1. Problem 1: Given a 3x3 matrix \(A=\begin{bmatrix}1 & 0 & 0 \\ 2 & -3 & 0 \\ 1 & 4 & 2\end{bmatrix}\), - We know the sum of two eigenvalues equals the trace of \(A\). - Find the determinant \(|A|\). - Find the eigenvalues of \(A^2\). Step 1: Calculate the trace of \(A\): $$\text{trace}(A) = 1 + (-3) + 2 = 0$$ Step 2: Since the sum of two eigenvalues equals the trace (which is 0), let the eigenvalues be \(\lambda_1, \lambda_2, \lambda_3\). Given the sum of two of them equals 0, the third eigenvalue must satisfy the sum of all three equals 0. Step 3: Find eigenvalues of \(A\) by solving \(|A - \lambda I| = 0\): \[ \begin{vmatrix} 1-\lambda & 0 & 0 \\ 2 & -3-\lambda & 0 \\ 1 & 4 & 2-\lambda \end{vmatrix} = 0 \] Since the matrix is upper-triangular in the first row and last column structure, calculate determinant as: $$ (1-\lambda) \begin{vmatrix} -3-\lambda & 0 \\ 4 & 2-\lambda \end{vmatrix} = 0 $$ Calculate inner determinant: $$ (-3-\lambda)(2-\lambda) - 0 = (-3-\lambda)(2-\lambda) $$ So characteristic polynomial is: $$ (1-\lambda)((-3 - \lambda)(2 - \lambda)) = 0 $$ Set each factor to 0: - \(1-\lambda=0 \Rightarrow \lambda=1\) - \((-3 - \lambda)(2 - \lambda)=0\) gives \(\lambda = -3\) or \( \lambda = 2 \) Step 4: Eigenvalues of \(A\) are \(1, -3, 2\). Trace is \(1 - 3 + 2 = 0\). The sum of any two eigenvalues can be 0 if choosing \(1\) and \(-1\) but here, as given, sum of two eigenvalues equals trace. Step 5: Find eigenvalues of \(A^2\) by squaring eigenvalues of \(A\): $$ \lambda_i^2 = 1^2 = 1, (-3)^2 = 9, 2^2 = 4 $$ 2. Problem 2: Given matrix \( A=\begin{bmatrix} 1 & 1 & 3 \\ 1 & 5 & 3 \\ 3 & 1 & A^{-1} \end{bmatrix} \), find eigenvalues of \(A\). Step 1: Finding eigenvalues requires solving characteristic polynomial \(|A - \lambda I|=0\). Step 2: Note the matrix element in position (3,3) is \(A^{-1}\), possibly a typo or symbolic. Step 3: Since data is incomplete for numerical eigenvalue calculation, eigenvalues cannot be explicitly found. 3. Problem 3: For matrix \(A=\begin{bmatrix} 0 & 5 & -1 \\ 5 & 1 & 6 \\ -1 & 6 & 2 \end{bmatrix}\) with eigenvalues \(3,6\) given, find the quadratic form corresponding to \(A\). Step 1: The quadratic form is given by: $$ Q(x) = x^T A x = \sum_{i=1}^3 \sum_{j=1}^3 a_{ij} x_i x_j $$ Step 2: Writing explicitly, $$ Q(x) = 0 \cdot x_1^2 + 5 x_1 x_2 - x_1 x_3 + 5 x_2 x_1 + 1 x_2^2 + 6 x_2 x_3 - x_3 x_1 + 6 x_3 x_2 + 2 x_3^2 $$ Step 3: Since quadratic forms are symmetric, combine terms: $$ Q(x) = x_1^T 0 x_1 + 2 \times 5 x_1 x_2 + 2 \times (-1) x_1 x_3 + 1 x_2^2 + 2 \times 6 x_2 x_3 + 2 x_3^2 $$ So, $$ Q(x) = 10 x_1 x_2 - 2 x_1 x_3 + x_2^2 + 12 x_2 x_3 + 2 x_3^2 $$ 4. Problem 4: Find index and signature of quadratic form $$ x_1^2 + 2 x_2^2 - 3 x_3^2 $$ Step 1: Identify coefficients and signs: - Positive coefficients: \(1, 2\) - Negative coefficient: \(-3\) Step 2: Index is the number of positive coefficients: 2 Step 3: Signature is \((p, q)\) where \(p\) is number of positive terms and \(q\) is number of negative terms: $$ \text{Signature} = (2, 1) $$ 5. Problem 5: Evaluate \( \lim_{x \to ?} \frac{x^{-1}}{x+1} \) Step 1: The limit's variable approaching value is missing. Assume \(x \to 0\), then: $$ \lim_{x\to 0} \frac{1/x}{x+1} = \lim_{x\to 0} \frac{1}{x (x+1)} $$ Step 2: As \(x \to 0\), denominator \(x(x+1) \to 0\), so limit diverges to \(\infty\) or \(-\infty\) depending on direction. 6. Problem 6: Evaluate: $$ \lim_{x \to 0} \frac{1 + \cos 2x}{(\pi - 2x)^2} $$ Step 1: As \(x \to 0\), numerator \(1+\cos 0 = 1 + 1 = 2\) Step 2: Denominator \((\pi - 0)^2 = \pi^2\) Step 3: So limit: $$ \frac{2}{\pi^2} $$ 7. Problem 7: Find \( dy/dx \) if \( y = \sin(\log x) \) Step 1: Use chain rule: $$ \frac{d}{dx} y = \cos(\log x) \times \frac{d}{dx} (\log x) = \cos(\log x) \times \frac{1}{x} $$ So, $$ \frac{dy}{dx} = \frac{\cos(\log x)}{x} $$ 8. Problem 8: Find \( y' \) if: $$ \cos(x y) = 1 + \sin y $$ Step 1: Differentiate both sides w.r.t \(x\) implicitly: $$ -\sin(x y) \cdot (y + x y') = \cos y \cdot y' $$ Step 2: Rewrite derivative noting \( y \) is function of \( x \): $$ -\sin(x y)(y + x y') = \cos y \cdot y' $$ Step 3: Group terms of \(y'\): $$ -\sin(x y) y - x \sin(x y) y' = \cos y y' $$ Step 4: Move terms: $$ - x \sin(x y) y' - \cos y y' = \sin(x y) y $$ Step 5: Factor \( y' \): $$ y' (- x \sin(x y) - \cos y) = \sin(x y) y $$ Step 6: Solve for \(y'\): $$ y' = \frac{\sin(x y) y}{- x \sin(x y) - \cos y} = - \frac{y \sin(x y)}{x \sin(x y) + \cos y} $$ 9. Problem 9: Find critical values of: $$ f(x) = 5 x^3 - 6 x $$ Step 1: Find derivative: $$ f'(x) = 15 x^2 - 6 $$ Step 2: Set derivative to zero for critical points: $$ 15 x^2 - 6 = 0 \Rightarrow x^2 = \frac{6}{15} = \frac{2}{5} $$ Step 3: So critical values at: $$ x = \pm \sqrt{\frac{2}{5}} $$