Question

If \( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix} \), then \( A^2 \) is equal to

• A. unit matrix
• B. null matrix
• C. A
• D. -A

Hint:

Matrices of this form often satisfy \(A^2 = I\) (involutory matrix).

Answer 1

The Correct Option is A

To find \( A^2 \), we need to multiply matrix \( A \) with itself. Given matrix \( A \) is:

\( A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix} \)

Now, let's calculate \( A^2 \) by multiplying matrix \( A \) with itself:

\( A^2 = A \times A \) \( = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix} \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ a & b & -1 \end{bmatrix} \)

We perform the matrix multiplication as follows:

• \(A[1,1] = (1 \times 1) + (0 \times 0) + (0 \times a) = 1\)
• \(A[1,2] = (1 \times 0) + (0 \times 1) + (0 \times b) = 0\)
• \(A[1,3] = (1 \times 0) + (0 \times 0) + (0 \times (-1)) = 0\)
• \(A[2,1] = (0 \times 1) + (1 \times 0) + (0 \times a) = 0\)
• \(A[2,2] = (0 \times 0) + (1 \times 1) + (0 \times b) = 1\)
• \(A[2,3] = (0 \times 0) + (1 \times 0) + (0 \times (-1)) = 0\)
• \(A[3,1] = (a \times 1) + (b \times 0) + (-1 \times a) = a - a = 0\)
• \(A[3,2] = (a \times 0) + (b \times 1) + (-1 \times b) = b - b = 0\)
• \(A[3,3] = (a \times 0) + (b \times 0) + (-1 \times (-1)) = 1\)

This gives us:

\( A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)

The resulting matrix is the identity matrix (unit matrix). Hence, the answer is the unit matrix.