Question

Help!!...........Let a,b,c be real numbers such that a-7b+8c=4 and 8a+4b-c=7.Find a^2-b^2+c^2
Help!!...........Let a,b,c be real numbers such that a-7b+8c=4 and 8a+4b-c=7.Find a^2-b^2+c^2
@Mathematics

Answer 1

Its 8a+4b-c=7

Answer 2

\[a^{2}-b^{2}+c^{2}=0\]

Answer 3

It's a fairly lengthy proof

Answer 4

I made a mistake somewhere - ignore the above
These are the steps I have so far:

a-7b+8c=4
8a+4b-c=7

using elimination you can get the following 3 equations in just 2 variables each:
(1) 13c-12b=5
(2) 13a+5b=12
(3) 12a+5c=13

(1)*(2), (1)*(3), and (2)*(3) then gives:
(4) 13*13ac+5*13bc-12*13ab-5*12b^2=5*12
(5) 12*13ac+5*13c^2-12*12ab-5*12bc=5*13
(6) 12*13a^2+5*13ac+5*12ab+5*5bc=12*13

then 5*(6)+13*(4)+12*(5) gives:
5*12*13(a^2-b^2+c^2) +(5*5*13+13*13*13+12*12*13)ac+(5*5*12-12*13*13-12*12*12)ab+(5*5*5+5*12*13-5*12*12)bc=5*12.13+5*12*13+5*12*13=3*5*12*13

this ca be re-written as:
5*12*13(a^2-b^2+c^2)+13(5*5+13*13+12*12)ac+12(5*5-13*13-12*12)ab+5(5*5+12*13-12*12)bc=3*5*12*13

giving:
a^2-b^2+c^2 = (3*5*12*13 - 5(5*5+12*13-12*12)bc - 12(5*5-13*13-12*12)ab - 13(5*5+13*13+12*12)ac) / 5*12*13

Answer 5

\[a^{2}-b^{2}+c^{2}=3-\frac{5(5^{2}-12^{2}+13^{2})bc+12(5^{2}-12^{2}-13^{2})ab+13(5^{2}+12^{2}+13^{2})ac}{5*12*13}\]

Answer 6

I'm sure I still have some mistakes - tricky question! :-)

Answer 7

I get a^2-b^2+c^2=1

Answer 8

solve a-7b+8c=4 and 8a+4b-c=7

you get

\[a=\frac{13}{12}-\frac{5}{12}t\]
\[b=\frac{-5}{12}+\frac{13}{12}t\]
\[c=t\]

then compute \[a^2-b^2+c^2\]

you will get 1

Answer 9

I still dont understand do you mind explaining without wrting those \frac.....

Answer 10

I solved the system a-7b+8c=4 and 8a+4b-c=7

( I row reduced the augmented matrix and solved in terms of the free variable)

Answer 11

Can you solve this sum without matrices as i have not learnt that topic yet......
:(

Answer 12

anything you can do with matrices you can do without (it just usually takes longer)

Answer 13

So,can you tell me how to do it without it

Answer 14

a-7b+8c=4
a=7b-8c+4

8a+4b-c=7
8(7b-8c+4)+4b-c=7
solve for b
b=(13c-5)/12

replace b in a=7b-8c+4 with b=(13c-5)/12

a=7[(13c-5)/12]-8c+4 =(13-5c)/12

now compute \[a^2-b^2+c^2\]

\[\left[(13-5c)/12\right]^2-\left[(13c-5)/12\right]^2+c^2\]