$, so $B\cdot B$ needs $3=2$ and is undefined; hence $B^2A^2$ does not exist. Correct approach: test conformability of every power before computing.

Mistake 2: Believing $(BA)^2=B^2A^2$ in general. …