Question

can someone please help me with this matrix?

Answer 1

I would suggest clicking the need help button first, then asking for assistance. I'm sure it will walk you through it.

Answer 2

I’m on my phone, so this is hard to type out, but you need to use row reduction to get the left three columns to be

1 0 0
0 1 0
0 0 1

When done properly, you’ll end up with the solutions for x, y and z in the last column in that order

Answer 3

You are allowed to:

1. Swap rows
2. Multiply rows by some constant
3. Add/subtract a multiple of a row to another row (ex: replace row 1 with “row 1 + 2row3”)

I would start off by swapping rows 2 and 3 to get

1 2 5 -9
0 -1 -3 5
2 -2 4 -12

Now I have 2 in the lower left, but I want 0 there. So I would replace row 3 with (row 3 - 2row1) to get

1 2 5 -9
0 -1 -3 5
0 -6 6 6

And keep going (I would next do another operation with row 3 to get that -6 to a 0)

Answer 4

*made a typo, should be 15 not 5 in last column

Answer 5

@vocaloid can you times row 3 by 0?

Answer 6

@ilovespaghetti the help button only directs you to a problem solved in the textbook with no explanation or steps, just the answer, which doesnt really help at all.

Answer 7

could i multiply by a fraction? so for R3 i could times it by 1/7 or does it have to be a whole number?

Answer 8

I should have mentioned this - multiplying an entire row by 0 is not useful, since you still need some nonzero number in there to represent the variable

think about an equation - multiplying both sides by 0 is not useful since it just makes both sides 0 = 0

multiplying both sides by a fraction is fine

Answer 9

1 2 5 -9
0 -1 -3 15
0 -6 6 6

to get that -6 in row 3 column 2 to a 0, I would replace row 3 with row 3 - 6(row2)

Answer 10

alternatively - to make the arithmetic easier, you could divide row 3 by -1 first to get 0 1 - 1 -1, then replace row 3 with row 3 + row 2 to get that 0 there too

Answer 11

god how do you know what steps to use?

Answer 12

@vocaloid for the first method you listed (R3 - 6R2), the 1 is supposed to be negative, so when subtracted 6 from -6, you get 12, right?

Answer 13

It’s a little hard to explain but you want to place the 0’s strategically to make the arithmetic easier

Because I want a 0 in row 3 column 2: row 3 has a negative 6 there, so I need to add 6 to get 0

Multiplying row 2 by 6 gets a negative 6 in the center of the matrix

So when we do row 3 - 6row(2) we get -6 - 6(-1) = 0 in that space

Answer 14

sorry for late reply, can you check over this work?
the program is still saying its wrong but im not seeing the mistake i made, but im sure you'll be able to @vocaloid

Answer 15

third step

R3 + 6R2 --> R3

looking at the last entry of row 3, -6 + 6(-3) is -24 not -12 (that's not necessarily the only mistake, just the first one I saw)

however, I think if you go to that step, and instead divide row 3 by -6 to get 0 1 1 -1, it'll make the arithmetic in future steps easier

Answer 16

alternatively, you could instead do the R3 + 6R2 --> R3 like you were doing before, then just divide the whole row by the appropriate constant to get 1 in row 3 column 3, to make future arithmetic easier

idk, it doesn't *really* matter either way as long as you follow the rules properly, it's just that certain steps make the numbers smaller and easier to work with

Answer 17

oh my god @vocaloid i think ive spent literal hours on this one question, but i finally solved it! i got through every single homework within my unit but for some reason i couldn't wrap my head around how this type of problem works...

my midterm on monday only requires us to do one step of a gaussian problem but i was determined to solve this it was driving me actually nuts

thanks for the help, i can sleep peacefully now knowing i figured it out lol