Here's a clear, educational, and well-structured solution to the problem.

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1. Understanding the Problem: Characteristic Equation and Eigenvalues

The problem begins by stating the equation $|\mathrm{A}-x \mathrm{I}|=0$. This is a fundamental concept in linear algebra known as the characteristic equation of the matrix $\mathrm{A}$. The roots of this equation, denoted by $x$, are called the eigenvalues of the matrix $\mathrm{A}$.

For a $2 \times 2$ matrix, there will always be two eigenvalues (which may be distinct or repeated, real or complex). We are given that these roots (eigenvalues) are $-1$ and $3$. Let's denote them as $\lambda_1 = -1$ and $\lambda_2 = 3$.

Why is this important? Eigenvalues are intrinsic properties of a matrix that describe how it transforms vectors. Many properties of a matrix, such as its trace and determinant, are directly related to its eigenvalues without needing to know the individual elements of the matrix.

2. Key Concepts: Properties of Eigenvalues for a $2 \times 2$ Matrix

For any $2 \times 2$ matrix $\mathrm{A}$, if its eigenvalues are $\lambda_1$ and $\lambda_2$, then there are two crucial properties that connect these eigenvalues to the matrix's fundamental scalar invariants:

1. Sum of Eigenvalues = Trace of the Matrix: The trace of a matrix $\mathrm{A}$, denoted as $\operatorname{tr}(\mathrm{A})$, is the sum of its diagonal elements. $\operatorname{tr}(\mathrm{A}) = \lambda_1 + \lambda_2$ Why? For a $2 \times 2$ matrix $\mathrm{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its characteristic equation is $(a-x)(d-x) - bc = 0$, which expands to $x^2 - (a+d)x + (ad-bc) = 0$. By Vieta's formulas, the sum of the roots ($x$) is $-( -(a+d) ) / 1 = a+d$, which is the trace.

2. Product of Eigenvalues = Determinant of the Matrix: The determinant of a matrix $\mathrm{A}$, denoted as $|\mathrm{A}|$, is a scalar value calculated from its elements. $|\mathrm{A}| = \lambda_1 \lambda_2$ Why? From the characteristic equation $x^2 - (a+d)x + (ad-bc) = 0$, the product of the roots ($x$) is $(ad-bc)/1$, which is the determinant of $\mathrm{A}$.

These properties are extremely powerful as they allow us to find the trace and determinant of a matrix solely from its eigenvalues, without needing to know the matrix elements.

3. Step 1: Calculate Trace and Determinant of A using Given Eigenvalues

We are given the eigenvalues $\lambda_1 = -1$ and $\lambda_2 = 3$. Let's apply the properties discussed above:

• Calculate the trace of A: $\operatorname{tr}(\mathrm{A}) = \lambda_1 + \lambda_2 = (-1) + 3 = 2$ So, the sum of the diagonal elements of matrix $\mathrm{A}$ is 2.

• Calculate the determinant of A: $|\mathrm{A}| = \lambda_1 \times \lambda_2 = (-1) \times 3 = -3$ So, the determinant of matrix $\mathrm{A}$ is -3.

4. Goal: Sum of Diagonal Elements of A² ($\operatorname{tr}(\mathrm{A}^2)$)

The problem asks for "the sum of the diagonal elements of the matrix $\mathrm{A}^2$". By definition, this is simply the trace of $\mathrm{A}^2$, i.e., $\operatorname{tr}(\mathrm{A}^2)$.

For a $2 \times 2$ matrix $\mathrm{A}$, there is a direct formula to calculate $\operatorname{tr}(\mathrm{A}^2)$ in terms of $\operatorname{tr}(\mathrm{A})$ and $|\mathrm{A}|$. This formula is: $\operatorname{tr}(\mathrm{A}^2) = (\operatorname{tr}(\mathrm{A}))^2 - 2|\mathrm{A}|$

Why this formula? (Derivation using Cayley-Hamilton Theorem) The Cayley-Hamilton Theorem states that every square matrix satisfies its own characteristic equation. For a $2 \times 2$ matrix $\mathrm{A}$, its characteristic equation is: $x^2 - (\operatorname{tr}(\mathrm{A}))x + |\mathrm{A}| = 0$ According to the Cayley-Hamilton Theorem, we can substitute the matrix $\mathrm{A}$ for $x$ (and the identity matrix $\mathrm{I}$ for the constant term to maintain matrix dimensions): $\mathrm{A}^2 - (\operatorname{tr}(\mathrm{A}))\mathrm{A} + |\mathrm{A}|\mathrm{I} = \mathrm{O}$ where $\mathrm{O}$ is the zero matrix. Rearranging this equation for $\mathrm{A}^2$: $\mathrm{A}^2 = (\operatorname{tr}(\mathrm{A}))\mathrm{A} - |\mathrm{A}|\mathrm{I}$ Now, let's take the trace of both sides: $\operatorname{tr}(\mathrm{A}^2) = \operatorname{tr}((\operatorname{tr}(\mathrm{A}))\mathrm{A} - |\mathrm{A}|\mathrm{I})$ Using the linearity property of the trace function ($\operatorname{tr}(k\mathrm{M} + l\mathrm{N}) = k\operatorname{tr}(\mathrm{M}) + l\operatorname{tr}(\mathrm{N})$) and knowing that $\operatorname{tr}(k\mathrm{I}) = k \times (\text{order of matrix})$: $\operatorname{tr}(\mathrm{A}^2) = (\operatorname{tr}(\mathrm{A}))\operatorname{tr}(\mathrm{A}) - |\mathrm{A}|\operatorname{tr}(\mathrm{I})$ Since $\mathrm{A}$ is a $2 \times 2$ matrix, $\mathrm{I}$ is a $2 \times 2$ identity matrix, so $\operatorname{tr}(\mathrm{I}) = 1+1 = 2$. $\operatorname{tr}(\mathrm{A}^2) = (\operatorname{tr}(\mathrm{A}))^2 - |\mathrm{A}|(2)$ $\operatorname{tr}(\mathrm{A}^2) = (\operatorname{tr}(\mathrm{A}))^2 - 2|\mathrm{A}|$ This derivation confirms the formula.

5. Step 2: Calculate $\operatorname{tr}(\mathrm{A}^2)$

Now we substitute the values of $\operatorname{tr}(\mathrm{A})$ and $|\mathrm{A}|$ that we found in Step 1 into the formula for $\operatorname{tr}(\mathrm{A}^2)$:

• $\operatorname{tr}(\mathrm{A}) = 2$
• $|\mathrm{A}| = -3$

$\operatorname{tr}(\mathrm{A}^2) = (\operatorname{tr}(\mathrm{A}))^2 - 2|\mathrm{A}|$ $\operatorname{tr}(\mathrm{A}^2) = (2)^2 - 2(-3)$ $\operatorname{tr}(\mathrm{A}^2) = 4 - (-6)$ $\operatorname{tr}(\mathrm{A}^2) = 4 + 6$ $\operatorname{tr}(\mathrm{A}^2) = 10$

Thus, the sum of the diagonal elements of $\mathrm{A}^2$ is 10.

Alternative Derivation (Direct Matrix Multiplication) Let $\mathrm{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. Then $\operatorname{tr}(\mathrm{A}) = a+d$ and $|\mathrm{A}| = ad-bc$. $\mathrm{A}^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a^2+bc & ab+bd \\ ca+dc & cb+d^2 \end{pmatrix}$ The sum of the diagonal elements of $\mathrm{A}^2$ is: $\operatorname{tr}(\mathrm{A}^2) = (a^2+bc) + (cb+d^2) = a^2+d^2+2bc$ We know that $(a+d)^2 = a^2+d^2+2ad$. So, $a^2+d^2 = (a+d)^2 - 2ad$. Substitute this into the expression for $\operatorname{tr}(\mathrm{A}^2)$: $\operatorname{tr}(\mathrm{A}^2) = ((a+d)^2 - 2ad) + 2bc$ $\operatorname{tr}(\mathrm{A}^2) = (a+d)^2 - 2ad + 2bc$ $\operatorname{tr}(\mathrm{A}^2) = (a+d)^2 - 2(ad-bc)$ Substituting $\operatorname{tr}(\mathrm{A}) = a+d$ and $|\mathrm{A}| = ad-bc$: $\operatorname{tr}(\mathrm{A}^2) = (\operatorname{tr}(\mathrm{A}))^2 - 2|\mathrm{A}|$ This algebraic derivation also confirms the formula and leads to the same result: $\operatorname{tr}(\mathrm{A}^2) = (2)^2 - 2(-3) = 4 + 6 = 10$

6. Key Takeaways and Tips

• Eigenvalues are powerful: They provide a concise way to understand properties of a matrix without explicitly knowing its elements.
• Characteristic Equation: Always recognize $|\mathrm{A}-x \mathrm{I}|=0$ as the characteristic equation, whose roots are the eigenvalues.
• Fundamental Relations for $2 \times 2$ Matrices:
  • Sum of eigenvalues = Trace of the matrix
  • Product of eigenvalues = Determinant of the matrix
• Trace of $A^2$ Formula: Remember the formula $\operatorname{tr}(\mathrm{A}^2) = (\operatorname{tr}(\mathrm{A}))^2 - 2|\mathrm{A}|$ for $2 \times 2$ matrices. This is a direct consequence of the Cayley-Hamilton Theorem.
• Common Mistakes:
  • Confusing sum and product of eigenvalues.
  • Algebraic errors when substituting values into formulas.
  • Forgetting the factor of 2 in $2|\mathrm{A}|$ or the negative sign in the formula for $\operatorname{tr}(\mathrm{A}^2)$.

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The final answer is $\boxed{10}$.