Q.If $A=\begin{bmatrix}8&4\\10&5\end{bmatrix}$, $B=\begin{bmatrix}5&-4\\10&-8\end{bmatrix}$ show that $(A+B)^2=A^2+AB+B^2$.

Given $A=\begin{bmatrix}8&4\\10&5\end{bmatrix}$, $B=\begin{bmatrix}5&-4\\10&-8\end{bmatrix}$.

Step 1 — compute BA (to see why the printed identity works):

$(BA)_{11}=5(8)+(-4)(10)=40-40=0$

$(BA)_{12}=5(4)+(-4)(5)=20-20=0$

$(BA)_{21}=10(8)+(-8)(10)=80-80=0$

$(BA)_{22}=10(4)+(-8)(5)=40-40=0$

$BA=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O$

Since in general $(A+B)^2=A^2+AB+BA+B^2$, and here $BA=O$, this reduces to exactly $A^2+AB+B^2$ — the identity to verify.

Step 2 — compute $(A+B)^2$ directly:

$A+B=\begin{bmatrix}13&0\\20&-3\end{bmatrix}$

$(A+B)^2$: Row 1: $[13(13)+0(20),\ 13(0)+0(-3)]=[169,\ 0]$; Row 2: $[20(13)+(-3)(20),\ 20(0)+(-3)(-3)]=[200,\ 9]$

$(A+B)^2=\begin{bmatrix}169&0\\200&9\end{bmatrix}$

Step 3 — compute $A^2$, $AB$, $B^2$ and add:

$A^2$: Row 1: $[8(8)+4(10),\ 8(4)+4(5)]=[104,\ 52]$; Row 2: $[10(8)+5(10),\ 10(4)+5(5)]=[130,\ 65]$; $A^2=\begin{bmatrix}104&52\\130&65\end{bmatrix}$ …

Compute AB and BA first — noticing BA=O explains why the printed identity (which omits a BA term) holds — then …

• Assuming $(A+B)^2=A^2+AB+B^2$ is a general algebraic identity — it only holds here because $BA=O$ for this particular A, B; in general the full expansion needs a BA term too.
• Sign errors handling the negative entries of B. …