2x+y+z=6
-3x-4y+2z=4
x+y-z=-2
solve the system of all variables using a matriz
+2 +1 +1 +6
-3 -4 +2 +4
+1 +1 -1 -2
+1 +1/2 +1/2 +3
-1 -4/3 +2/3 +4/3
+1 +1/1 -1/1 -2
+1 +1/2 +1/2 +3
+0 +2/3 +7/6 +13/3
+0 -1/2 +3/2 +5
+1 +1/2 +1/2 +3
+0 +1/1 +7/4 +13/2
+0 -1/2 +3/2 +5
+1 +1/2 +1/2 +3
+0 +1/1 +7/4 +13/2
+0 -1/2 +3/2 +5
(middle times 1/2--> 0 1/2 7/8 13/4 subtract from 1, add to 3 for new rows 1 and 3 )
+1 +0 -3/8 -1/4
+0 +1 +7/4 +13/2
+0 +0 19/8 +39/4
+1 +0 -3/8 -1/4
+0 +1 +7/4 +13/2
+0 +0 +1/1 +78/19
(row 3 times 7/4, subtract from row 2 for new row 2)
+1 +0 -3/8 -1/4
+0 +1 + 0 -13/19
+0 +0 +1/1 +78/19
(Row 3 times 3/8, add to row 1 for new row 1)
+1 +0 +0 +49/38
+0 +1 +0 -13/19
+0 +0 +1 +78/19
Damon, you must have made an arithmetic error somewhere, but I'm too lazy to find it, lol
here is my manipulation:
1 1 -1 -2
2 1 1 6
-3 -4 2 4 re-arranged the rows
=
1 1 -1 -2
0 1 -3 -10 did 2x#1 - #2
0 -1 -1 -2 did 3x#1 + #3
=
1 1 -1 -2
0 1 -3 -10
0 0 -4 -12 did #2 + #3
here we can stop, since line #3 says
-4z = -12
z = 3
now back in line 2
y - 3z = -10
y - 9 = -10
y = -1
and finally in #1 which says
x + y - z = -2
x - 1 - 3 = -2
x = 2
so x=2, y=-1, and z=3
which verifies in all 3 equations
+2 +1 +1 +6
-3 -4 +2 +4
+1 +1 -1 -2
+1 +1/2 +1/2 +3
-1 -4/3 +2/3 +4/3
+1 +1/1 -1/1 -2
+1 +1/2 +1/2 +3
+0 -5/6 +7/6 +13/3 correction
+0 -1/2 +3/2 +5
You will have to follow through with the correction
To solve the system of equations using a matrix, we need to represent the given system in matrix form. Let's define the coefficient matrix, variable matrix, and constant matrix:
Coefficient matrix (A):
| 2 1 1 |
| -3 -4 2 |
| 1 1 -1 |
Variable matrix (X):
| x |
| y |
| z |
Constant matrix (B):
| 6 |
| 4 |
| -2 |
The system of equations can be written as AX = B, where X represents the variable matrix.
To solve for X, we can use the inverse of matrix A. The solution for X can be found using the formula X = A^(-1) * B.
Let's proceed with finding the inverse of matrix A:
Step 1: Calculate the determinant of matrix A.
We can use the following formula for a 3x3 matrix:
det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)
det(A) = 2((-4)(-1) - (2)(1)) - 1((-3)(-1) - (2)(1)) + 1((-3)(2) - (-4)(1))
= 2(-4 + 2) - 1(3 - 2) + 1(-6 + 4)
= 2(-2) - 1(1) + 1(-2)
= -4 + (-1) - 2
= -7
Step 2: Find the cofactor matrix of A.
The cofactor matrix is formed by calculating the determinant of each 2x2 matrix in A.
Cofactor matrix (C):
| 4 -3 3 |
| 1 2 -1 |
|-4 -1 2 |
Step 3: Find the adjugate (adjoint) matrix of A.
The adjugate matrix is obtained by transposing the cofactor matrix.
Adjugate matrix (Adj):
| 4 1 -4 |
|-3 2 -1 |
| 3 -1 2 |
Step 4: Calculate the inverse of matrix A.
The inverse of A is given by A^(-1) = (1/det(A)) * Adj.
A^(-1) = (1/-7) * | 4 1 -4 |
|-3 2 -1 |
| 3 -1 2 |
= | -4/7 -1/7 4/7 |
| 3/7 -2/7 1/7 |
|-3/7 1/7 -2/7 |
Now that we have the inverse matrix, let's find the solution:
X = A^(-1) * B
| x | | -4/7 -1/7 4/7 | | 6 |
| y | = | 3/7 -2/7 1/7 | * | 4 |
| z | |-3/7 1/7 -2/7 | |-2 |
Simplifying this equation, we find:
x = (-4/7)*(6) + (-1/7)*(4) + (4/7)*(-2) = -24/7 - 4/7 - 8/7 = -36/7
y = (3/7)*(6) + (-2/7)*(4) + (1/7)*(-2) = 18/7 - 8/7 - 2/7 = 8/7
z = (-3/7)*(6) + (1/7)*(4) + (-2/7)*(-2) = -18/7 + 4/7 - 4/7 = -18/7
Therefore, the solution to the given system of equations is:
x = -36/7
y = 8/7
z = -18/7