Let $D$ be a non-zero $n \times n$ real matrix with $n \ge 2$, Which of the following implications are valid?

1. $det(D) = 0 \implies rank(D)=0$.

2. $det(D) = 1 \implies rank(D)\ne1$.

3. $rank(D)=1 \implies det(D) \ne 0$ and

4. $rank(D) = n \implies det(D) \ne 1$.

1. (

2. (

3. (

4. (

1. $det(D) = 0 \implies rank(D)=0$.

2. $det(D) = 1 \implies rank(D)\ne1$.

3. $rank(D)=1 \implies det(D) \ne 0$ and

4. $rank(D) = n \implies det(D) \ne 1$.

**Solution: TO SAY AN OPTION IS WRONG IT IS ENOUGH TO GIVE AN EXAMPLE**1. (

**false**) Consider the matrix $D = \begin{bmatrix} 1&0 \\ 2&0 \end{bmatrix}$ then $det(D)= 0$ but $rank(D) = 1 \ne 0$.**Result : $D$ is invertible iff $det(D) \ne 0$ iff $rank(D) = n$. Equivalently****$det(D) = 0$ iff $rank(D) < n$.**2. (

**True**) Given that $det(D) = 1$ and hence by the above result $rank(D) = n$ and it is given that $n \ge 2$. Therefore $rank(D) \ne 1$.3. (

**false**) Given $rank(D) = 1$ which is strictly less than $n$. From the above-given result we have $det(D) = 0$.4. (

**false**) Consider the matrix $D = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$ then $rank(D)= 2 = n$ but $det(D) = 1$.

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