William continues to provide, excellent support to my son - helping him to go beyond being able to answer a complex question with a by rote, step by step approach - to having a much fuller understanding of the concepts. This mean that he feels much better equipped to deal with what the exam might throw at him. Thanks very much Will : )

Susan, Parent from Hertfordshire

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Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after lessons. In my tutorials, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

Let M be a 3x3 matrix s.t.
M=
|a b c|
|g h i|
|d e f|
Then Det(M)= a(Det(e,f,h,i))-b(Det(d,f,g,i))+c(Det(d,e,g,h).
Given that the determinant of a 2x2 matrix such as (e,f,h,i) is = ei-fh.
The solution is; Det(M)=a(ei-fh)-b(di-fg)+c(dh-eg).
Since the inverse of a matrix, M^-1 = 1/Det(M) * Adj(M), the inverse does not exist when Det(M)=0.

Answered by Oskar D.

Studies Economics and Mathematics (MA) at Edinburgh

The easiest way to complete questions of these types is to first sketch an Argand diagram. With 5âˆš 3 on the x (real) axis and -5 on the y (imaginary) axis, the modulus would be calculated simply by using pythagoras's theorem. Thus, the modulus of z would be equal to âˆš((5âˆš3)Â² + 5Â²) = âˆš(75+25) = âˆš100 = 10.
The argument is then found as the angle between the real axis and the vector of the complex number. This can once again be calculated with trigonometry. As we know the magnitude of all three edges of the triangle, any of sin cos and tan operations can be used. In this example, i will compute it using tan. thus, tan(Î¸)=opp/adj = Imimaginary/real components = -5/(5âˆš3)
therefore arg(z)=arctan(-1/âˆš3), which gives a value of -30Â° or -Ï€/6
once we have both the modulus and the argument of the complex number, expressing it in modulus-argument form is straightforward. the complex number z= |z|((cos(arg(z) + isin(arg(z))) = 10(cos(-Ï€/6) + isin(-Ï€/6) ).

First, let's imagine the point 3 + 4j as a point on an Argand diagram, with coordinates 3,4. The polar form of an imaginary number is in the form re^(jÎ¸), where r is the modulus of the number (the distance between the point on the graph and the origin), and Î¸ is the argument (the angle the point makes with the horizontal). In order to find r, we can simply use Pythagoras' Theorem, giving us the answer r = 5. To find Î¸, we must use trigonometry, identifying the angle Î¸ as the inverse tangent of (4/3), which is equal to 0.927. Therefore the angle Î¸ is 0.927. This means the polar form of 3 + 4j is 5e^0.927jÎ¸

First you prove the case n=1 is true. Then you assume that n=k is true, then calculate what n=k+1 is. This should prove true, so that by induction, you have proved that the statement is true for all natural numbers.

Answered by Praveenaa K.

Studies Mathematics at Bristol

Firstly you need to check if the matrices conform. Matrices are an array of elements within a pair of brackets, there are some number of columns, usually denoted m, and some number of rows, denoted n. These two numbers, m and n, give the order of the matrix, how large it is, suppose you had a 2x3 matrix, then it has two columns and 3 rows. Matrix multiplication can only occur if the two matrices conform, that is given two matrices A and B, the operation AB (AxB) can only occur if the number of rows of B match the number of columns of A. So if A is a mx2 matrix, for AB to exist, B must be some 2xn matrix. An interesting thing to note, m and n in this example can be any number and the matrix produced by the multiplication will be mxn.
The multiplication process itself initially looks abstract and at times in unwieldy but it is relatively simple, it is also quite difficult to explain without an example. Let A and B be conforming matrices such that AB exists, to perform the operation you firstly go across the first row of A and down the first column of B, then take the sum of all corresponding pairs of elements, that is the number that is first in the row of A multiplied by the number that is first in column of B and so on for the second, third etc. In this case the number you get is the element that is in the first row and first column for AB. You then repeat this with each pair of rows in A and columns in B, so say you did this with the second row of A and the third column of B, then you get the element that is in the second row and third column of AB. Then your final result should be the matrix AB.

Answered by Mark D.

Studies Mathematics at Leeds

First we find the auxilary equation by substituting y with m^0, y' with m^1 and y'' with m^2. We get m^2 + 4m + 13 and find the roots using the differential equation, m = (-4 +- (16-4x1x13)^0.5)/(2x1).
From this we get complex roots m = -2 + 3i and m = -2 - 3i. Now we solve the homogenous form using these roots, y = e^(-2x) (Acos(3x) + Bsin(3x)).
So we have solved the differential equation for when the right hand side is equal to zero but we must solve it for when the RHS is equal to sin(x) so we need to take y = psin(x) + qcos(x) and find y' and y'' to substitute into the LHS. So y' = -psin(x) + qcos(x) and y'' = -pcos(x) - qsin(x).
By comparing coefficients of the substitued LHS and the RHS we get, (-pcos(x) - qsin(x)) + 4(-psin(x) + qcos(x)) + 13(pcos(x) + qsin(x)) == sin(x). After comparing coefficients and solving the resulting simultaneous equation we find, p = -1/40 and q = 3/40.
Now we just put all the parts together to obtain the general solution to the equation, y = e^(-2x) (Acos(3x) + Bsin(3x)) - (1/40)cos(x) + (3/40)sin(x).

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