Question

If \(A = \begin{bmatrix} 5 & 0 & -2 \\ 0 & 1 & 0 \\ -4 & 0 & -1 \end{bmatrix}\) and \(I\) is the \(3 \times 3\) unit matrix, then the rank of \(I - A\) is:

• A. 0
• B. 1
• C. 2
• D. 3

Hint:

Compute \(I - A\) directly, then check if its nonzero rows are independent.

Answer 1

The Correct Option is C

Step 1: Write down \(I - A\). Since \(I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\) and \(A = \begin{bmatrix} 5 & 0 & -2 \\ 0 & 1 & 0 \\ -4 & 0 & -1 \end{bmatrix}\), subtracting entrywise gives \[I - A = \begin{bmatrix} -4 & 0 & 2 \\ 0 & 0 & 0 \\ 4 & 0 & 2 \end{bmatrix}\]

Step 2: Reduce using row operations. Add row 1 to row 3: \(R_3 \to R_3 + R_1\) gives \((4,0,2)+(-4,0,2) = (0,0,4)\). The matrix becomes \[\begin{bmatrix} -4 & 0 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 4 \end{bmatrix}\]

Step 3: The middle row is entirely zero, so it contributes nothing to the rank. The remaining two rows, \((-4,0,2)\) and \((0,0,4)\), are not proportional to each other, so they are linearly independent.

Step 4: Since there are exactly 2 linearly independent rows, the rank of \(I-A\) is \(2\). The answer is option (C), \(\boxed{2}\).