The system of equations has one solution.x = 80 x = 80 x = 80 y = 50 y = 50 y = 50 z = 70 z = 70 z = 70
Solve by Gauss-Jordan Elimination: Begin by writing the augmented matrix of the system of equations. [ 1 1 1 ∣ 200 0 − 1 1 ∣ 20 150 575 260 ∣ 58950 ] \begin{bmatrix}1 & 1 & 1 &\bigm |& 200\\0 & -1 & 1 &\bigm |& 20\\150 & 575 & 260 &\bigm |& 58950\end{bmatrix} 1 0 150 1 − 1 575 1 1 260 200 20 58950
Then use a series of elementary row operations to convert the matrix into reduced-row echelon form. The three elementary row operations are:
1. Swap the positions of any two rows.
2. Multiply any row by a nonzero scalar.
3. Multiply a row by a nonzero scalar and add it to any other row.
First, switch the rows in the matrix such that the row with the leftmost non-zero entry with the greatest magnitude is at the top.
[ 150 575 260 ∣ 58950 0 − 1 1 ∣ 20 1 1 1 ∣ 200 ] \begin{bmatrix}150 & 575 & 260 &\bigm |& 58950\\0 & -1 & 1 &\bigm |& 20\\1 & 1 & 1 &\bigm |& 200\end{bmatrix} 150 0 1 575 − 1 1 260 1 1 58950 20 200
Multiply row 1 1 1 by scalar 1 150 \frac{1}{150} 150 1 to make the leading term 1 1 1 .
[ 1 23 6 26 15 ∣ 393 0 − 1 1 ∣ 20 1 1 1 ∣ 200 ] \begin{bmatrix}1 & \frac{23}{6} & \frac{26}{15} &\bigm |& 393\\0 & -1 & 1 &\bigm |& 20\\1 & 1 & 1 &\bigm |& 200\end{bmatrix} 1 0 1 6 23 − 1 1 15 26 1 1 393 20 200
Multiply row 1 1 1 by scalar − 1 -1 − 1 and add it to row 3 3 3 .
[ 1 23 6 26 15 ∣ 393 0 − 1 1 ∣ 20 0 − 17 6 − 11 15 ∣ − 193 ] \begin{bmatrix}1 & \frac{23}{6} & \frac{26}{15} &\bigm |& 393\\0 & -1 & 1 &\bigm |& 20\\0 & - \frac{17}{6} & - \frac{11}{15} &\bigm |& -193\end{bmatrix} 1 0 0 6 23 − 1 − 6 17 15 26 1 − 15 11 393 20 − 193
Switch the rows in the matrix such that the row with the next leftmost non-zero entry with the next greatest magnitude is the next row from the top.
[ 1 23 6 26 15 ∣ 393 0 − 17 6 − 11 15 ∣ − 193 0 − 1 1 ∣ 20 ] \begin{bmatrix}1 & \frac{23}{6} & \frac{26}{15} &\bigm |& 393\\0 & - \frac{17}{6} & - \frac{11}{15} &\bigm |& -193\\0 & -1 & 1 &\bigm |& 20\end{bmatrix} 1 0 0 6 23 − 6 17 − 1 15 26 − 15 11 1 393 − 193 20
Multiply row 2 2 2 by scalar − 6 17 - \frac{6}{17} − 17 6 to make the leading term 1 1 1 .
[ 1 23 6 26 15 ∣ 393 0 1 22 85 ∣ 1158 17 0 − 1 1 ∣ 20 ] \begin{bmatrix}1 & \frac{23}{6} & \frac{26}{15} &\bigm |& 393\\0 & 1 & \frac{22}{85} &\bigm |& \frac{1158}{17}\\0 & -1 & 1 &\bigm |& 20\end{bmatrix} 1 0 0 6 23 1 − 1 15 26 85 22 1 393 17 1158 20
Multiply row 2 2 2 by scalar − 23 6 - \frac{23}{6} − 6 23 and add it to row 1 1 1 .
[ 1 0 63 85 ∣ 2242 17 0 1 22 85 ∣ 1158 17 0 − 1 1 ∣ 20 ] \begin{bmatrix}1 & 0 & \frac{63}{85} &\bigm |& \frac{2242}{17}\\0 & 1 & \frac{22}{85} &\bigm |& \frac{1158}{17}\\0 & -1 & 1 &\bigm |& 20\end{bmatrix} 1 0 0 0 1 − 1 85 63 85 22 1 17 2242 17 1158 20
Multiply row 2 2 2 by scalar 1 1 1 and add it to row 3 3 3 .
[ 1 0 63 85 ∣ 2242 17 0 1 22 85 ∣ 1158 17 0 0 107 85 ∣ 1498 17 ] \begin{bmatrix}1 & 0 & \frac{63}{85} &\bigm |& \frac{2242}{17}\\0 & 1 & \frac{22}{85} &\bigm |& \frac{1158}{17}\\0 & 0 & \frac{107}{85} &\bigm |& \frac{1498}{17}\end{bmatrix} 1 0 0 0 1 0 85 63 85 22 85 107 17 2242 17 1158 17 1498
Multiply row 3 3 3 by scalar 85 107 \frac{85}{107} 107 85 to make the leading term 1 1 1 .
[ 1 0 63 85 ∣ 2242 17 0 1 22 85 ∣ 1158 17 0 0 1 ∣ 70 ] \begin{bmatrix}1 & 0 & \frac{63}{85} &\bigm |& \frac{2242}{17}\\0 & 1 & \frac{22}{85} &\bigm |& \frac{1158}{17}\\0 & 0 & 1 &\bigm |& 70\end{bmatrix} 1 0 0 0 1 0 85 63 85 22 1 17 2242 17 1158 70
Multiply row 3 3 3 by scalar − 63 85 - \frac{63}{85} − 85 63 and add it to row 1 1 1 .
[ 1 0 0 ∣ 80 0 1 22 85 ∣ 1158 17 0 0 1 ∣ 70 ] \begin{bmatrix}1 & 0 & 0 &\bigm |& 80\\0 & 1 & \frac{22}{85} &\bigm |& \frac{1158}{17}\\0 & 0 & 1 &\bigm |& 70\end{bmatrix} 1 0 0 0 1 0 0 85 22 1 80 17 1158 70
Multiply row 3 3 3 by scalar − 22 85 - \frac{22}{85} − 85 22 and add it to row 2 2 2 .
[ 1 0 0 ∣ 80 0 1 0 ∣ 50 0 0 1 ∣ 70 ] \begin{bmatrix}1 & 0 & 0 &\bigm |& 80\\0 & 1 & 0 &\bigm |& 50\\0 & 0 & 1 &\bigm |& 70\end{bmatrix} 1 0 0 0 1 0 0 0 1 80 50 70
Once the matrix is in reduced-row echelon form, convert the matrix back into linear equations to find the solution. 1 ⋅ x + 0 ⋅ y + 0 ⋅ z = 80 x = 80 \begin{aligned}1 \cdot x+ 0 \cdot y+ 0 \cdot z = 80 \\ x = 80\end{aligned} 1 ⋅ x + 0 ⋅ y + 0 ⋅ z = 80 x = 80 0 ⋅ x + 1 ⋅ y + 0 ⋅ z = 50 y = 50 \begin{aligned}0 \cdot x+ 1 \cdot y+ 0 \cdot z = 50 \\ y = 50\end{aligned} 0 ⋅ x + 1 ⋅ y + 0 ⋅ z = 50 y = 50 0 ⋅ x + 0 ⋅ y + 1 ⋅ z = 70 z = 70 \begin{aligned}0 \cdot x+ 0 \cdot y+ 1 \cdot z = 70 \\ z = 70\end{aligned} 0 ⋅ x + 0 ⋅ y + 1 ⋅ z = 70 z = 70
Solve by matrix inversion: In cases where the coefficient matrix of the system of equations is invertible, we can use the inverse to solve the system. Use this method with care as matrix inversion can be numerically unstable for ill-conditioned matrices.
Express the linear equations in the form A × X = B A \times X = B A × X = B where A A A is the coefficient matrix, X X X is the matrix of unknowns, and B B B is the constant matrix.[ 1 1 1 0 − 1 1 150 575 260 ] × [ x y z ] = [ 200 20 58950 ] \left[\begin{matrix}1 & 1 & 1\\0 & -1 & 1\\150 & 575 & 260\end{matrix}\right] \times \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}200\\20\\58950\end{matrix}\right] 1 0 150 1 − 1 575 1 1 260 × x y z = 200 20 58950
The product of A A A and its inverse A − 1 A^{-1} A − 1 is the identity matrix. Any matrix multiplied by the identity matrix remains unchanged, so this yields the matrix of unknowns on the left hand side of the equation, and the solution matrix on the right. A × X = B A − 1 × A × X = A − 1 × B I × X = A − 1 × B X = A − 1 × B \begin{aligned} A \times X &= B\\ A^{-1} \times A \times X &= A^{-1} \times B \\ I \times X &= A^{-1} \times B \\ X &= A^{-1} \times B \end{aligned} A × X A − 1 × A × X I × X X = B = A − 1 × B = A − 1 × B = A − 1 × B
Using a computer algebra system, calculate A − 1 A^{-1} A − 1 . [ 167 107 − 63 107 − 2 535 − 30 107 − 22 107 1 535 − 30 107 85 107 1 535 ] \left[\begin{matrix}\frac{167}{107} & - \frac{63}{107} & - \frac{2}{535}\\- \frac{30}{107} & - \frac{22}{107} & \frac{1}{535}\\- \frac{30}{107} & \frac{85}{107} & \frac{1}{535}\end{matrix}\right] 107 167 − 107 30 − 107 30 − 107 63 − 107 22 107 85 − 535 2 535 1 535 1
Multiply both sides of the equation by the inverse. [ 167 107 − 63 107 − 2 535 − 30 107 − 22 107 1 535 − 30 107 85 107 1 535 ] × [ 1 1 1 0 − 1 1 150 575 260 ] × [ x y z ] = [ 167 107 − 63 107 − 2 535 − 30 107 − 22 107 1 535 − 30 107 85 107 1 535 ] × [ 200 20 58950 ] \left[\begin{matrix}\frac{167}{107} & - \frac{63}{107} & - \frac{2}{535}\\- \frac{30}{107} & - \frac{22}{107} & \frac{1}{535}\\- \frac{30}{107} & \frac{85}{107} & \frac{1}{535}\end{matrix}\right] \times \left[\begin{matrix}1 & 1 & 1\\0 & -1 & 1\\150 & 575 & 260\end{matrix}\right] \times \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}\frac{167}{107} & - \frac{63}{107} & - \frac{2}{535}\\- \frac{30}{107} & - \frac{22}{107} & \frac{1}{535}\\- \frac{30}{107} & \frac{85}{107} & \frac{1}{535}\end{matrix}\right] \times \left[\begin{matrix}200\\20\\58950\end{matrix}\right] 107 167 − 107 30 − 107 30 − 107 63 − 107 22 107 85 − 535 2 535 1 535 1 × 1 0 150 1 − 1 575 1 1 260 × x y z = 107 167 − 107 30 − 107 30 − 107 63 − 107 22 107 85 − 535 2 535 1 535 1 × 200 20 58950 [ x y z ] = [ 80 50 70 ] \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}80\\50\\70\end{matrix}\right] x y z = 80 50 70
x = 80 x = 80 x = 80 y = 50 y = 50 y = 50 z = 70 z = 70 z = 70