So I am gonna stack it so u can see what I have and also the equation
3 |1|1 = -4
-1|2|-2 = 1
2|-1|-1 = -1
Looks weird I know. the answer is (5,7,4)
so what I have done with symbolabs was get rid of the (-4,1,-1) which gives me to (5,7,4) but I dont know how to set it up to have the right format.
My instructor even had trouble doing this one as well as myself
By format I mean solve by matrix. the original text is
-3x+y+z=-4
-x+2y-2z=1
2x-y-z=-1
He is only math 102 instructor that taught this subject no one even bother with it plus he has never taught 102
I need it to be in 1's and 0's.
In that case, ya got me. Not sure how you would go about converting the matrix to a form involving just 1's and 0's.
Maybe you could clarify a bit... What does the problem actually say? You need to use row manipulations (elimination) to get something on the left with zero entries except for the main diagonal? If so, then just determine
A-1
and AA-1 is the identity matrix.
yes like this
100
010
001
nvm I found it...I had to really dumb it down for the calculator to give more accurate details
I have the url but unforuntately I dont have access to it
thxs oobleck
google gauss jordan reduction
In the end
1 0 0 -1
0 1 0 -1/2
0 0 1 -1/2
To solve this problem, it seems like you are given a system of equations in matrix form. You have a matrix with three rows and three columns, where each entry represents a coefficient in the equations. You also have a column vector on the right, representing the right-hand side of the equations.
To solve the system of equations, you can use the matrix equation AX = B, where A is the matrix of coefficients, X is the column vector of variables, and B is the right-hand side column vector.
In your case, the matrix A is:
| 3 1 1 |
|-1 2 -2 |
| 2 -1 -1 |
And the column vector B is:
|-4 |
| 1 |
|-1 |
The matrix equation AX = B can be written as:
| 3 1 1 | | x | | -4 |
|-1 2 -2 | * | y | = | 1 |
| 2 -1 -1 | | z | | -1 |
To solve for X = (x, y, z), you can use matrix inversion. First, calculate the inverse of matrix A, denoted as A^(-1). Then, multiply both sides of the equation by A^(-1) to isolate X.
To find the inverse of matrix A, you can use various methods, such as Gaussian elimination or determinant and adjugate matrix. Once you find A^(-1), multiply both sides of the equation by A^(-1) to obtain X.
Finally, substitute the values of x, y, and z back into the equation to verify that (5, 7, 4) are indeed the solutions.
Keep in mind that matrix inversion can be computationally intensive, especially for larger matrices. Alternatively, you can also use numerical methods like Gaussian elimination or matrix factorization techniques like LU decomposition to solve the system of equations.
two matrices
one matrix
Not sure what you so with symbolabs, but to solve the equation
AX = B
you multiply on the left by A-1, giving
X = A-1B
See it written out at
https://www.wolframalpha.com/input/?i=%7B%7B-3,1,1%7D,%7B-1,2,-2%7D,%7B2,-1,-1%7D%7D*%7B%7Bx%7D,%7By%7D,%7Bz%7D%7D+%3D+%7B%7B-4%7D,%7B1%7D,%7B-1%7D%7D
Shame on your instructor if he had trouble doing it!