Question:

Step 1: Relate the given matrix equation to the characteristic polynomial using the Cayley-Hamilton theorem. The given matrix equation is $A^3 = 4A^2 - A - 21I$. Rearranging the terms, we get: $$A^3 - 4A^2 + A + 21I = 0$$ By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. For a $3 \times 3$ matrix $A$, the characteristic polynomial $\det(A-\lambda I)$ is given by: $$\det(A-\lambda I) = -\lambda^3 + \text{Tr}(A)\lambda^2 - (M_{11}+M_{22}+M_{33})\lambda + \det(A)$$ Setting this to zero and replacing $\lambda$ with $A$, we obtain the characteristic equation for the matrix $A$: $$A^3 - \text{Tr}(A)A^2 + (M_{11}+M_{22}+M_{33})A - \det(A)I = 0$$ Comparing the coefficients of the powers of $A$ in the given equation $A^3 - 4A^2 + A + 21I = 0$ with the Cayley-Hamilton equation, we deduce: 1. $\text{Tr}(A) = 4$ 2. $M_{11}+M_{22}+M_{33} = 1$ (where $M_{ii}$ are the principal minors of $A$) 3. $-\det(A) = 21 \implies \det(A) = -21$ Step 2: Calculate the Trace of $A$ and determine the value of $b$. The trace of a matrix is the sum of its diagonal elements. For the given matrix $A=\begin{bmatrix}2&a&0\\1&3&1\\0&5&b\end{bmatrix}$: $$\text{Tr}(A) = 2+3+b = 5+b$$ From Step 1, we know $\text{Tr}(A) = 4$. Equating these, we get: $$5+b = 4 \implies b = -1$$ Step 3: Calculate the sum of principal minors and determine the value of $a$. The principal minors are the determinants of the submatrices obtained by removing the $i$-th row and $i$-th column. $M_{11} = \det \begin{pmatrix} 3 & 1 \\ 5 & b \end{pmatrix} = 3b - 5$ $M_{22} = \det \begin{pmatrix} 2 & 0 \\ 0 & b \end{pmatrix} = 2b - 0 = 2b$ $M_{33} = \det \begin{pmatrix} 2 & a \\ 1 & 3 \end{pmatrix} = 6 - a$ The sum of principal minors is: $$M_{11}+M_{22}+M_{33} = (3b-5) + 2b + (6-a) = 5b - a + 1$$ From Step 1, we know $M_{11}+M_{22}+M_{33} = 1$. Equating these, we get: $$5b - a + 1 = 1 \implies 5b - a = 0$$ Substitute the value of $b=-1$ from Step 2: $$5(-1) - a = 0 \implies -5 - a = 0 \implies a = -5$$ Step 4: Calculate the expression $2a+3b$. Now that we have the values $a=-5$ and $b=-1$, we can calculate the required expression: $$2a+3b = 2(-5) + 3(-1)$$ $$2a+3b = -10 - 3$$ $$2a+3b = -13$$ Step 5: State the final answer. The value of $2a+3b$ is $-13$. The final answer is $\boxed{-13}$. <div class="key-concept"><strong>Key Concept:</strong> From $A^3-4A^2+A+21I=0$, the characteristic polynomial of $A$ is $\lambda^3-4\lambda^2+\lambda+21=0$. Use $\text{tr}(A)=2+3+b=5+b$ and constant term $=-\det(A)=-21$.</div> <div class="trap-box"><strong>Trap:</strong> $\text{tr}(A)=4\Rightarrow5+b=4+b$... From solution: $\text{tr}(A)=4\Rightarrow2+3+b=4$... wait: coefficient of $\lambda^2$ is $-\text{tr}(A)=-4$, so $\text{tr}(A)=4\Rightarrow b=-1$. $\det(A)=-21$: $\det=-16+a=-21\Rightarrow a=-5$. $2a+3b=-10-3=-13$.</div>