Further Maths (Matrices and solving simultaneous equations)

[olee]

Homework Statement

FP1 (MEI June 07)

Homework Equations

The Attempt at a Solution

(i) Find the value of n
I multiplied the top row of A by the left column of B and got the equation:
-5-16+k=k-n
moved stuff around and ended up with n=21

(ii)Write down the inverse matrix A^-1 and write down the conditions on k for this to exist.
Well, I know that k can not be 21 otherwise the determinant will be 0 which cannot be true in this circumstance.
I'm not too sure how to do A^-1. I know that the matrix itself must be what B originally was, but I'm not sure how to find out what k is (and thus the determinant).

(iii)Using the result from part (ii), or otherwise, solve the following simultaneous equations.

x-2y+z=1
2x+2y+z=12
3x+2y-z=3

I solved this using Gaussian elimination. I know how to solve this using the inverse of matrices, but I couldn't quite do part (ii) so i was unable to use that method.

 

[rock.freak667]

(ii)Write down the inverse matrix A^-1 and write down the conditions on k for this to exist.
Well, I know that k can not be 21 otherwise the determinant will be 0 which cannot be true in this circumstance.
I'm not too sure how to do A^-1. I know that the matrix itself must be what B originally was, but I'm not sure how to find out what k is (and thus the determinant).

Well you know that det(A)≠ 0, so you can find the condition for k using the determinant. (This will directly show you why k≠ 21)

Otherwise if k=n, then wouldn't AB=0, meaning that either A or B is the zero matrix?

Thus I think you should write A^-1 in terms of k

(iii)Using the result from part (ii), or otherwise, solve the following simultaneous equations.

x-2y+z=1
2x+2y+z=12
3x+2y-z=3

I solved this using Gaussian elimination. I know how to solve this using the inverse of matrices, but I couldn't quite do part (ii) so i was unable to use that method.

If you put the equations into the form CX=D, the matrix C would look similar to A wouldn't it?

and since AB=constant*I

what does that say about A and B?

 

[olee]

Well you know that det(A)≠ 0, so you can find the condition for k using the determinant. (This will directly show you why k≠ 21)

Otherwise if k=n, then wouldn't AB=0, meaning that either A or B is the zero matrix?

Thus I think you should write A^-1 in terms of k



If you put the equations into the form CX=D, the matrix C would look similar to A wouldn't it?

and since AB=constant*I

what does that say about A and B?

Thanks for the reply.

I've decided to do this for part (ii)

i'm unsure on how to solve for k in order to do part (iii) without Gaussian methods.

 

[rock.freak667]

Thanks for the reply.

I've decided to do this for part (ii)

i'm unsure on how to solve for k in order to do part (iii) without Gaussian methods.

Well look at the system of equations

x-2y+z=1
2x+2y+z=12
3x+2y-z=3

if you write this in a matrix form, what would it be?

(Compare the 3x3 matrix to the matrix A)

 

[olee]

Well look at the system of equations

x-2y+z=1
2x+2y+z=12
3x+2y-z=3

if you write this in a matrix form, what would it be?

(Compare the 3x3 matrix to the matrix A)

like this

 

[rock.freak667]

like this

no no just put these equations

x-2y+z=1
2x+2y+z=12
3x+2y-z=3

in a matrix form. Forget about the previous parts from before. What does the system look like in a matrix form?

 

[olee]

no no just put these equations

x-2y+z=1
2x+2y+z=12
3x+2y-z=3

in a matrix form. Forget about the previous parts from before. What does the system look like in a matrix form?

i think..

 

[rock.freak667]

i think..

Let's start simpler with 2 equations with 2 unknowns

x+y=1
2x+y=2

if we wanted to put this in a matrix form we'd get

\left(<br /> \begin{array}{cc}<br /> 1 &amp; 1\\<br /> 2 &amp; 1<br /> \end{array}<br /> \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 1 \\ 2 \end{array} \right)


Essentially, in the matrix, on the left side, the top line is the coefficients of the first matrix. The second line, the coefficients of the second matrix.

On the right side matrix, the first line is what ever is on the right side of the equal sign and so on.


Can you do something similar for the given set of equations?

 

[olee]

Let's start simpler with 2 equations with 2 unknowns

x+y=1
2x+y=2

if we wanted to put this in a matrix form we'd get

\left(<br /> \begin{array}{cc}<br /> 1 &amp; 1\\<br /> 2 &amp; 1<br /> \end{array}<br /> \right) \left( \begin{array}{c} x \\ y \end{array} \right) = \left( \begin{array}{c} 1 \\ 2 \end{array} \right)


Essentially, in the matrix, on the left side, the top line is the coefficients of the first matrix. The second line, the coefficients of the second matrix.

On the right side matrix, the first line is what ever is on the right side of the equal sign and so on.


Can you do something similar for the given set of equations?

 

[rock.freak667]

your second column should be -2,1,2

Check the original question, you typed it incorrectly and I quoted you so you used the wrong equations. Re-do it and then compare the 3x3 matrix to the matrix A.

 

[olee]

your second column should be -2,1,2

Check the original question, you typed it incorrectly and I quoted you so you used the wrong equations. Re-do it and then compare the 3x3 matrix to the matrix A.

.: k=1

thanks!

but i was wondering, is this method true with all matrix questions like this? or is it specific to this one?

 

[rock.freak667]

.: k=1

thanks!

but i was wondering, is this method true with all matrix questions like this? or is it specific to this one?

In these types of exams usually in Further math, if the questions have i,ii,iii,... in them, they are usually linked.

If you are asking if the matrix method can be used to solve a system of equations, then yes.

 

[olee]

thanks mate! i appreciate it!