Matrix Determinant Calculation
We need to calculate the determinant of $2A$, where $A$ is a $2 \times 2$ matrix and $\text{det}(A) = 2$. We use the properties of determinants for this calculation.
Determinant Property for Scalar Multiplication
For any $n \times n$ matrix $A$ and any scalar $k$, the determinant of the matrix $kA$ is related to the determinant of $A$ by the formula:
$ \text{det}(kA) = k^n \text{det}(A) $
Here, $n$ represents the dimension (number of rows or columns) of the square matrix.
Applying the Property to the Problem
Given information:
- The matrix $A$ is $2 \times 2$, so its dimension is $n=2$.
- The scalar multiplier is $k=2$.
- The determinant of matrix $A$ is $\text{det}(A) = 2$.
Calculating det(2A)
Substitute the values into the determinant property formula:
$ \text{det}(2A) = 2^n \times \text{det}(A) $
Since $n=2$, the equation becomes:
$ \text{det}(2A) = 2^2 \times \text{det}(A) $
Now, substitute the given value of $\text{det}(A) = 2$:
$ \text{det}(2A) = 2^2 \times 2 $
$ \text{det}(2A) = 4 \times 2 $
$ \text{det}(2A) = 8 $
Final Result
The determinant of $2A$ is 8.
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 ?
