Collection of Maths Problems

Systems II.

Task number: 2384

• Variant 1

  \( \mathbf A=\begin{pmatrix} 2 & 3 & 4 & 5 \\ 3 & 0 & 2 & -3 \\ 7 & 6 & 10 & 7 \\ \end{pmatrix} \)

• Hint

  Use the echelon form.

• Resolution

  \( \begin{pmatrix} 2 & 3 & 4 & 5 \\ 3 & 0 & 2 & -3 \\ 7 & 6 & 10 & 7 \\ \end{pmatrix} \sim…\sim \begin{pmatrix} 2 & 3 & 4 & 5 \\ 0 & 9 & 8 & 21 \\ 0 & 0 & 0 & 0 \\ \end{pmatrix} \)

  Variables \(x_1\) and \(x_2\) are basic, while \(x_3\) and \(x_4\) are free.

  We choose at least on free variable nonzero and compute the values of basic variables by the backward substitution. For \(x_4=0\) and \(x_3=9\) we get e.g.: \(\mathbf x=(-6,-8{,}9,0)^T\),

• Variant 2

  \( \mathbf A=\begin{pmatrix} 2 & -3 & 13 & 18 \\ 6 & -9 & 7 & 10 \\ 2 & -3 & -3 & -4 \\ \end{pmatrix} \)

• Result

  \(\mathbf x=(3{,}2,0{,}0)^T\),

• Variant 3

  \( \mathbf A=\begin{pmatrix} 1 & 2 & -1 & 1 \\ 2 & -1 & 4 & 10 \\ 1 & 0 & 3 & -5 \\ 2 & 5 & 2 & 2 \\ \end{pmatrix}, \)

• Result

  The system has only trivial solution \(\mathbf x=(0{,}0,0{,}0)^T\).

• Variant 4

  \( \mathbf A=\begin{pmatrix} 1 & -1 & 1 & 2 \\ 1 & 8 & 7 & -7 \\ 1 & 2 & 3 & -1 \\ 1 & 5 & 5 & -4 \\ \end{pmatrix} \)

• Result

  \(\mathbf x=(-1{,}1,0{,}1)^T\).