If \( \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc \), then the value of \( k \) is:
We are given the following determinant and are tasked with finding its value: \[ \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc \]. Taking \(a\), \(b\), and \(c\) out of the matrix from columns \(C_1\), \(C_2\), and \(C_3\), respectively: \[ abc \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = kabc \] Dividing both sides by \(abc\), we get: \[ \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = k \] Using column operations \(C_2 \to C_2 + C_1\) and \(C_3 \to C_3 + C_1\), the determinant simplifies to: \[ \begin{vmatrix} -1 & 0 & 0 \\ 1 & 0 & 2 \\ 1 & 2 & 0 \end{vmatrix} = k \] Expanding the determinant along the first row: \[ -1(0 \times 0 - 2 \times 2) = k \] .
Simplifying further: \[ -1(-4) = k \] \[ k = 4 \] \(\therefore k = 4\) Given the problem setup, the value of \(k\) is 4, and thus the correct option is <strong>(D) 4</strong>.
Let I be the identity matrix of order 3 × 3 and for the matrix $ A = \begin{pmatrix} \lambda & 2 & 3 \\ 4 & 5 & 6 \\ 7 & -1 & 2 \end{pmatrix} $, $ |A| = -1 $. Let B be the inverse of the matrix $ \text{adj}(A \cdot \text{adj}(A^2)) $. Then $ |(\lambda B + I)| $ is equal to _______
Let $A = \{ z \in \mathbb{C} : |z - 2 - i| = 3 \}$, $B = \{ z \in \mathbb{C} : \text{Re}(z - iz) = 2 \}$, and $S = A \cap B$. Then $\sum_{z \in S} |z|^2$ is equal to
If $ y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\27 & 28 & 27 \\1 & 1 & 1 \end{vmatrix} $, $ x \in \mathbb{R} $, then $ \frac{d^2y}{dx^2} + y $ is equal to
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: