If $A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & \frac{1}{12} \end{bmatrix}$, then determinant of $A^{-1}$ is_______
To find the determinant of \(A^{-1}\), we first need to understand the relationship between a matrix and its inverse. For any invertible matrix \(A\), the determinant of the inverse, \(\det(A^{-1})\), is the reciprocal of the determinant of \(A\). Thus, \(\det(A^{-1}) = \frac{1}{\det(A)}\).
The given matrix \(A\) is a diagonal matrix:
\[A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & \frac{1}{12} \end{bmatrix}\]
The determinant of a diagonal matrix is the product of its diagonal elements:
\(\det(A) = 3 \times 4 \times \frac{1}{12} = \frac{12}{12} = 1\).
Now, substituting the determinant of \(A\) into the formula for the determinant of the inverse gives:
\(\det(A^{-1}) = \frac{1}{\det(A)} = \frac{1}{1} = 1\).
The calculated determinant of \(A^{-1}\) is 1. Checking this against the expected range (1,1), we confirm it correctly falls within the specified range.
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 ?
