Matrix Determinant Properties Explained
We are given information about two square matrices, A and B, of the same size.
- The determinant of their product is $\det[\textbf{A} \cdot \textbf{B}] = -80$.
- The determinant of matrix A is $\det[\textbf{A}] = 5$.
The goal is to find the value of $\det[\textbf{B}^T]$.
Calculating the Determinant of B
We use a fundamental property of determinants: the determinant of a matrix product equals the product of the determinants.
Mathematically, this is stated as:
$ \det[\textbf{A} \cdot \textbf{B}] = \det[\textbf{A}] \cdot \det[\textbf{B}] $
Substitute the known values into this equation:
$ -80 = 5 \cdot \det[\textbf{B}] $
To find $\det[\textbf{B}]$, we rearrange the equation:
$ \det[\textbf{B}] = \frac{-80}{5} $
$ \det[\textbf{B}] = -16 $
Determining the Determinant of B Transpose
Another key property of determinants is that the determinant of a matrix transpose is always equal to the determinant of the original matrix.
This property is:
$ \det[\textbf{B}^T] = \det[\textbf{B}] $
Since we calculated $\det[\textbf{B}]$ to be $-16$, we can directly find $\det[\textbf{B}^T]$:
$ \det[\textbf{B}^T] = -16 $
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 ?
