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 : \(A_2 \rightleftharpoons 2A\)
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,