Collection of Maths Problems

Systems III.

Task number: 2387

• Variant 1

  \( \begin{array}{rcrcrcr} 2 x_1 &-& 3x_2 &+& 2x_3 &=\ & 1 \\ x_1 &-& 2x_2 &+& x_3 &=\ & 0 \\ 5 x_1 &-& 9x_2 &+& 5x_3 &=\ & 1 \\ \end{array} \)

• Hint

  Build the augmented matrix of the system, then reduce it to the row echelon form. By backward substitution we express all basic variables in terms of the free variables (parameters).

• Resolution

  \( \begin{pmatrix} \begin{array}{ccc|c} 2 & -3 & 2 & 1 \\ 1 & -2 & 1 & 0 \\ 5 & -9 & 5 & 1 \\ \end{array} \end{pmatrix}\sim \begin{pmatrix} \begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ \end{array} \end{pmatrix}\sim \begin{pmatrix} \begin{array}{ccc|c} 1 & -2 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ \end{array} \end{pmatrix} \)
  \(x_3\) is free, we use parameter \(x_3=p\) instead. From the second row we get \(x_2=1\). From the first \(x_1-2+p=0\) follows that \(x_1=2-p\).

  Test of correctness:

  \( \begin{array}{rcrcrcrcr} 2 (2-p) &-& 3{\cdot} 1 &+& 2p &=\ & 4-2p-3+2p &=\ & 1 \\ 2-p &-& 2{\cdot} 1 &+& p &=\ & 2-p-2+p &=\ & 0 \\ 5 (2-p) &-& 9{\cdot} 1 &+& 5p &=\ & 10-5p-9+5p&=\ & 1 \\ \end{array} \)

• Result

  The solution is \(\mathbf x=(2{,}1,0)^T+p(-1{,}0,1)^T\).

• Variant 2

  \( \begin{array}{rcrcrcrcr} & & x_2 & & &+& x_4 &=\ & 1 \\ 3 x_1 &-& 2x_2 &-& 3x_3 &+& 4x_4 &=\ & -2 \\ x_1 &+& x_2 &-& x_3 &+& x_4 &=\ & 2 \\ x_1 & & &-& x_3 & & &=\ & 1 \end{array} \)

• Result

  \(\mathbf x=(1,\frac{3}{2},0,-\frac{1}{2})^T+p(1{,}0,1{,}0)^T\).

• Variant 3

  \( \begin{array}{rcrcrcrcr} - x_1 &+& x_2 &-& 3x_3 &+& 4x_4 &=\ & 1 \\ 2x_1 &-& x_2 &+& 4x_3 &-& 7x_4 &=\ & 0 \\ - x_1 &-& x_2 &+& x_3 &+& 2x_4 &=\ & -3 \\ - x_1 &+& 2x_2 &-& 5x_3 &+& 5x_4 &=\ & 3 \\ \end{array} \)

• Result

  \(\mathbf x=(1{,}2,0{,}0)^T+p(-1{,}2,1{,}0)^T+q(3,-1{,}0,1)^T\)

• Variant 4

  \( \begin{array}{rcrcrcrcr} - x_1 &+& 2x_2 &+& x_3 &+& 4x_4 &=\ & 3 \\ -2x_1 &+& 4x_2 &+& x_3 &+& 7x_4 &=\ & 5 \\ x_1 &-& 2x_2 &+& x_3 &-& 2x_4 &=\ & -1 \\ - x_1 &+& 2x_2 &+& 2x_3 &+& 5x_4 &=\ & 4 \\ \end{array} \)

• Result

  \(\mathbf x=(-2{,}0,1{,}0)^T+p(2{,}1,0{,}0)^T+q(3{,}0,-1{,}1)^T\)