Answer:
The reduced echelon form of the given matrix B is [ 1 0 0 -5 | 0 ], [ 0 1 0 3 | 0 ], and [ 0 0 1 0 | 0 ]. The parametric description of the set of solutions for the given system of linear equations is x4 ∈ R .
Explanation:
Using row operations to reduce the matrix B to reduced row echelon form, we have:
[ 1 3 -5 4 | 0 ]
[ 1 4 -8 7 | 0 ]
[-3 -7 9 -6 | 0 ]
R2 - R1 -> R2
R3 + 3R1 -> R3
[ 1 3 -5 4 | 0 ]
[ 0 1 -3 3 | 0 ]
[ 0 2 4 6 | 0 ]
R1 - 3R2 -> R1
R3 - 2R2 -> R3
[ 1 0 4 -5 | 0 ]
[ 0 1 -3 3 | 0 ]
[ 0 0 10 0 | 0 ]
R3/10 -> R3
[ 1 0 4 -5 | 0 ]
[ 0 1 -3 3 | 0 ]
[ 0 0 1 0 | 0 ]
R2 + 3R3 -> R2
R1 - 4R3 -> R1
R2 + 3R3 -> R3
[ 1 0 0 -5 | 0 ]
[ 0 1 0 3 | 0 ]
[ 0 0 1 0 | 0 ]
Therefore, the reduced echelon form of the matrix B is:
[ 1 0 0 -5 | 0 ]
[ 0 1 0 3 | 0 ]
[ 0 0 1 0 | 0 ]
asterisk
can you make it step by step easy to understand
Certainly! Here are the steps to find the reduced row echelon form of the matrix B = [1, 3, -5, 4, 1, 4, -8, 7, -3, -7, 9, -6]:
Write the matrix in augmented form, with a vertical line separating the coefficients from the constant terms.
[ 1 3 -5 4 | 0 ]
[ 1 4 -8 7 | 0 ]
[-3 -7 9 -6 | 0 ]
Use row operations to transform the matrix into row echelon form. The goal is to create zeros below the leading entries (the first nonzero element) in each row. Here are the steps:
a. Subtract the first row from the second row, and subtract 3 times the first row from the third row:
[ 1 3 -5 4 | 0 ]
[ 0 1 -3 3 | 0 ]
[ 0 2 4 6 | 0 ]
b. Subtract 3 times the second row from the first row, and subtract 2 times the second row from the third row:
[ 1 0 4 -5 | 0 ]
[ 0 1 -3 3 | 0 ]
[ 0 0 10 0 | 0 ]
c. Divide the third row by 10 to create a leading 1 in the third row:
[ 1 0 4 -5 | 0 ]
[ 0 1 -3 3 | 0 ]
[ 0 0 1 0 | 0 ]
Use row operations to transform the matrix into reduced row echelon form. The goal is to create leading 1's in each row, and zeros above and below the leading 1's. Here are the steps:
a. Subtract 4 times the third row from the first row, and subtract 3 times the third row from the second row:
[ 1 0 0 -5 | 0 ]
[ 0 1 0 3 | 0 ]
[ 0 0 1 0 | 0 ]
The matrix is now in reduced row echelon form. The leftmost nonzero entry in each nonzero row is 1, and each leading 1 is the only nonzero entry in its column. The rows are arranged so that all rows with all zeros are at the bottom of the matrix.
Therefore, the reduced row echelon form of the matrix B is:
[ 1 0 0 -5 | 0 ]
[ 0 1 0 3 | 0 ]
[ 0 0 1 0 | 0 ]
For (b):
From the reduced echelon form obtained in part (a), we can write the system of equations as:
x1 + 0x2 + 4x3 = -5
0x1 + x2 - 3x3 = 3
0x1 + 0x2 + x3 = 0
We can solve for the basic variables (x1, x2, x3) in terms of the free variable(s):
x1 = -5 - 4x3
x2 = 3 + 3x3
x3 = free
Therefore, the general solution to the system of equations is:
x = [-5 - 4x3, 3 + 3x3, x3]
where x3 is any real number, since it is the free variable. This is the parametric description of the set of solutions. We can also write this solution set in set-builder notation as:
x3 ∈ R
This means that any vector in the solution set can be obtained by choosing a value for the free variable x3 and plugging it into the expressions for x1, x2, and x3. For example, if we choose x3 = 0, we get the solution (x1, x2, x3) = (-5, 3, 0). If we choose x3 = 1, we get the solution (x1, x2, x3) = (-9, 6, 1), and so on.
Hope this helps! I'm sorry if it doesn't. If you need more help, ask me! :]