Write the system as a matrix equation and solve using inverses.
x1 + 2x2 - x3 = -3
-2x1 - x2 + 3x3 = 0
-4x1 + 4x2 - x3 = -12
A. x1 = 2, x2 = -3, x3 = 1
B. x1 = 1, x2 = -2, x3 = 1
C. x1 = 1, x2 = -2, x3 = 0
D. x1 = 1, x2 = -3, x3 = 1
this one seems difficult...plz help
You have an equation AX=B
where
A =
(1 2 -3)
(-2 -1 3)
(-4 4 -1)
and
B =
(-3)
(0)
(-12)
A-1 = 1/3 times
(11 10 -3)
(14 13 -3)
(12 12 -3)
So, X = A-1B = (1,-2,0)
fabulous
To write the given system of equations as a matrix equation, we can form a coefficient matrix of the variables x1, x2, and x3 and set it equal to a column matrix of the constants.
The coefficient matrix, A, is defined as:
| 1 2 -1 |
| -2 -1 3 |
| -4 4 -1 |
The column matrix of constants, B, is defined as:
| -3 |
| 0 |
| -12|
The matrix equation is:
Ax = B
To solve this equation using inverses, we can multiply both sides of the equation by the inverse of A:
(A^(-1))(Ax) = (A^(-1))B
Since the inverse of a matrix A exists only if the determinant of A is nonzero, we need to calculate the determinant of A first.
det(A) = | 1 2 -1 |
| -2 -1 3 |
| -4 4 -1 |
= (1)((-1)(-1) - (4)(3)) - (2)((-2)(-1) - (-4)(3)) + (-1)((-2)(4) - (-4)(-1))
= (-1 - 12) - (4 + 12) + (8 + 4)
= -13 - 16 + 12
= -17
Since the determinant is nonzero (-17 ≠ 0), the inverse of A exists.
Now, we can calculate the inverse of A using the formula:
A^(-1) = (1/det(A)) * adj(A)
adj(A) represents the adjugate of matrix A, where each element is the transpose of the cofactor.
To find the cofactor of each element of A, we calculate the determinant of the 2x2 matrix formed by excluding the row and column of the element.
Cofactor of a11 = | -1 3 | = -1(3) - (-1)(-1) = 2
Cofactor of a12 = | -2 3 | = -2(3) - (-4)(-1) = 6
Cofactor of a13 = | -2 -1 | = -2(-1) - (-4)(3) = -10
Cofactor of a21 = | 4 -1 | = 4(3) - (-1)(-1) = 15
Cofactor of a22 = | -4 -1 | = -4(3) - (-4)(-1) = 8
Cofactor of a23 = | -4 4 | = -4(-1) - (-4)(3) = -4
Cofactor of a31 = | 4 -1 | = 4(-1) - 4(3) = -16
Cofactor of a32 = | -4 -1 | = -4(-1) - (-4)(-1) = 0
Cofactor of a33 = | -4 4 | = -4(-1) - (-4)(-1) = 0
Now, we can calculate the adjugate of A by swapping the rows and columns:
adj(A) = | 2 6 -10 |
| 15 8 -4 |
|-16 0 0 |
Finally, we can find the inverse of A:
A^(-1) = (1/det(A)) * adj(A) = (1/-17) * | 2 6 -10 |
| 15 8 -4 |
|-16 0 0 |
= | -2/17 -6/17 10/17 |
|-15/17 -8/17 4/17 |
| 16/17 0 0 |
Now, we can substitute this inverse of A into our equation:
(A^(-1))(Ax) = (A^(-1))B
| -2/17 -6/17 10/17 | * | x1 | = | -3 |
| -15/17 -8/17 4/17 | | x2 | | 0 |
| 16/17 0 0 | | x3 | |-12 |
Multiplying these matrices, we get:
(-2/17)x1 + (-6/17)x2 + (10/17)x3 = (-3)
(-15/17)x1 + (-8/17)x2 + (4/17)x3 = (0)
(16/17)x1 + (0)x2 + (0)x3 = (-12)
Simplifying the third equation, we get:
(16/17)x1 = -12
x1 = (-12)(17/16)
x1 = -51/4
Substituting this value of x1 into the second equation, we have:
(-15/17)(-51/4) + (-8/17)x2 + (4/17)x3 = (0)
255/68 + (-8/17)x2 + (4/17)x3 = 0
Multiplying through by the least common denominator, 68, we get:
255 + (-32)x2 + (16)x3 = 0
(-32)x2 + (16)x3 = -255
Let's define a new variable, y = x2 - x3. We can solve for y using the equation above:
(-32)y = -255
y = (-255)/(-32)
y = 255/32
Now, we can substitute the value of y back into our equation and solve for x2 and x3:
x2 - x3 = 255/32
x2 = (255/32) + x3
Substituting this value of x2 into the first equation, we have:
(-2/17)(-51/4) + (-6/17)(255/32) + (10/17)x3 = (-3)
51/34 - (765/544) + (10/17)x3 = (-3)
Multiplying through by the least common denominator, 544, we get:
816/544 - 765/544 + (32/17)x3 = (-1632)/544
51/34 - (765/544) + (32/17)x3 = (-3)
Combining like terms, we have:
(-243)/544 + (32/17)x3 = (-3)
Multiplying through by the least common denominator, 544, we get:
(-243) + (32/17)(x3) = (-1632)
(32/17)(x3) = (-1632) - (-243)
(32/17)(x3) = (-1632) + 243
Simplifying, we have:
(32/17)(x3) = (-1389)
Now, let's solve for x3:
x3 = (-1389)(17/32)
x3 = (-1389)(17/32)
x3 = -7353/16
Substituting this value of x3 into the equation for x2, we have:
x2 = (255/32) + (-7353/16)
x2 = (255*2)/(32*2) + (-7353)/(16)
x2 = 510/64 - 7353/16
x2 = (-11523 - 29472)/64
x2 = -40995/64
Therefore, the solution to the system of equations is:
x1 = -51/4
x2 = -40995/64
x3 = -7353/16
Now, let's simplify these fractions:
x1 = -12.75
x2 = -640.859375
x3 = -459.5625
The correct answer is not provided in the options given.
It is important to note that while solving equations using inverses is a valid method, it can be time-consuming and may involve complex calculations. In practice, using techniques such as Gaussian elimination or matrix decomposition methods like LU decomposition are often preferred for larger systems of equations.