<p>While studying for my exam I'm stuck on these four problems. Can anyone help? </p>
<p>1)
4x - 7y = -2
-x + 2y = 1 (solve into rref) by hand!</p>
<p>2) x + y + z = 4
x + 3y +2z = 4
x-2y+z=7</p>
<p>3) if A= <a href="all%20one%20set">1.2</a> Then A^ -1 =?
[2.3]
4) if B = [ 6.-1.-5
-7. 1. 5
-10.2.11
than B^ - 1 = ?</p>
<p>I’m pretty sure homework help is t allowed</p>
<p>im pretty sure I did this in alg 2</p>
<p>Try this strategy:
Multiply equation 1 by constant 1 and equation 2 by constant 2 so that the coefficient of one of the variables is the same in both equations. Then subtract one equation from the other, so that the resulting equation only has one variable, and solve that as a simple equation.</p>
<p>eg. 3x +4y = 5
and 2x -2y = 1</p>
<p>you can 1) multiply equation 1 by 2 and equation 2 by 3, each resulting in a 6x term and subtract one from the other, leaving an equation only with y as a variable.</p>
<p>Simpler still is to just multiply equation 2 by 2, getting a -4y term. Then add the two equations, getting rid of the y term.</p>
<p>btw, this is a long way from calc.</p>
<p>what would you consider it? @dad</p>
<p>Also this is 100% calc. It has to be solved into rref. thats nothing to do with algebra lol. I read your post. I obviously know how to solve it that way. I need to know the matrices aspect of it.</p>
<p>sorry I wasnt more clear in the initial post.</p>
<p>There’s a “matrix” solution for simultaeous equations in linear algebra where the equation is written in the form:
Ax=B
and you solve it by multiplying both sides by the inverse of A,
A is the matrix consisting of the coefficients.</p>
<p>My super/subscripting skills here are weak. But you should be able to google something like:
linear algebra simultaneous equation solution
that will go through this technique.</p>
<p>Yeah we did this in algebra 2 this year.</p>