ISI2019-MMA-15

Question

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

• A. $3$ for any real number $t$
• B. $2$ for any real number $t$
• C. $2$ or $3$ depending on the value of $t$
• D. $1,2$ or $3$ depending on the value of $t$

Answer 1

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.

Answer 2

Option C is correct

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

Comments

yes rank is 2 or 3 depending on the value of t as the determinant is 4t - 2t^2 -2.

now largest submatrix is 2*2 which have non-zero determinant. so, 1 can't be the rank.