To determine the determinant of a \(3 \times 3\) matrix \(A\) with rank 2, we must understand the relationship between the rank of a matrix and its determinant.
The rank of a matrix is the maximum number of linearly independent row vectors or column vectors in the matrix. For a matrix to have full rank, the rank must be equal to the smallest dimension \(n\) of the matrix. Here, the smallest dimension of a \(3 \times 3\) matrix is 3.
Given:
- Matrix \(A\) is \(3 \times 3\).
- The rank of matrix \(A\) is 2.
The determinant of a square matrix is zero if and only if the matrix is singular, i.e., it does not have full rank. For a \(3 \times 3\) matrix, having a full rank means a rank of 3. Since matrix \(A\) has rank 2 (i.e., not full rank for a \(3 \times 3\) matrix), it is singular.
When a matrix is singular, its determinant is always zero.
Therefore, the determinant of the matrix \(A\) is 0.
Conclusion: Since the rank of matrix \(A\) is less than its full rank, it is singular, and thus, the determinant is 0.
Correct Answer: 0
Important Questions from Determinants
- An $n \times n$ square matrix $A$ satisfies $A^T = A^{-1}$. The determinant of this matrix may take which of the following value(s)?
The determinant of the matrix $\begin{pmatrix} 1 & p & q + r \\ 1 & q & r + p \\ 1 & r & p + q \end{pmatrix}$ is ______.
Let $f(x) = \begin{vmatrix} x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3 \end{vmatrix}$
where $p$ is a constant.
The value of $\frac{d^3}{dx^3}f(x)$ at $x = 0$ is
How many matrices of different orders are possible with elements comprising all prime numbers less than 30?
If A is a square matrix of order 3 and det A = 5, then what is det [(2A) -1 ] equal to ?
