Q.If $A=\begin{bmatrix}3&4\\-4&3\end{bmatrix}$ and $B=\begin{bmatrix}2&1\\-1&2\end{bmatrix}$, show that $(A+B)(A-B)=A^2-B^2$.

Given $A=\begin{bmatrix}3&4\\-4&3\end{bmatrix}$, $B=\begin{bmatrix}2&1\\-1&2\end{bmatrix}$.

Step 1 — check AB and BA (to see why the identity holds):

$AB$: Row 1: $[3(2)+4(-1),\ 3(1)+4(2)]=[2,\ 11]$; Row 2: $[-4(2)+3(-1),\ -4(1)+3(2)]=[-11,\ 2]$; $AB=\begin{bmatrix}2&11\\-11&2\end{bmatrix}$

$BA$: Row 1: $[2(3)+1(-4),\ 2(4)+1(3)]=[2,\ 11]$; Row 2: $[-1(3)+2(-4),\ -1(4)+2(3)]=[-11,\ 2]$; $BA=\begin{bmatrix}2&11\\-11&2\end{bmatrix}$

Since $AB=BA$ here, the general expansion $(A+B)(A-B)=A^2-AB+BA-B^2$ collapses exactly to $A^2-B^2$ — the identity to verify.

Step 2 — compute $(A+B)(A-B)$ directly:

$A+B=\begin{bmatrix}5&5\\-5&5\end{bmatrix}$, $A-B=\begin{bmatrix}1&3\\-3&1\end{bmatrix}$

$(A+B)(A-B)$: Row 1: $[5(1)+5(-3),\ 5(3)+5(1)]=[-10,\ 20]$; Row 2: $[-5(1)+5(-3),\ -5(3)+5(1)]=[-20,\ -10]$

$(A+B)(A-B)=\begin{bmatrix}-10&20\\-20&-10\end{bmatrix}$

Step 3 — compute $A^2-B^2$: …

Check that AB=BA for these particular matrices (which is what makes the identity hold), then verify …

• Assuming $(A+B)(A-B)=A^2-B^2$ holds for ANY two matrices — it only holds here because AB happens to equal BA; in general it expands to $A^2-AB+BA-B^2$.
• Sign slips expanding the product of two matrices with several negative entries. …