Question

If \(A\) is square matrix such that \(A^2 = A\), then \((2I + A)^4 - 65A\) is equal to (Where \(I\) is identity matrix)

• A. \(8I\)
• B. \(16I\)
• C. \(16I + A\)
• D. \(8I + A\)

Hint:

Since \(A^2=A\), square \(2I+A\) twice and use \(A^2=A\) each time. You get \(16I+65A\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A matrix with \(A^2=A\) is called idempotent. Then every power \(A^n\) equals \(A\). Also \(I\) commutes with every matrix, so we can expand \((2I+A)^n\) like an ordinary binomial.

Step 2: Square the bracket first.
\[ (2I+A)^2 = 4I^2 + 4IA + A^2 = 4I + 4A + A \]
Here we used \(I^2=I\), \(IA=A\) and \(A^2=A\). So \[ (2I+A)^2 = 4I+5A \]

Step 3: Square again to get the fourth power.
\[ (2I+A)^4 = (4I+5A)^2 = 16I + 40A + 25A^2 \]
Using \(A^2=A\), this becomes \[ 16I + 40A + 25A = 16I + 65A \]

Step 4: Subtract 65A.
\[ (2I+A)^4 - 65A = 16I + 65A - 65A = 16I \]

Step 5: Check the options.
Option 1, \(8I\), is too small. Options 3 and 4 keep an extra \(A\), but the \(A\) terms cancel fully. So only option 2 fits. A quick test with \(A=I\) gives \(3^4-65=16\), which equals \(16I\) for the identity, so it matches.

Final Answer:
The value is \(16I\). \[ \boxed{16I} \]