Q.If $A=\begin{bmatrix}1&2&0\\5&4&2\\0&7&-3\end{bmatrix}$, find the product $(A+I)(A-I)$.

Given $A=\begin{bmatrix}1&2&0\\5&4&2\\0&7&-3\end{bmatrix}$, $I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$.

$A+I=\begin{bmatrix}2&2&0\\5&5&2\\0&7&-2\end{bmatrix}$, $A-I=\begin{bmatrix}0&2&0\\5&3&2\\0&7&-4\end{bmatrix}$

Multiply (A+I)(A−I):

Row 1: $[2(0)+2(5)+0(0),\ 2(2)+2(3)+0(7),\ 2(0)+2(2)+0(-4)]=[10,\ 10,\ 4]$

Row 2: $[5(0)+5(5)+2(0),\ 5(2)+5(3)+2(7),\ 5(0)+5(2)+2(-4)]=[25,\ 39,\ 2]$

Row 3: $[0(0)+7(5)+(-2)(0),\ 0(2)+7(3)+(-2)(7),\ 0(0)+7(2)+(-2)(-4)]=[35,\ 7,\ 22]$

$(A+I)(A-I)=\begin{bmatrix}10&10&4\\25&39&2\\35&7&22\end{bmatrix}$ …

Form A+I and A−I by adjusting only the diagonal of A, then multipl …

• Assuming $(A+I)(A-I)=A^2-I^2$ works like ordinary algebra for ANY two matrices $(X+Y)(X-Y)$ — it only collapses this way here because I commutes with every matrix, not in general for two arbitrary matrices. …