Question

Let \[ A= \begin{bmatrix} 1 & 0 & -1 & -3\\ 0 & 1 & 1 & k-1\\ 0 & 0 & k-1 & 1 \end{bmatrix} \] and \(k\in \mathbb{R}\). Then the value of \(k\), if exists, for which the rank of \(A\) is \(2\), is

• A. \(1\)
• B. Does not exist
• C. \(\dfrac{1}{3}\)
• D. \(1,\dfrac{1}{3}\)

Hint:

For rank to reduce, a row must become a linear combination of the other rows. If the row contains a fixed non-zero entry that cannot become zero, then rank reduction may be impossible.

Answer 1

The Correct Option is B

Step 1: Understand the given matrix.
The matrix is \[ A= \begin{bmatrix} 1 & 0 & -1 & -3\\ 0 & 1 & 1 & k-1\\ 0 & 0 & k-1 & 1 \end{bmatrix} \] This is a \(3\times 4\) matrix.
The first row is \[ R_1=[1,0,-1,-3] \] The second row is \[ R_2=[0,1,1,k-1] \] The third row is \[ R_3=[0,0,k-1,1] \]

Step 2: Check the independence of the first two rows.
The first row has leading entry in the first column, and the second row has leading entry in the second column.
So, \[ R_1 \text{ and } R_2 \] are linearly independent.
Therefore, \[ \operatorname{rank}(A)\geq 2 \]

Step 3: Condition for rank to be exactly \(2\).
For \[ \operatorname{rank}(A)=2, \] the third row must be a linear combination of the first two rows.
So, suppose \[ R_3=aR_1+bR_2 \] Then, \[ [0,0,k-1,1] = a[1,0,-1,-3]+b[0,1,1,k-1] \] This gives \[ [0,0,k-1,1] = [a,b,-a+b,-3a+b(k-1)] \]

Step 4: Compare corresponding entries.
Comparing the first component, \[ a=0 \] Comparing the second component, \[ b=0 \] Therefore, \[ aR_1+bR_2=0 \] So the third row must be the zero row for rank to be \(2\).
Thus, we need \[ [0,0,k-1,1]=[0,0,0,0] \] This gives \[ k-1=0 \] and \[ 1=0 \] But \[ 1=0 \] is impossible.

Step 5: Final conclusion.
Hence, there is no real value of \(k\) for which the rank of \(A\) is \(2\).
Therefore, \[ \boxed{\text{Does not exist}} \]