Question #79914

Check whether the following system of equations has a solution.
4x+2y+8z+6w = 3
2x+2y+2z+2w = 1
x+3z+2w = 3

Expert's answer

Answer on Question #79914 - Math – Linear Algebra

Question

Check whether the following system of equations has a solution.


4x+2y+8z+6w=34x + 2y + 8z + 6w = 32x+2y+2z+2w=12x + 2y + 2z + 2w = 1x+3z+2w=3x + 3z + 2w = 3


Solution

Find a rank of the coefficient matrix A:


A=(428622221032)(10.521.522221032)(10.521.5012100.510.5)(10.521.501210000).A = \begin{pmatrix} 4 & 2 & 8 & 6 \\ 2 & 2 & 2 & 2 \\ 1 & 0 & 3 & 2 \end{pmatrix} \sim \begin{pmatrix} 1 & 0.5 & 2 & 1.5 \\ 2 & 2 & 2 & 2 \\ 1 & 0 & 3 & 2 \end{pmatrix} \sim \begin{pmatrix} 1 & 0.5 & 2 & 1.5 \\ 0 & 1 & -2 & -1 \\ 0 & -0.5 & 1 & 0.5 \end{pmatrix} \sim \begin{pmatrix} 1 & 0.5 & 2 & 1.5 \\ 0 & 1 & -2 & -1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.Rank(A)=2.\operatorname{Rank}(A) = 2.


Write an augmented matrix of the system of the equations and find her rank:


B=(428632222110323)(10.521.50.752222110323)(10.521.50.7501210.500.510.52.25)B = \begin{pmatrix} 4 & 2 & 8 & 6 & 3 \\ 2 & 2 & 2 & 2 & 1 \\ 1 & 0 & 3 & 2 & 3 \end{pmatrix} \sim \begin{pmatrix} 1 & 0.5 & 2 & 1.5 & 0.75 \\ 2 & 2 & 2 & 2 & 1 \\ 1 & 0 & 3 & 2 & 3 \end{pmatrix} \sim \begin{pmatrix} 1 & 0.5 & 2 & 1.5 & 0.75 \\ 0 & 1 & -2 & -1 & -0.5 \\ 0 & -0.5 & 1 & 0.5 & 2.25 \end{pmatrix} \sim(10.521.50.7501210.500002)(10.521.50.7501210.500001).\sim \begin{pmatrix} 1 & 0.5 & 2 & 1.5 & 0.75 \\ 0 & 1 & -2 & -1 & -0.5 \\ 0 & 0 & 0 & 0 & 2 \end{pmatrix} \sim \begin{pmatrix} 1 & 0.5 & 2 & 1.5 & 0.75 \\ 0 & 1 & -2 & -1 & -0.5 \\ 0 & 0 & 0 & 0 & 1 \end{pmatrix}.Rank(B)=3.\operatorname{Rank}(B) = 3.

Rank(A)Rank(B)\operatorname{Rank}(A) \neq \operatorname{Rank}(B). By Kronecker-Capelli theorem, the given system of linear equations is incompatible, the system has no solution.

Answer: the system has no solution.

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