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The rank of the matrix $\begin{bmatrix} 0 &1 &t \\ 2& t & -1\\ 2& 2 & 0 \end{bmatrix}$ equals 

  1. $3$ for any real number $t$
  2. $2$ for any real number $t$
  3. $2$ or $3$ depending on the value of $t$
  4. $1,2$ or $3$ depending on the value of $t$
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2 Answers

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Convert the given matrix in Row-Echelon form.

$\begin{bmatrix} 0 & 1 & t \\ 2 & t & -1 \\ 2 & 2 & 0 \end{bmatrix} \implies \begin{bmatrix} 2 & 2 & 0 \\ 2 & t & -1 \\ 0 & 1 & t \end{bmatrix} \implies \begin{bmatrix} 2 & 2 & 0 \\ 0 & 2 - t & 1 \\ 0 & 1 & t \end{bmatrix}$

Now, to make the second and third rows equal, we should have $$2 - t = t \implies t = 1$$

If $t = 1$, then rank of matrix will be $2$ since there are only two linearly independent rows in the matrix.

For all other values of $t$, the rank will be $3$.

So, Option $(C)$ is correct.
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0 votes
Option C is correct

Rank can be 2.. and max rank is 3 when determinant is not zero

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