Show that w= [3,-5] can be written as a linear combination of the vectors [2,7] [1,-1] [-2,1] (All vertical vectors btw)
like this : \[ W= \left[ {\begin{array}{cc} 3 \\ -5\\ \end{array} } \right] , A = \left[ {\begin{array}{cc} 2 \\ 7\\ \end{array} } \right], B = \left[ {\begin{array}{cc} 1 \\ -1\\ \end{array} } \right], C = \left[ {\begin{array}{cc} -2\\ 1\\ \end{array} } \right] \]
you wanto write \( W\) as linearcombination of \(A, B, C\), right ?
yep
lets see.. guess we need to find it trial and error
there is way to do it by just putting it int a matrix but the thing is there is 2 equations and 3 unknowns so that is why I am confused
W = Ax + By + Cz you need to find x, y, z
so yes, two equations, and 3 unknowns
3 = 2x+y-2z --------(1) -5 = 7x-y+z---------(2)
you will get infinite solutions, just pick one ?
oh okay how do you know it will be infinite soln though?
look at each of the equation u have, it has 3 variables eh ?
yep
that means, its a plane in 3 dimensions. 2 planes meet in a line
a line has infinite points. so infinite solutions
lets find them..
few of them atleast :)
okay
:)
3 = 2x+y-2z --------(1) -5 = 7x-y+z---------(2) (1) + (2) gives us, -2 = 9x-z ----------(3)
yep
thats ur solution line, every point on that line (3) is a solution. which means, u can use any point on that line, that gives u linear combination of A B C which equals W
say, x =1, put this in (3), and solve z -2 = 9(1) - z z = 11
put x=1, z=11 in (1) and solve y 3 = 2(1) +y-2(11) y = 23
so, when x =1, we have y=23, z=11
check if this combination works
W = Ax+By+Cz
it workss
we could have taken x=0, that would have simplified the calculaiton, but its ok.. u see that all points on that line can be used right ?
im still ed how adding 1 and 2 gives you the line of intersection though?
confused*
ohk, thats simple, i just got rid of one variable by 'elimination method'
yeah
so, if i understand ur question correctly, you are asking, why adding both the equations gives line of intersection of planes ?
yeah :)
thats really very good question:) before i answer that, let me ask you a similar q, u knw that intersection of two lines is a point, right ?
for ex, take below two lines :- x+y = 2 x+2y = 1
yep and the intersection of two panes is a line
since they're not parallel, they will meet at a point for sure.
now tell me, how u wud find that intersection point ?
tell me how to find intersecting point of below two lines, x+y = 2 x+2y = 1
equal them to eachother
yes, why it works ?
by adding the two planes vertically, I did the same previously
oh
you're trying to figure out, why adding them vertically gives the line of intersection ?
yes
we can find answer to that, once we find answer to below :- find intersection point of below two lines x+y = 2 x+2y = 1
follow me closely if u can, to find intersection point, lets take the first equation x+y = 2
we can add/subtract same thing to both sides, right ? so lets add "x+2y" to both sides
*subtract actually.. x+y = 2 subtract "x+2y" from both sides x+y - (x+2y) = 2 -(x+2y)
yep
simplify left side
notice that, till now, 2nd equation dint come into picture
x+y = 2 subtract "x+2y" from both sides x+y - (x+2y) = 2 -(x+2y) x+y -x-2y = 2-(x+2y) -y = 2-(x+2y)
yep
Now, look at 2nd equation, x+2y=1, so put that value on right side. (this step actually determines the solution we get lies on second equation also)
ok
x+y = 2 subtract "x+2y" from both sides x+y - (x+2y) = 2 -(x+2y) x+y -x-2y = 2-(x+2y) -y = 2-(x+2y) from 2nd equation, x+2y=1 -y = 2-1 y = -1
since you knw y, u can find x x = 3 so, (3, -1) lies on both sides
oohhh ok
if u are convinced (3,-1) lies on both lines, then u will understand elimination method. in elimination method also, when we add equations vertically, exact same thing happens
so, you started linear algebra is it
yep.
i did this few months back gilbert strang's course... it was very good, if u get extra time, go thru lectures... they're very helpful :)
ooh okay thanks so much for your help!
np :)
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