Question:

Step 1: From the given relation $A^{-1} = 5A - A^2$, multiply both sides by $A$: $$A A^{-1} = A(5A - A^2)$$ $$I = 5A^2 - A^3$$ Rearranging this equation, we obtain: $$A^3 - 5A^2 + I = 0$$ Step 2: By the Cayley-Hamilton theorem, every matrix satisfies its own characteristic polynomial. For a $3 \times 3$ matrix $A$, the characteristic polynomial is of the form $p(\lambda) = \det(A - \lambda I) = -\lambda^3 + \text{tr}(A)\lambda^2 - C_1 \lambda + |A|$. Thus, $A$ satisfies the equation: $$A^3 - \text{tr}(A)A^2 + C_1 A - |A|I = 0$$ Comparing this equation with the derived equation $A^3 - 5A^2 + I = 0$: By comparing the coefficients of $A^2$, we find $\text{tr}(A) = 5$. By comparing the constant terms, we find $-|A|I = I$, which implies $|A| = -1$. Step 3: Calculate the trace of $A$ from its definition. Given $A = \begin{bmatrix} a & b & 1 \\ 2 & 1 & 3 \\ 1 & c & 2 \end{bmatrix}$, the trace is the sum of the diagonal elements: $$\text{tr}(A) = a + 1 + 2 = a + 3$$ Since we determined $\text{tr}(A) = 5$, we set $a + 3 = 5$, which implies $a = 2$. Step 4: Calculate the determinant of $A$ using the value $a=2$. Substitute $a=2$ into the matrix $A$: $$A = \begin{bmatrix} 2 & b & 1 \\ 2 & 1 & 3 \\ 1 & c & 2 \end{bmatrix}$$ The determinant $|A|$ is calculated as: $$|A| = 2(1 \cdot 2 - 3 \cdot c) - b(2 \cdot 2 - 3 \cdot 1) + 1(2 \cdot c - 1 \cdot 1)$$ $$|A| = 2(2 - 3c) - b(4 - 3) + 1(2c - 1)$$ $$|A| = 4 - 6c - b + 2c - 1$$ $$|A| = 3 - 4c - b$$ Since we determined $|A| = -1$, we set $3 - 4c - b = -1$. Rearranging this equation yields $b + 4c = 4$. Step 5: Consider the expression $Tr(A) = a+b+c$. We know that $Tr(A) = a+3$. For the statement $Tr(A) = a+b+c$ to be true, it must be that $a+3 = a+b+c$, which simplifies to $b+c=3$.