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.
For real number a, b (a > b > 0), let
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \leq a^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 1 \right\} = 30\pi\)
and
\(\text{{Area}} \left\{ (x, y) : x^2 + y^2 \geq b^2 \text{{ and }} \frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1 \right\} = 18\pi\)
Then the value of (a – b)2 is equal to _____.
Simplify each of the following expressions:
(i) (3 + √3)(2 + √2)
(ii) (3 + √3)(3 - √3)
(iii) (√5 + √2 )2
(iv) (√5 - √2)(√5 + √2)