Question

Let A be a symmetric matrix such that |A| = 2 and $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 3 & \alpha \\ 3 & \beta \end{bmatrix}$. If the sum of the diagonal

Answer 1

164/9

Answer 2

The problem statement is incomplete. It ends with "If the sum of the diagonal". Assuming the question asks for the sum of the diagonal elements of matrix A, we proceed as follows:

Let the given matrices be $M_1 = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$ and $M_2 = \begin{bmatrix} 3 & \alpha \\ 3 & \beta \end{bmatrix}$. Matrix $A$ is given by $A = M_1 M_2$.

1. Calculate Matrix A: $A = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 3 & \alpha \\ 3 & \beta \end{bmatrix} = \begin{bmatrix} (2)(3) + (1)(3) & (2)(\alpha) + (1)(\beta) \\ (1)(3) + (2)(3) & (1)(\alpha) + (2)(\beta) \end{bmatrix}$ $A = \begin{bmatrix} 6 + 3 & 2\alpha + \beta \\ 3 + 6 & \alpha + 2\beta \end{bmatrix} = \begin{bmatrix} 9 & 2\alpha + \beta \\ 9 & \alpha + 2\beta \end{bmatrix}$

2. Use the condition that A is a symmetric matrix:

A matrix $A = [a_{ij}]$ is symmetric if $a_{ij} = a_{ji}$ for all $i, j$. For a 2x2 matrix, this means the off-diagonal elements must be equal. From matrix A, we have $a_{12} = 2\alpha + \beta$ and $a_{21} = 9$. Therefore, $2\alpha + \beta = 9$. (Equation 1)

3. Use the condition that $|A| = 2$ (Determinant of A is 2):

The determinant of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is $ad - bc$. $|A| = (9)(\alpha + 2\beta) - (2\alpha + \beta)(9) = 2$ Factor out 9: $9[(\alpha + 2\beta) - (2\alpha + \beta)] = 2$ $9[\alpha + 2\beta - 2\alpha - \beta] = 2$ $9[\beta - \alpha] = 2$ $\beta - \alpha = \frac{2}{9}$ This can be written as $-\alpha + \beta = \frac{2}{9}$. (Equation 2)

4. Solve the system of linear equations for $\alpha$ and $\beta$:

We have two equations:

1. $2\alpha + \beta = 9$
2. $-\alpha + \beta = \frac{2}{9}$

Subtract Equation 2 from Equation 1: $(2\alpha + \beta) - (-\alpha + \beta) = 9 - \frac{2}{9}$ $2\alpha + \beta + \alpha - \beta = \frac{81 - 2}{9}$ $3\alpha = \frac{79}{9}$ $\alpha = \frac{79}{27}$

Substitute the value of $\alpha$ back into Equation 1 to find $\beta$: $2\left(\frac{79}{27}\right) + \beta = 9$ $\frac{158}{27} + \beta = 9$ $\beta = 9 - \frac{158}{27}$ $\beta = \frac{9 \times 27 - 158}{27}$ $\beta = \frac{243 - 158}{27}$ $\beta = \frac{85}{27}$

5. Calculate the sum of the diagonal elements of A:

The diagonal elements of A are $a_{11} = 9$ and $a_{22} = \alpha + 2\beta$. Sum of diagonal elements = $a_{11} + a_{22} = 9 + (\alpha + 2\beta)$ Substitute the values of $\alpha$ and $\beta$: Sum $= 9 + \left(\frac{79}{27} + 2\left(\frac{85}{27}\right)\right)$ Sum $= 9 + \left(\frac{79}{27} + \frac{170}{27}\right)$ Sum $= 9 + \left(\frac{79 + 170}{27}\right)$ Sum $= 9 + \frac{249}{27}$ Simplify the fraction $\frac{249}{27}$ by dividing the numerator and denominator by 3: $\frac{249 \div 3}{27 \div 3} = \frac{83}{9}$ Sum $= 9 + \frac{83}{9}$ Sum $= \frac{9 \times 9 + 83}{9}$ Sum $= \frac{81 + 83}{9}$ Sum $= \frac{164}{9}$