Hi! My name is Anya and I’m a 2nd year Mathematics student at the University of Bristol. I’m a friendly and patient individual, excited for the opportunity to aid pupils in the advancement of their mathematical fluency and confidence, and ultimately in achieving their potential! Tutorials will be tailored to cater to the needs of the pupil and to focus on areas that might require particular attention. Most importantly I believe that maths can be stimulating and fun. I will work together with the pupil to establish a strong understanding of the topic and to fully grasp concepts and methods covered, as it is important not to just memorise the course and the correct solutions to everything. Then we will work on examples and past paper questions to further develop and consolidate this understanding, and also refine exam technique. Please do not hesitate to drop me a ‘Webmail’ or book a ‘Meet The Tutor Session’. I look forward to meeting you! Thanks AnyaHi! My name is Anya and I’m a 2nd year Mathematics student at the University of Bristol. I’m a friendly and patient individual, excited for the opportunity to aid pupils in the advancement of their mathematical fluency and confidence, and ultimately in achieving their potential! Tutorials will be tailored to cater to the needs of the pupil and to focus on areas that might require particular attention. Most importantly I believe that maths can be stimulating and fun. I will work together with the pupil to establish a strong understanding of the topic and to fully grasp concepts and methods covered, as it is important not to just memorise the course and the correct solutions to everything. Then we will work on examples and past paper questions to further develop and consolidate this understanding, and also refine exam technique. Please do not hesitate to drop me a ‘Webmail’ or book a ‘Meet The Tutor Session’. I look forward to meeting you! Thanks Anya

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Let

** y = ln(x)**

Then

** x = e^y**

*[ ln(x) = ln(e^y) = y ln(e) = y*1 = y ]*

Taking the derivative of this we get

** dx/dy = e^y**

Taking the reciprocal we get

** dy/dx = 1/e^y = 1/x**

So the derivative of ln(x) is

** d(ln(x))/dx = 1/x**

Let

** y = ln(x)**

Then

** x = e^y**

*[ ln(x) = ln(e^y) = y ln(e) = y*1 = y ]*

Taking the derivative of this we get

** dx/dy = e^y**

Taking the reciprocal we get

** dy/dx = 1/e^y = 1/x**

So the derivative of ln(x) is

** d(ln(x))/dx = 1/x**

36 and 48 are comprised of the following prime factors:

** **36 = 2 x 2 x 3 x 3

** **48 = 2 x 2 x 2 x 2 x 3

To find the highest common factor (HCF) we consider the prime factors that are the same for both 36 and 48:

** **36 = **2** x **2** x **3** x 3

** **48 = **2** x **2 **x 2 x 2 x **3**

The prime factors that are the same are two 2s and one 3, so we multiply these to find the HCF:

** **2 x 2 x 3 = 12.

So the HCF of 36 and 48 is 12.

To find the lowest common multiple (LCM) we consider which number consists of the most of each prime factor:

** **36 = (2 x 2) x (**3 x 3**)

** **48 = (**2 x 2 x 2 x 2**) x (3)

36 consists of the most 3s (two 3s); 48 consists of the most 2s (four 2s). So we multiply these two 3s and four 2s:

** **(2 x 2 x 2 x 2) x (3 x 3) = 144

So the LCM of 36 and 48 is 144.

36 and 48 are comprised of the following prime factors:

** **36 = 2 x 2 x 3 x 3

** **48 = 2 x 2 x 2 x 2 x 3

To find the highest common factor (HCF) we consider the prime factors that are the same for both 36 and 48:

** **36 = **2** x **2** x **3** x 3

** **48 = **2** x **2 **x 2 x 2 x **3**

The prime factors that are the same are two 2s and one 3, so we multiply these to find the HCF:

** **2 x 2 x 3 = 12.

So the HCF of 36 and 48 is 12.

To find the lowest common multiple (LCM) we consider which number consists of the most of each prime factor:

** **36 = (2 x 2) x (**3 x 3**)

** **48 = (**2 x 2 x 2 x 2**) x (3)

36 consists of the most 3s (two 3s); 48 consists of the most 2s (four 2s). So we multiply these two 3s and four 2s:

** **(2 x 2 x 2 x 2) x (3 x 3) = 144

So the LCM of 36 and 48 is 144.

(i) Find the DETERMINANT of A:

1. Multiply the (1,1)th element by the determinant of the 2x2 matrix that remains when the row and the column containing the element are crossed out.

2. Repeat for the elements (1,2)th and (1,3)th elements.

3. Alternately add and subtract the results.

(ii) Find the ADJUGATE of A (also known as the classical adjoint of A):

1. Form M, the matrix of minors of A, where the minor of an element of a 3x3 matrix is the determinant of the 2x2 matrix that remains when the row and the column containing the element are crossed out.

2. Form C, the matrix of cofactors, by changing the signs of alternating elements of M starting with the top left element as positive.

3. Form adj(A) = C^T, the transpose of the matrix of cofactors, by reflecting the matrix about the main diagonal (i.e. swapping the (i,j)th element with the (j,i)th element.)

(iii) Find the INVERSE of A, given by

** A^(-1)=1/det(A) * adj(A)**

where det(A) us the determinant of A and adj(A) is the adjugate of A.

NB: The same method can also be applied to larger square matrices but the process is long so instead we use the Guass-Jordon Method of Elimination, which is also applicable to 3x3 matrices.

(i) Find the DETERMINANT of A:

1. Multiply the (1,1)th element by the determinant of the 2x2 matrix that remains when the row and the column containing the element are crossed out.

2. Repeat for the elements (1,2)th and (1,3)th elements.

3. Alternately add and subtract the results.

(ii) Find the ADJUGATE of A (also known as the classical adjoint of A):

1. Form M, the matrix of minors of A, where the minor of an element of a 3x3 matrix is the determinant of the 2x2 matrix that remains when the row and the column containing the element are crossed out.

2. Form C, the matrix of cofactors, by changing the signs of alternating elements of M starting with the top left element as positive.

3. Form adj(A) = C^T, the transpose of the matrix of cofactors, by reflecting the matrix about the main diagonal (i.e. swapping the (i,j)th element with the (j,i)th element.)

(iii) Find the INVERSE of A, given by

** A^(-1)=1/det(A) * adj(A)**

where det(A) us the determinant of A and adj(A) is the adjugate of A.

NB: The same method can also be applied to larger square matrices but the process is long so instead we use the Guass-Jordon Method of Elimination, which is also applicable to 3x3 matrices.