Linear Algebra and Calculus

Eigenvalues of matrix

An eigenvalue \lambda of a square matrix A is a scalar for which A\mathbf{v}=\lambda\mathbf{v} has a non-zero solution \mathbf{v}, called an eigenvector. Equivalently, \det(A-\lambda I)=0, which is the characteristic equation. For a 2\times2 matrix it reads \lambda^2-\operatorname{tr}(A)\lambda+\det A=0, so the two eigenvalues sum to the trace and multiply to the determinant. Several shortcuts are standard in exams. If A is invertible and \lambda is an eigenvalue of A, then 1/\lambda is an eigenvalue of A^{-1}, because A^{-1}\mathbf{v}=\mathbf{v}/\lambda. If A=V D V^{-1}, where D is diagonal and the columns of V are eigenvectors, then the given eigenpairs determine A completely. For a rank-one matrix, the only non-zero eigenvalue equals the trace. If a column of A has zeros everywhere except on the diagonal, that diagonal entry is an eigenvalue, and expanding the characteristic determinant factors out (a_{ii}-\lambda). For a 2\times2 matrix, substituting a candidate value into \lambda^2-\operatorname{tr}(A)\lambda+\det A shows directly whether it is an eigenvalue. Example: A=\begin{pmatrix}2&1\\1&2\end{pmatrix} has trace 4 and determinant 3, so its eigenvalues are 1 and 3.

Matrix determinant

The determinant \det A of a square matrix decides invertibility: A is invertible exactly when \det A\neq0, and then \det(A^{-1})=1/\det A. For 2\times2 matrices \det A=ad-bc. For larger matrices, use cofactor expansion along the row or column with the most zeros. Row operations change the determinant in controlled ways. Adding a multiple of one row to another leaves it unchanged, swapping two rows changes the sign, and scaling a row by c multiplies the determinant by c. Hence an upper or lower triangular matrix has determinant equal to the product of its diagonal entries. For a homogeneous system M\mathbf{x}=\mathbf{0} in two unknowns, a non-zero solution exists if and only if \det M=0, so the parameter values are obtained by solving this equation. A square matrix with zero determinant is singular, has rank smaller than its order, and has no inverse. It still has a determinant and a transpose, since every square matrix has both. Example: \begin{vmatrix}1&2\\3&4\end{vmatrix}=-2, and the system 3x+ky=0, x+2y=0 has a non-zero solution when 6-k=0, i.e. k=6. Numerical integration is also tested in this group. The composite trapezoidal rule with step h approximates \int_a^b f\,dx\approx\frac{h}{2}\left[f_0+f_n+2\sum_{i=1}^{n-1}f_i\right], and the error shrinks as h decreases. For example, \int_0^1x^2\,dx with h=0.5 gives 0.375, close to 1/3.

Infinite geometric series

A geometric series a+ar+ar^2+\cdots, with first term a and common ratio r, converges if and only if |r|<1. Its sum is S=\frac{a}{1-r}. The key step is recognising a geometric pattern: each term must be a fixed multiple of the previous one, and the ratio is found as t_{n+1}/t_n. Series whose terms are powers of an expression, such as a trigonometric function evaluated at a point, become geometric once that value is substituted. When the ratio is 1/\sqrt2 at a standard angle, the sum is \frac{1}{1-1/\sqrt2}=\frac{\sqrt2}{\sqrt2-1}, which is rationalised to 2+\sqrt2. If |r|\ge1 the series diverges and the formula must not be used. Example: \sum_{n\ge0}\left(\tfrac25\right)^n=\frac{1}{1-2/5}=\frac53.

Definite integral evaluation

A definite integral \int_a^b f(x)\,dx equals F(b)-F(a), where F is an antiderivative of f. Substitution is a common technique. When the variable changes, the limits must change too, or the integral must be converted back before the original limits are applied. Integration by parts uses \int u\,dv=uv-\int v\,du, with u chosen so that its derivative simplifies; a factor such as 2t is a typical choice for u. The integral of a linear function over an interval is the area of a trapezoid. For y=ax+b on [0,L], the area is \int_0^L(ax+b)\,dx=\frac{aL^2}{2}+bL. Thus, over a fixed interval, a line with larger slope and intercept encloses more area. Example: \int_0^1(2x+3)\,dx=1+3=4.

Function extrema via derivative

To find the maximum or minimum of a differentiable function on a domain, first locate critical points where f'(x)=0 or f' is undefined. Then classify each point with the second derivative test: f''<0 gives a local maximum and f''>0 gives a local minimum. Endpoints of the domain and the behaviour at its boundary must also be examined, because the second derivative test applies only to interior points. A critical point outside the restricted domain is discarded. For f(x)=x+c/x with x>0, the derivative vanishes at x=\sqrt c, where f''=2c/x^3>0, so the minimum value is 2\sqrt c. This agrees with the arithmetic-geometric mean inequality. Example: for f(x)=x^3-3x on x>0, f'=3x^2-3 vanishes at x=1 with f''(1)=6>0, so x=1 gives a local minimum with value -2.

Intersection of curves

Points of intersection of two curves are the common solutions of their equations. The usual method is to solve simultaneously, either by substitution or by subtracting one equation from the other to eliminate a variable. A line and a parabola usually reduce to a simple quadratic in one variable. Once the intersection points are known, the distance between them is \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}. For a line meeting a circle, the chord length is found from the perpendicular distance d from the centre to the line and the radius r, using L=2\sqrt{r^2-d^2}, with d computed from d=|ax_0+by_0+c|/\sqrt{a^2+b^2} for the line ax+by+c=0. Example: the parabola y=x^2 meets y=2 at (\pm\sqrt2,2), which are 2\sqrt2 apart.

Polar-to-Cartesian coordinate conversion

A point given in polar form (r,\theta) has Cartesian coordinates x=r\cos\theta and y=r\sin\theta. The reverse conversion uses r=\sqrt{x^2+y^2} and \tan\theta=y/x, with the quadrant of (x,y) deciding the correct angle. Exact values at standard angles are needed, such as \cos\frac{\pi}{4}=\sin\frac{\pi}{4}=\frac{1}{\sqrt2}, which makes the two coordinates equal. A quick check is that x^2+y^2 must equal r^2, which rejects options with the wrong distance from the origin. Example: (2,\pi/2) converts to (0,2).

Triangle area via determinant

For a triangle with vertices (x_1,y_1), (x_2,y_2) and (x_3,y_3), the area is \frac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|. This equals half the absolute value of the 3\times3 determinant with rows (x_i,y_i,1). The absolute value makes the area independent of the order of the vertices. When one side lies on an axis, base times height over two is faster. Because of the absolute value, an equation \text{area}=\text{given} often gives two values of the parameter. Example: the triangle with vertices (0,0), (4,0) and (0,3) has area 6.

Limits of multivariable functions

A limit of a function of two variables exists only if every approach path gives the same value. A practical method is to reduce the expression to a single ratio when it depends only on t=x_1/x_2, and then apply one-variable limit rules. The standard limit \lim_{t\to1}\frac{t-1}{\ln t}=1 follows from L'Hôpital's rule, since the ratio of derivatives t tends to 1, or from \ln t\approx t-1 near t=1. The factor x_2 cancels once the expression is written in terms of t. Example: \lim_{t\to1}\frac{t^2-1}{t-1}=2.