Step 1: Using the given condition \( |A| = 2 \). We know that: \[ |kA| = k^m \cdot |A| \quad \text{where } m = \text{order of the matrix}. \] From the given, \( |2A| = 2^2 |A| = 4 |A| \), so \[ |2 \, adj(2A)| = |2^7 \, adj(A)| = 2^7 \cdot |adj(A)|. \]
Step 2: Simplifying the expression. Since we know that \( |adj(A)| = |A|^{m-1} \), we can substitute \( |A| = 2 \): \[ |adj(A)| = |A|^2 = 2^2 = 4. \] Thus, \[ |2 \, adj(2 \, adj(2A))| = 2^7 \cdot 4 = 2^7 \cdot 2^2 = 2^{9} = 32^n. \] Equating powers of 2, we get \( n = 5 \).
Step 3: Solving for \( \alpha \). We now use the equation for \( |A| = 2 \): \[ (6-1) - 2(2\alpha - 1) + 3(\alpha - 3) = 2. \] Simplifying: \[ 5 - 4\alpha + 2 + 3\alpha - 9 = 2 \quad \Rightarrow \quad 5 - 4\alpha + 2 + 3\alpha - 9 = 2 \quad \Rightarrow \quad \alpha = -4. \]
Step 4: Final Calculation. Finally, we calculate \( 3n + \alpha \): \[ 3(5) + (-4) = 15 - 4 = 11. \]
If \[ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix} \] then find \( A^{-1} \). Hence, solve the system of linear equations: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]
Let $ A \in \mathbb{R} $ be a matrix of order 3x3 such that $$ \det(A) = -4 \quad \text{and} \quad A + I = \left[ \begin{array}{ccc} 1 & 1 & 1 \\2 & 0 & 1 \\4 & 1 & 2 \end{array} \right] $$ where $ I $ is the identity matrix of order 3. If $ \det( (A + I) \cdot \text{adj}(A + I)) $ is $ 2^m $, then $ m $ is equal to:
A square loop of sides \( a = 1 \, {m} \) is held normally in front of a point charge \( q = 1 \, {C} \). The flux of the electric field through the shaded region is \( \frac{5}{p} \times \frac{1}{\varepsilon_0} \, {Nm}^2/{C} \), where the value of \( p \) is: