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

Step 1: Use the meaning of rank 2.
For a $3\times3$ matrix, if the rank is less than 3, its determinant must be zero. Since $\text{Rank}(A)=2$, we set $|A|=0$ to find $K$.
Step 2: Expand the determinant of A.
Expanding along the first row gives \[ 1(4+2K) + 2(8+4) + 2(2K-2) = 0 \]
Step 3: Simplify and solve for K.
This becomes $4 + 2K + 24 + 4K - 4 = 0$, so $6K + 24 = 0$, giving $K = -4$.
Step 4: Now find the rank of B.
For $B$, check its determinant. If it is not zero, the rank is the full value 3.
Step 5: Compute the determinant of B.
Expanding along the first row, \[ 2(8-10) - 4(6-5) + 3(6-4) = -4 - 4 + 6 = -2 \] Since $-2 \neq 0$, matrix $B$ has rank 3.
Step 6: Add the results.
Now combine: $K + \text{Rank}(B) = -4 + 3$. \[ \boxed{-1} \]