Expand along the first row: |A| = 1 x (1 x 0 - 4 x 6) - 2 x (0 x 0 - 4 x 5) + 3 x (0 x 6 - 1 x 5) = 1 x (0 - 24) - 2 x (0 - 20) + 3 x (0 - 5) = -24 + 40 - 15 = 1. So the value of the determinant is 1.

(1) The value of a determinant remains unchanged if its rows and columns are interchanged (that is, |A'| = |A|). (2) If any two rows (or two columns) of a determinant are interchanged, the sign of the determinant changes. (3) If any two rows (or columns) of a determinant are identical (or proportional), the value of the determinant is zero. Also, if each element of a row (or column) is multiplied by a constant k, the determinant is multiplied by k.

The area of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Substituting: area = (1/2)|1(0 - 3) + 6(3 - 0) + 4(0 - 0)| = (1/2)|-3 + 18 + 0| = (1/2)(15) = 7.5 square units.

Area = (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|

The minor of an element aij of a determinant is the value of the determinant obtained by deleting the ith row and the jth column in which that element lies; it is denoted by Mij. The cofactor of the element aij is Aij = (-1)^(i + j) x Mij, that is, the minor with a sign attached according to the position of the element. Cofactors are used to expand a determinant and to find the adjoint of a matrix.

Aij = (-1)^(i+j) Mij

First find the determinant: |A| = (2)(4) - (3)(1) = 8 - 3 = 5, which is non-zero, so the inverse exists. For a 2 x 2 matrix [[a, b],[c, d]], the adjoint is [[d, -b],[-c, a]], so adj A = [[4, -3],[-1, 2]]. Then the inverse is A inverse = (1/|A|) adj A = (1/5)[[4, -3],[-1, 2]] = [[4/5, -3/5],[-1/5, 2/5]].

A inverse = (1/|A|) adj A

A square matrix A is called singular if its determinant is zero (|A| = 0), and non-singular if its determinant is non-zero (|A| is not equal to 0). The condition for a square matrix to be invertible (to have an inverse) is that it must be non-singular, that is, |A| must not be zero, because the inverse A inverse = (1/|A|) adj A is defined only when |A| is not zero.

A inverse exists if and only if |A| is not 0

Write D = determinant of coefficients = |[[2, 3],[1, 2]]| = (2)(2) - (3)(1) = 4 - 3 = 1. Dx (replace first column by constants) = |[[8, 3],[5, 2]]| = (8)(2) - (3)(5) = 16 - 15 = 1. Dy (replace second column by constants) = |[[2, 8],[1, 5]]| = (2)(5) - (8)(1) = 10 - 8 = 2. By Cramer's rule, x = Dx/D = 1/1 = 1 and y = Dy/D = 2/1 = 2. So x = 1, y = 2.

x = Dx/D, y = Dy/D

For an n x n matrix, |adj A| = |A|^(n - 1). Here n = 3, so |adj A| = |A|^2 = 5^2 = 25. Also, for a scalar k, |kA| = k^n |A|. Here |2A| = 2^3 x |A| = 8 x 5 = 40. So |adj A| = 25 and |2A| = 40.

|adj A| = |A|^(n-1); |kA| = k^n |A|

A function f(x) is continuous at a point x = a if the following three conditions are all satisfied: (1) f(a) is defined (the function has a value at a); (2) the limit of f(x) as x approaches a exists (that is, the left-hand limit equals the right-hand limit); and (3) this limit is equal to the value of the function, that is, limit of f(x) as x approaches a = f(a).

limit(x->a) f(x) = f(a)

For continuity at x = 5, the left-hand limit, the right-hand limit and f(5) must all be equal. Left-hand value (using kx + 1): f(5) = 5k + 1. Right-hand limit (using 3x - 5): as x approaches 5, 3(5) - 5 = 10. For continuity, 5k + 1 = 10, so 5k = 9, giving k = 9/5.

LHL = RHL = f(a)

If a function is differentiable at a point, then it is necessarily continuous at that point. However, the converse is not always true: a function that is continuous at a point need not be differentiable there. For example, f(x) = |x| is continuous at x = 0 but not differentiable at x = 0 (it has a sharp corner). So differentiability implies continuity, but continuity does not imply differentiability.

Let y = sin(x^2). Using the chain rule, dy/dx = cos(x^2) x d/dx(x^2) = cos(x^2) x 2x = 2x cos(x^2). The chain rule states that the derivative of a composite function f(g(x)) is f'(g(x)) x g'(x).

d/dx f(g(x)) = f'(g(x)) g'(x)

Differentiate both sides with respect to x, treating y as a function of x. d/dx(x^2) + d/dx(y^2) = d/dx(25). This gives 2x + 2y (dy/dx) = 0. Solving for dy/dx: 2y (dy/dx) = -2x, so dy/dx = -2x/(2y) = -x/y.

dy/dx = -x/y

Take the natural logarithm of both sides: ln y = ln(x^x) = x ln x. Now differentiate both sides with respect to x. On the left: (1/y)(dy/dx). On the right, using the product rule on x ln x: (1)(ln x) + x(1/x) = ln x + 1. So (1/y)(dy/dx) = ln x + 1, which gives dy/dx = y(ln x + 1) = x^x (1 + ln x).

dy/dx = x^x (1 + ln x)

The derivative of sin inverse (x) with respect to x is 1/sqrt(1 - x^2), valid for -1 < x < 1. The derivative of tan inverse (x) with respect to x is 1/(1 + x^2), valid for all real x. These standard results are obtained by implicit differentiation of y = sin inverse (x) and y = tan inverse (x).

d/dx(sin inverse x) = 1/sqrt(1 - x^2)

Differentiate each with respect to the parameter t: dx/dt = -a sin t and dy/dt = a cos t. Then dy/dx = (dy/dt)/(dx/dt) = (a cos t)/(-a sin t) = -cos t/sin t = -cot t. So dy/dx = -cot t.

dy/dx = (dy/dt)/(dx/dt)

Rolle's theorem: if a function f is continuous on the closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one point c in (a, b) such that f'(c) = 0. Lagrange's mean value theorem: if a function f is continuous on [a, b] and differentiable on (a, b), then there exists at least one point c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).

f'(c) = (f(b) - f(a))/(b - a)

First derivative using the product rule: dy/dx = e^x sin x + e^x cos x = e^x(sin x + cos x). Differentiate again using the product rule: d2y/dx2 = e^x(sin x + cos x) + e^x(cos x - sin x) = e^x(sin x + cos x + cos x - sin x) = e^x(2 cos x) = 2 e^x cos x.

d2y/dx2 = 2 e^x cos x

The derivative dy/dx represents the instantaneous rate of change of y with respect to x. When a quantity changes with time, its derivative with respect to time gives the rate of change. The area of a circle is A = pi r^2, so the rate of change of area with respect to the radius is dA/dr = 2 pi r. If the radius changes with time, dA/dt = 2 pi r (dr/dt).

dA/dr = 2 pi r

Area A = pi r^2. Differentiate with respect to time t: dA/dt = 2 pi r (dr/dt). Given dr/dt = 3 cm/s and r = 10 cm, so dA/dt = 2 pi (10)(3) = 60 pi cm^2/s. Thus the area is increasing at the rate of 60 pi square cm per second (about 188.5 cm^2/s).

dA/dt = 2 pi r (dr/dt)