Q.If $A=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$, Show that $A^2-4A$ is a scalar matrix.

Given $A=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix}$.

Compute $A^2=A\cdot A$:

Row 1: $[1(1)+2(2)+2(2),\ 1(2)+2(1)+2(2),\ 1(2)+2(2)+2(1)]=[9,\ 8,\ 8]$

Row 2: $[2(1)+1(2)+2(2),\ 2(2)+1(1)+2(2),\ 2(2)+1(2)+2(1)]=[8,\ 9,\ 8]$

Row 3: $[2(1)+2(2)+1(2),\ 2(2)+2(1)+1(2),\ 2(2)+2(2)+1(1)]=[8,\ 8,\ 9]$

$A^2=\begin{bmatrix}9&8&8\\8&9&8\\8&8&9\end{bmatrix}$

Form 4A:

$4A=\begin{bmatrix}4&8&8\\8&4&8\\8&8&4\end{bmatrix}$

Subtract:

$A^2-4A=\begin{bmatrix}9-4&8-8&8-8\\8-8&9-4&8-8\\8-8&8-8&9-4\end{bmatrix}=\begin{bmatrix}5&0&0\\0&5&0\\0&0&5\end{bmatrix}$ …

Compute A² by matrix multiplication, subtract 4A, and show every off-diagonal entry vanishes whil …

• Forgetting the definition of a scalar matrix (diagonal, all diagonal entries equal) and stopping once the off-diagonal zeros appear without checking the diagonal entries match. …