  # Augmented MATRIX HELP!

(1 pt) The following system has an infinite number of solutions. Form an augmented matrix, then write the matrix in the reduced form. Write the reduced form of the matrix below and then write the solution in terms of z.
3x + 2y + 4z = 4
6x - 4y + 3z = 3
9x − 2y + 7z = 7

The required augmented matrix is:
a11 a12 a13 b1
a21 a22 a23 b2
a31 a32 a33 b3
where
a11 = , a12 = , a13 = , b1 =
a21 = , a22 = , a23 = , b2 =
a31 = , a32 = , a33 = , b3 = ,
and x = y=

### Comments for Augmented MATRIX HELP!

 Mar 25, 2011 Augmented MATRIX HELP by: Staff The question: (1 pt) The following system has an infinite number of solutions. Form an augmented matrix, and then write the matrix in the reduced form. Write the reduced form of the matrix below and then write the solution in terms of z. 3x + 2y + 4z = 4 6x - 4y + 3z = 3 9x - 2y + 7z = 7 The required augmented matrix is: a11 a12 a13 b1 a21 a22 a23 b2 a31 a32 a33 b3 where a11 = , a12 = , a13 = , b1 = a21 = , a22 = , a23 = , b2 = a31 = , a32 = , a33 = , b3 = , and x = y= The answer: Three simultaneous equations 3x + 2y + 4z = 4 6x - 4y + 3z = 3 9x − 2y + 7z = 7 Augmented Matrix 3 2 4 : 4 6 -4 3 : 3 9 -2 7 : 7 Row reduction multiply the 1st row by 1/3 1 2/3 4/3 : 4/3 6 -4 3 : 3 9 -2 7 : 7 add -6 times the 1st row to the 2nd row 1 2/3 4/3 : 4/3 0 -8 -5 : -5 9 -2 7 : 7 add -9 times the 1st row to the 3rd row 1 2/3 4/3 : 4/3 0 -8 -5 : -5 0 -8 -5 : -5 multiply the 2nd row by -1/8 1 2/3 4/3 : 4/3 0 1 5/8 : 5/8 0 -8 -5 : -5 Using the augmented matrix in reduced form, we arrive at the following (this is the best we can do): 1 2/3 4/3 : 4/3 0 1 5/8 : 5/8 0 -8 -5 : -5 Convert matrix back to equations: x + (2/3)y + (4/3)z = 4/3 y + (5/8)z = 5/8 -8y – 5z = -5 Notice that the 2nd and 3rd equations are not unique. They are the same equation. equation 2 = equation 3 2nd equation: y + (5/8)z = 5/8 Multiply each side of the 2nd equation by -8 and it is transformed into the 3rd equation. -8y - 5z = -5 The solution in terms of z (which is the independent variable) The solution for y in terms of z: y + (5/8)z = 5/8 y = 5/8 - (5/8)z The solution for x in terms of z: x + (2/3)y + (4/3)z = 4/3 (substitute 5/8 - (5/8)z for the variable y) x + (2/3)( 5/8 - (5/8)z) + (4/3)z = 4/3 x + (10/24) - (10/24)z + (4/3)z = 4/3 x + (10/24) - (10/24)z + (32/24)z = 32/24 x + (10/24) - (10/24) + (22/24)z = 32/24 - (10/24) x + 0 + (22/24)z = 22/24 x + (22/24)z = 22/24 x + (22/24)z - (22/24)z = 22/24 - (22/24)z x + 0 = 22/24 - (22/24)z x = 22/24 - (22/24)z x = 11/12 - (11/12)z the final answer (in terms of z) is: x = 11/12 - (11/12)z y = 5/8 - (5/8)z CHECK THE WORK using the final answer: Arbitrarily select a value for z For example, let z = 2 Solve for x using z = 2 x = 11/12 - (11/12)z x = 11/12 - (11/12)*2 x = 11/12 - (22/12) x = -11/12 Solve for y using z = 2 y = 5/8 - (5/8)z y = 5/8 - (5/8)*2 y = 5/8 - (10/8) y = - 5/8 Substitute the values of x = -11/12, y = - 5/8, and z = 2 in the original three equations: 3x + 2y + 4z = 4 3*(-11/12) + 2*(- 5/8) + 4*2 = 4, OK 6x - 4y + 3z = 3 6*(-11/12) - 4*(- 5/8) + 3*2 = 3, OK 9x - 2y + 7z = 7 9*(-11/12) - 2*(- 5/8) + 7*2 = 7, OK Although only one value of z has been tested, all three of the original equations did compute properly for those values. Thanks for writing. Staff www.solving-math-problems.com