If \(A = \begin{bmatrix}1&0&0\\0&1&0\\a&b&-1\end{bmatrix}\), where a and b are any real numbers, then \(A^2\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → I, the identity matrix of order 3
Multiplying rows of A by columns of A:
Row 1: \(\begin{bmatrix}1&0&0\end{bmatrix}\)
Row 2: \(\begin{bmatrix}0&1&0\end{bmatrix}\)
Row 3: \(\begin{bmatrix}a(1) + b(0) + (-1)a&a(0) + b(1) + (-1)b&(-1)(-1)\end{bmatrix} = \begin{bmatrix}0&0&1\end{bmatrix}\)
So
\( \displaystyle A^2 = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix} = I \)
for every value of a and b.