# A Level Further maths - Matrices

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Can someone please help me with part iii of this Q I have done the rest of the parts but unsure how to get to p

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#3

I only had a glance but im pretty sure it has to do with substituting 5 for K and using it with the xyz matrix

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(Original post by

I only had a glance but im pretty sure it has to do with substituting 5 for K and using it with the xyz matrix

**SmorL**)I only had a glance but im pretty sure it has to do with substituting 5 for K and using it with the xyz matrix

Are you looking at part i

I am stuck on part iii I’ve completed i

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#5

(Original post by

Sorry Im unsure what u mean by this ?

Are you looking at part i

I am stuck on part iii I’ve completed i

**davidjohn03**)Sorry Im unsure what u mean by this ?

Are you looking at part i

I am stuck on part iii I’ve completed i

Try and find the two row multipliers (row reduction) which will make the left hand side of one equation zero . They'll be similar to how you solved i and ii. Then look at the right hand side of that equation and the value of p which makes that zero as well will mean the equations are consistent (not inconsistent) so there are solutions.

Last edited by mqb2766; 3 weeks ago

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(Original post by

Ive not worked it through, but I guess the equations will be linearly dependent / consistent.

Try and find the two row multipliers (row reduction) which will make the left hand side of one equation zero . They'll be similar to how you solved i and ii. Then look at the right hand side of that equation and the value of p which makes that zero as well will mean the equations are consistent (not inconsistent) so there are solutions.

**mqb2766**)Ive not worked it through, but I guess the equations will be linearly dependent / consistent.

Try and find the two row multipliers (row reduction) which will make the left hand side of one equation zero . They'll be similar to how you solved i and ii. Then look at the right hand side of that equation and the value of p which makes that zero as well will mean the equations are consistent (not inconsistent) so there are solutions.

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#7

(Original post by

Is there any other way of finding the solution as we haven't done anything like row reduction in class before

**davidjohn03**)Is there any other way of finding the solution as we haven't done anything like row reduction in class before

Note you could solve for the two row multipliers by solving a 2*2 system in x and y. Then use the same multipliers to directly get p on the right hand side.

Last edited by mqb2766; 3 weeks ago

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(Original post by

What did you do for the first two parts? This one shoud be similar.

Note you could solve for the two row multipliers by solving a 2*2 system in x and y. Then use the same multipliers to directly get p on the right hand side.

**mqb2766**)What did you do for the first two parts? This one shoud be similar.

Note you could solve for the two row multipliers by solving a 2*2 system in x and y. Then use the same multipliers to directly get p on the right hand side.

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#9

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Ive found the inverse in the first part and then used matrices multiplication in the 2nd part

**davidjohn03**)Ive found the inverse in the first part and then used matrices multiplication in the 2nd part

3x + 2y +5z = a(4x + y + 5z) + b(8x + 5y + 13z)

So use the x and y coefficients to solve for a and b. The z coefficients should be matched by definition. In some cases you can determine a and b by inspection.

Then use them to combine the right hand sides to get the value of p to make the equations consistent.

Last edited by mqb2766; 3 weeks ago

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(Original post by

Ok probably the easiest way is just to note that when k=5, the left hand sides are linearly dependent (det is zero in part i). This means that equation 2 can be expressed as a linear combination of equations 1 and 3, i.e.

3x + 2y +5z = a(4x + y + 5z) + b(8x + 5y + 13z)

So use the x and y coefficients to solve for a and b. The z coefficients should be matched by definition. In some cases you can determine a and b by inspection.

Then use them to combine the right hand sides to get the value of p to make the equations consistent.

**mqb2766**)Ok probably the easiest way is just to note that when k=5, the left hand sides are linearly dependent (det is zero in part i). This means that equation 2 can be expressed as a linear combination of equations 1 and 3, i.e.

3x + 2y +5z = a(4x + y + 5z) + b(8x + 5y + 13z)

So use the x and y coefficients to solve for a and b. The z coefficients should be matched by definition. In some cases you can determine a and b by inspection.

Then use them to combine the right hand sides to get the value of p to make the equations consistent.

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