Matrices...a quick way to solve w/out a calculator...?

Question

we are learning matrices in intermediate algebra....a student mentioned to the professor that there was an easy way of solving the system using matrices....he told her if she can find it let him know....please i need help....

Answer 1

There is a way to do it:
If you want to get the value, do the following:

Be the matrix of this form:

a b c
d e f
g h i

Expand the matrix like this:

a b c a b
d e f d e
g h i g h

now, starting at the top left, go in diagonal down multiplying, like this:
a*e*i
then go with b, same way
b*f*g
and then to c
c*d*h

Add these three:

(a*e*i) + (b*f*g) + (c*d*h)

Then, start at the upper right corner, diagonal again, like this:
b*d*i
Then by a:
a*f*h
and then c:
c*e*g

Add these again:

(b*d*i) + (a*f*h) + (c*e*g)

Substract these last ones from the first ones:

((a*e*i) + (b*f*g) + (c*d*h)) - ((b*d*i) + (a*f*h) + (c*e*g))

That's the way to do it without a calculator.

Answer 2

Since ^1/3 power means the cubed root of something, and negative means you do 1/what-it-would-be-if-it-were-positive: 8^-1/3=1/cubed-root-of-8=1/2 o the answer to the first one is 1/2 Second one: This one is really hard. Since 6^anything-real=something-real, x must be an imaginary number, or maybe a complex number( a mix of a real and imaginary, in the form of a+bi). If you don't know what this means, I will tell you: sqrt(-1)=i=imaginary unit because anything normal times itself is positive. Third one: 16/2=8 2=4th root of 16 16*1/(16^1/4)=8 So you have 16 times 1 over 16 to the 1/4 power. Remember that when you have 1 over something, it means that it is negative. So: the answer is: 16^(1+ (-1/4)) 16^(1-1/4) 16^3/4, so the answer is x=3/4

Answer 3

Sorry...I only learned with a calculator...but a quick way to solve it without on would be great.

Answer 4

yep. MATLAB!