Question

If \( A=\begin{bmatrix}1&-2&2\\ 2&1&-2\\ 2&K&4\end{bmatrix} \), \( B=\begin{bmatrix}2&4&3\\ 3&4&5\\ 1&2&2\end{bmatrix} \) and \( \text{Rank}(A)=2 \), then \( K + \text{Rank}(B) = \)

• A. \( 1 \)
• B. \( 0 \)
• C. \( -1 \)
• D. \( -2 \)

Hint:

For any \( n \times n \) square matrix, if its determinant is non-zero, its rank is exactly \( n \). If the determinant is zero, its rank is strictly less than \( n \).

Answer 1

The Correct Option is C

Concept: If a \( 3 \times 3 \) square matrix has a rank less than 3, its determinant must be equal to zero (\( |A| = 0 \)). We can utilize this rule to isolate the value of \( K \) from matrix \( A \), and then evaluate the rank of matrix \( B \) via row reduction or its determinant.

Step 1: Finding the value of \( K \) from \( \text{Rank}(A) = 2 \).
Since \( \text{Rank}(A) = 2 < 3 \), we set \( \det(A) = 0 \): \[ \begin{vmatrix}1&-2&2\\ 2&1&-2\\ 2&K&4\end{vmatrix} = 0 \] Expanding along the first row: \[ 1(4 - (-2K)) - (-2)(8 - (-4)) + 2(2K - 2) = 0 \] \[ (4 + 2K) + 2(12) + 4K - 4 = 0 \] \[ 4 + 2K + 24 + 4K - 4 = 0 \implies 6K + 24 = 0 \implies K = -4 \]

Step 2: Finding the Rank of matrix \( B \).
Let us calculate the determinant of matrix \( B \): \[ \det(B) = \begin{vmatrix}2&4&3\\ 3&4&5\\ 1&2&2\end{vmatrix} \] Expanding along the first row: \[ 2(8 - 10) - 4(6 - 5) + 3(6 - 4) = 2(-2) - 4(1) + 3(2) = -4 - 4 + 6 = -2 \] Since \( \det(B) = -2 \neq 0 \), matrix \( B \) is non-singular, which means its rank is equal to its order: \[ \text{Rank}(B) = 3 \]

Step 3: Calculating the final requested value.
\[ K + \text{Rank}(B) = -4 + 3 = -1 \]