Let A be a symmetric matrix such that |A|=2 and A=pmatrix2/3 & 11 & 2pmatrix-type relation…

Question

Let $A$ be a symmetric matrix such that $|A|=2$ and $A=\begin{pmatrix}2/3 & 1\1 & 2\end{pmatrix}$-type relation determines $A=\begin{pmatrix}a&b\b&c\end{pmatrix}$ with $2a+b=1$, $2b+c=2$, $ac-b^2=2$. If the sum of the diagonal elements of $A$ is $s$, then $\dfrac{\beta s}{\alpha^2}$ (with $\alpha,\beta$ defined from $A$'s entries as in the source) is equal to \_\_\_\_\_\_\_\_.

Answer: 5

Accepted Answer

Solving $2a+b=1$, $2b+c=2$, and $ac-b^2=2$ simultaneously:

From $2a+b=1$: $a=\dfrac{1-b}2$. From $2b+c=2$: $c=2-2b$. Substituting into $ac-b^2=2$:
$$\frac{1-b}2(2-2b)-b^2=2 \Rightarrow (1-b)^2-b^2=2\Rightarrow1-2b=2\Rightarrow b=-\frac12.$$

Then $a=\dfrac34,\ c=3$.

With $\alpha=3a+\dfrac{3b}2=\dfrac94-\dfrac34=\dfrac32$ and $\beta=3b+\dfrac{3c}2=-\dfrac32+\dfrac92=3$, and $s=a+c=\dfrac{15}4$:
$$\frac{\beta s}{\alpha^2}=\frac{3\times\frac{15}4}{\left(\frac32\right)^2}=\frac{\frac{45}4}{\frac94}=5.$$

Solutions · 1