(A)Let \( A = \begin{bmatrix} a & b b & c \end{bmatrix} \). Since \( |A| = 2 \): \[ ac - b^2 = 2. \] (B)From the given equation: \[ \begin{bmatrix} 3 & -2 2 & 1 \end{bmatrix} \begin{bmatrix} a & b b & c \end{bmatrix} = \begin{bmatrix} 1 & 2 2 & 7 \end{bmatrix}. \] Expanding row-wise gives equations: \[ 3a - 2b = 1, \quad 3b - 2c = 2, \quad 2a + b = 2, \quad 2b + c = 7. \] (C)Solve these equations to find: \[ a = \frac{3}{4}, \, b = \frac{5}{4}, \, c = \frac{9}{2}. \] Sum of diagonal elements: \[ s = a + c = \frac{3}{4} + \frac{9}{2} = \frac{21}{4}. \] (D)Given \( \alpha = 3 \) and \( \beta = 15 \), compute: \[ \frac{\beta s}{\alpha^2} = \frac{15 \times \frac{21}{4}}{9} = 5. \]
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: \[ x - 2y = 10, \] \[ 2x - y - z = 8, \] \[ -2y + z = 7. \]
In the given circuit the sliding contact is pulled outwards such that the electric current in the circuit changes at the rate of 8 A/s. At an instant when R is 12 Ω, the value of the current in the circuit will be A.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
For $ \alpha, \beta, \gamma \in \mathbb{R} $, if $$ \lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2} - 3}{\sin 2x - \beta x} = 3, $$ then $ \beta + \gamma - \alpha $ is equal to:
A matrix is a rectangular array of numbers, variables, symbols, or expressions that are defined for the operations like subtraction, addition, and multiplications. The size of a matrix is determined by the number of rows and columns in the matrix.