The row echelon form of a system of linear equations is given. (a) Write the system of equations corresponding to the given matrix. Use x, y, or x, y, z, or x_1,x_2,x_3, x_4 (b) Determine whether the system is consistent. If it is consistent, give the solution. begin{matrix}1 & 0 & 3 & 0 &1 0 & 1 & 4 & 3&20&0&1&2&30&0&0&0&0 end{matrix}

The row echelon form of a system of linear equations is given. (a) Write the system of equations corresponding to the given matrix. Use x, y, or x, y, z, or x_1,x_2,x_3, x_4 (b) Determine whether the system is consistent. If it is consistent, give the solution. begin{matrix}1 & 0 & 3 & 0 &1 0 & 1 & 4 & 3&20&0&1&2&30&0&0&0&0 end{matrix}

Question
Matrices
asked 2020-12-22
The row echelon form of a system of linear equations is given.
(a) Write the system of equations corresponding to the given matrix.
Use x, y, or x, y, z, or \(x_1,x_2,x_3, x_4\)
(b) Determine whether the system is consistent. If it is consistent, give the solution.
\(\begin{matrix}1 & 0 & 3 & 0 &1 \\ 0 & 1 & 4 & 3&2\\0&0&1&2&3\\0&0&0&0&0 \end{matrix}\)

Answers (1)

2020-12-23
a. The system represented by the matrix is
\(\begin{cases}x_1+3x_3=1\\x_2+4x_3+3x_4=2\\x_3+2x_4=3\\0=0\end{cases}\)
b. This is a consistent system (because the last equation is true for all ordered quadruplets).
The system will not have a single solution, as it is, in effect, a system of THREE equations in FOUR variables.
It is a dependent system.
To solve, we take \(x_4\) to be any real number \(t\in\mathbb{R}\), and back-substitute into the other two equations:
Back substitute into (3): \(\begin{bmatrix}x_3+2(t)=0\\ x_3=3-2t \end{bmatrix}\)
Back substitute into (2): \(\begin{bmatrix}x_2+4(3-2t)+3(t)=2\\ x_2+12-8t+3t=2\\x_2+12-5t=-10+5t \end{bmatrix}\)
Back substitute into (1): \(\begin{bmatrix}x_1+3(3-2t)-2(t)=1\\ x_1+9-6t=1\\ x_1=-8+6t \end{bmatrix}\)
Result:
a. \(\begin{cases}x_1+3x_3=1\\x_2+4x_3+3x_4=2\\x_3+2x_4=3\\0=0\end{cases}\)
b. Consistent, solution set: \(\left\{(-8+6t, -10+5t,3-2t,t),\ t\in\mathbb{R}\right\}\)
0

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