Let A =\(\left[\begin{matrix} 2 & 1 & 0 \\ 1 & 2 & -1 \\ 0 & -1 & 2 \end{matrix} \right]\). If |adj(adj(adj 2A)) | = (16)n, then n is equal to
We are given the matrix \( A \), and we need to calculate the value of \( n \) in the equation \( \left| \text{adj}(\text{adj}(\text{adj}(2A))) \right| = (16)^n \).
Step 1: Find the determinant of \( A \).
From the given matrix \( A \), we calculate the determinant \( |A| \): \[ |A| = 2[3] - 1[2] = 4. \] Step 2: Use the properties of the adjugate matrix.
We know the following properties of the adjugate matrix: \[ | \text{adj}(A) | = |A|^{n-1}. \] Therefore, \[ | \text{adj}(\text{adj}(\text{adj}(2A))) | = 2A |(n-1)3 = |2A|^8 = 16^n. \] Step 3: Simplify the equation.
We can now solve the equation: \[ (3^2) 4 = 16n = 16^n. \] Simplifying further: \[ (23 \times 32)^8 = 16^n \quad \Rightarrow 2^{40} = 16^n \quad \Rightarrow 16^{10} = 16^n \quad \Rightarrow n = 10. \] Final Answer: \( n = 10 \).
Let \(A=\) [\(a_{ij}\)]\(_{2\times2}\) be a matrix and \(A^2 = I\) where \(a_{ij} \neq0\). If a sum of diagonal elements and b=det(A), then \(3a^2+4b^2\) is
If \(A=\frac{1}{2}\begin{bmatrix}1 & \sqrt{3} \\ -\sqrt{3} & 1\end{bmatrix}\), then :
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)