The determinant of the matrix $\begin{pmatrix} 1 & p & q + r \\ 1 & q & r + p \\ 1 & r & p + q \end{pmatrix}$ is ______.
We need to calculate the determinant of the matrix:
$ A = \begin{pmatrix} 1 & p & q + r \\ 1 & q & r + p \\ 1 & r & p + q \end{pmatrix} $We can simplify this calculation using properties of determinants, specifically column operations.
Let $C_1$, $C_2$, and $C_3$ denote the columns of the matrix. Replace the third column ($C_3$) with the sum of the third and second columns ($C_3 + C_2$). This operation preserves the determinant's value.
The elements of the new third column become:
The matrix after the column operation is:
$ \begin{pmatrix} 1 & p & p + q + r \\ 1 & q & p + q + r \\ 1 & r & p + q + r \end{pmatrix} $Observe that the entire third column consists of the term $(p + q + r)$. We can factor this common term out of the determinant:
$ \det(A) = (p + q + r) \det \begin{pmatrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{pmatrix} $In the matrix $\begin{pmatrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{pmatrix}$, the first column ($C_1$) and the third column ($C_3$) are identical.
A key property of determinants states that if a matrix has two identical columns (or rows), its determinant is zero.
Therefore, the determinant of the matrix $\begin{pmatrix} 1 & p & 1 \\ 1 & q & 1 \\ 1 & r & 1 \end{pmatrix}$ is 0.
Substituting this result back into the factored expression:
$ \det(A) = (p + q + r) \times 0 $ $ \det(A) = 0 $The determinant of the given matrix is 0.
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 ?
If (a - b) (b - c) (c - a) = 2 and abc = 6, then what is the value of \(\begin{vmatrix}a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix}?\)