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:

Question

Let $ A = \begin{bmatrix} 1 & \sqrt{2} \\ -2 & 0 & 1 \end{bmatrix} $ and $ P = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} $, $ \theta > 0 $. If $ B = P A P^T $, $ C = P^T B P $, and the sum of the diagonal elements of $ C $ is $ \frac{m}{n} $, where $ \gcd(m, n) = 1 $, then $ m + n $ is:

Answer 1

Step 1: Form the matrix $I - A = \begin{bmatrix} -4 & 0 & 2 \\ 0 & 0 & 0 \\ 4 & 0 & 2 \end{bmatrix}$ by subtracting $A$ from the identity matrix entrywise.

Step 2: Since the middle row is all zeros, the full $3\times 3$ determinant vanishes (a determinant with a zero row is always zero), so the rank cannot be $3$.

Step 3: Check a $2\times 2$ minor formed from rows 1 and 3, columns 1 and 3: $\begin{vmatrix} -4 & 2 \\ 4 & 2 \end{vmatrix} = (-4)(2) - (2)(4) = -8 - 8 = -16$, which is nonzero.

Step 4: A nonzero $2\times 2$ minor exists while every $3\times 3$ minor (the determinant itself) is zero, so the rank is exactly $2$.

\[\boxed{\text{rank}(I-A) = 2}\]

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:

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The Correct Option is C

Solution and Explanation

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