We are given two equations that describe the matrix \( A \):
\( A^2 = 3A + \alpha I \quad \cdots (1)\)
\( A^4 = 21A + \beta I \quad \cdots (2)\)
Where \( A \) is a matrix, and \( I \) is the identity matrix.
We need to express \( A^4 \) in terms of \( A^2 \). To begin, we square equation (1) to get the expression for \( A^2 \):
\( A^2 = 3A + \alpha I \)
Now, multiply both sides of this equation by \( A \) to find the expression for \( A^3 \):
\( A^3 = A \cdot A^2 = A(3A + \alpha I) = 3A^2 + \alpha A\)
Now, substitute \( A^2 = 3A + \alpha I \) into this equation to simplify:
\( A^3 = 3(3A + \alpha I) + \alpha A = 9A + 3\alpha I + \alpha A\)
Now, expand to get the expression for \( A^4 \):
\( A^4 = (9 + \alpha)A^2 + 3\alpha A\)
Now, substitute the expression for \( A^2 \) from equation (1) into this new expression for \( A^4 \):
\( A^4 = (9 + \alpha)(3A + \alpha I) + 3\alpha A\)
Now, expand this equation:
\( A^4 = A(27 + 6\alpha) + \alpha(9 + \alpha)I\)
From the equation for \( A^4 \), we now have:
\( A^4 = A(27 + 6\alpha) + \alpha(9 + \alpha)I\)
To make this consistent with equation (2), compare the coefficients of \( A \) and \( I \):
For the coefficient of \( A \), we have:
\( 27 + 6\alpha = 21 \Rightarrow \alpha = -1\)
For the coefficient of \( I \), we have:
\( \beta = \alpha(9 + \alpha) = -8\)
Thus, the values of \( \alpha \) and \( \beta \) are:
\( \alpha = -1, \quad \beta = -8\)
Therefore, the correct values for \( \alpha \) and \( \beta \) are \( \alpha = -1 \) and \( \beta = -8 \), as derived from the matrix equations.
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,