I have a matrix with
brackets around
[7 6 ][13]
[14 -6] [8]
and i know there are two separate brackets but on this thing you cant just do one big one..
so if you don't understand the equation that was suppose to go into matricies was
x + y = 13
2x - y = 8
x = 7
y = 6
then i am suppose to figure out these 3 equations to that matrix
-2R1 + R2 >R2
-1/3 R2 > R2
-1R2 + R1 > R1
and my question is for those 3 equations i don't know how to figure them out and they are suppose to be resulting in a matrix with the answers. ?? can some one help me out please
Beats me what you are supposed to do.
Perhaps solve for the inverse as follows:
|+1 +1 | |x| |R1|
|+2 -1 | |y| |R2|
meaning coefficient matrix of numbers times the x,y column equals the R1,R2 column
perform algebra on the rows to get identity matrix on the left so the solution is x, y in terms of R1,R2 (in other words find the inverse matrix)
form matrix augmented
+1 +1 R1
+2 -1 R2
add the two rows to get zero in first row second element
+3 +0 (R1+R2)
+2 -1 R2
divide first row by three to get one in upper left
+1 +0 (R1+R2)/3
+2 -1 R2
subtract twice the first from the second to get zero in lower left
+1 +0 (R1+R2)/3
+0 -1 R2 - (2/3)(R1+R2)
multiply second row by -1 to get 0,1 in second row
+1 +0 (R1+R2)/3
+0 +1 -R2 + (2/3)(R1+R2)
That means that
x = (R1+R2)/3
y = (2/3)(R1+R2) - R2
check that with R1 = 13 and R2 = 8
x = (13+8)/3 = 21/3 = 7 check
y = (2/3)(21) - 8 = 14-8 = 6 check
To solve the given system of equations using matrix operations, we can represent the coefficients and constants in matrix form.
Let's define the following matrices:
A = [1 1 ; 2 -1] (coefficients matrix)
X = [x ; y] (variable matrix)
B = [13 ; 8] (constants matrix)
The system of equations can now be written as AX = B.
We need to perform row operations on matrix [A | B] to transform it into row-echelon form or reduced row-echelon form. These row operations will also be applied to matrix [X | B].
Let's perform the given row operations step by step:
1. -2R1 + R2 > R2:
Multiply the first row of matrix A by -2 and add it to the second row of A. Simultaneously, perform the same operation on matrix B.
[A | B] = [1 1 | 13]
[2 -1 | 8]
Multiply the first row of matrix A by -2:
[A | B] = [ 1 1 | 13]
[ 0 -3 | -18]
2. -1/3 R2 > R2:
Divide the second row of matrix A by -3. Perform the same row operation on matrix B as well.
[A | B] = [ 1 1 | 13]
[ 0 1 | 6]
3. -1R2 + R1 > R1:
Multiply the second row of matrix A by -1 and add it to the first row of A. Simultaneously, perform the same operation on matrix B.
[A | B] = [ 1 0 | 7]
[ 0 1 | 6]
At this point, matrix A is in reduced row-echelon form, and the augmented matrix [A | B] represents the system of equations in its simplified form.
To obtain the values of x and y, we can read off the solutions from the augmented matrix:
x = 7
y = 6
Therefore, the solution to the given system of equations is x = 7 and y = 6.
Note: The matrix with the answers will be in the form [x ; y], where x and y are the values of the variables obtained from the row operations.