2x - 3y + 2z = -3

-3x + 2y + z = 1

4x + y - 3z = 4

asked by guest
on Mar 29, 2025 at 9:19 pm



You asked:

Solve the system of equations:2x−3y+2z=−3−(3x)+2y+z=14x+y−3z=4\begin{aligned}2 x - 3 y + 2 z &= -3\\-\left( 3 x \right) + 2 y + z &= 1\\4 x + y - 3 z &= 4\end{aligned}

MathBot Answer:

The system of equations has one solution.x=23x = \frac{2}{3} y=3121y = \frac{31}{21} z=121z = \frac{1}{21}

Solve by substitution:

Solve 2x−3y+2z=−32 x - 3 y + 2 z = -3 for xx. x=3y2−z−32x = \frac{3 y}{2} - z - \frac{3}{2}

Substitute 3y2−z−32\frac{3 y}{2} - z - \frac{3}{2} for xx in each of the remaining equations and simplify. −3x+2y+z=1−3(3y2−z−32)+2y+z=15y−8z=7\begin{aligned}- 3 x + 2 y + z &= 1 \\ - 3 \left(\frac{3 y}{2} - z - \frac{3}{2}\right) + 2 y + z &= 1 \\ 5 y - 8 z &= 7 \end{aligned}4x+y−3z=44(3y2−z−32)+y−3z=4y−z=107\begin{aligned}4 x + y - 3 z &= 4 \\ 4 \left(\frac{3 y}{2} - z - \frac{3}{2}\right) + y - 3 z &= 4 \\ y - z &= \frac{10}{7} \end{aligned}

Solve 5y−8z=75 y - 8 z = 7 for yy. y=8z5+75y = \frac{8 z}{5} + \frac{7}{5}

Substitute 8z5+75\frac{8 z}{5} + \frac{7}{5} for yy in y−z=107y - z = \frac{10}{7} and simplify. y−z=107(8z5+75)−z=107z=121\begin{aligned}y - z &= \frac{10}{7} \\ \left(\frac{8 z}{5} + \frac{7}{5}\right) - z &= \frac{10}{7} \\ z &= \frac{1}{21} \end{aligned}

Use substitution of the numerical value of zz to get the values of xx and y y. y=8z5+75y=8⋅1⋅121⋅15+75y=3121\begin{aligned}y &= \frac{8 z}{5} + \frac{7}{5} \\ y &= 8 \cdot 1 \cdot \frac{1}{21} \cdot \frac{1}{5} + \frac{7}{5} \\ y &= \frac{31}{21}\end{aligned}x=3y2−z−32x=−32−1⋅121+3⋅31⋅121⋅12x=23\begin{aligned}x &= \frac{3 y}{2} - z - \frac{3}{2} \\ x &= - \frac{3}{2} - 1 \cdot \frac{1}{21} + 3 \cdot 31 \cdot \frac{1}{21} \cdot \frac{1}{2} \\ x &= \frac{2}{3}\end{aligned}

Solve by Gauss-Jordan Elimination:

Begin by writing the augmented matrix of the system of equations. [2−32∣−3−321∣141−3∣4]\begin{bmatrix}2 & -3 & 2 &\bigm |& -3\\-3 & 2 & 1 &\bigm |& 1\\4 & 1 & -3 &\bigm |& 4\end{bmatrix}

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.

[41−3∣4−321∣12−32∣−3]\begin{bmatrix}4 & 1 & -3 &\bigm |& 4\\-3 & 2 & 1 &\bigm |& 1\\2 & -3 & 2 &\bigm |& -3\end{bmatrix}

Multiply row 11 by scalar 14\frac{1}{4} to make the leading term 11.

[114−34∣1−321∣12−32∣−3]\begin{bmatrix}1 & \frac{1}{4} & - \frac{3}{4} &\bigm |& 1\\-3 & 2 & 1 &\bigm |& 1\\2 & -3 & 2 &\bigm |& -3\end{bmatrix}

Multiply row 11 by scalar 33 and add it to row 22.

[114−34∣10114−54∣42−32∣−3]\begin{bmatrix}1 & \frac{1}{4} & - \frac{3}{4} &\bigm |& 1\\0 & \frac{11}{4} & - \frac{5}{4} &\bigm |& 4\\2 & -3 & 2 &\bigm |& -3\end{bmatrix}

Multiply row 11 by scalar −2-2 and add it to row 33.

[114−34∣10114−54∣40−7272∣−5]\begin{bmatrix}1 & \frac{1}{4} & - \frac{3}{4} &\bigm |& 1\\0 & \frac{11}{4} & - \frac{5}{4} &\bigm |& 4\\0 & - \frac{7}{2} & \frac{7}{2} &\bigm |& -5\end{bmatrix}

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.

[114−34∣10−7272∣−50114−54∣4]\begin{bmatrix}1 & \frac{1}{4} & - \frac{3}{4} &\bigm |& 1\\0 & - \frac{7}{2} & \frac{7}{2} &\bigm |& -5\\0 & \frac{11}{4} & - \frac{5}{4} &\bigm |& 4\end{bmatrix}

Multiply row 22 by scalar −27- \frac{2}{7} to make the leading term 11.

[114−34∣101−1∣1070114−54∣4]\begin{bmatrix}1 & \frac{1}{4} & - \frac{3}{4} &\bigm |& 1\\0 & 1 & -1 &\bigm |& \frac{10}{7}\\0 & \frac{11}{4} & - \frac{5}{4} &\bigm |& 4\end{bmatrix}

Multiply row 22 by scalar −14- \frac{1}{4} and add it to row 11.

[10−12∣91401−1∣1070114−54∣4]\begin{bmatrix}1 & 0 & - \frac{1}{2} &\bigm |& \frac{9}{14}\\0 & 1 & -1 &\bigm |& \frac{10}{7}\\0 & \frac{11}{4} & - \frac{5}{4} &\bigm |& 4\end{bmatrix}

Multiply row 22 by scalar −114- \frac{11}{4} and add it to row 33.

[10−12∣91401−1∣1070032∣114]\begin{bmatrix}1 & 0 & - \frac{1}{2} &\bigm |& \frac{9}{14}\\0 & 1 & -1 &\bigm |& \frac{10}{7}\\0 & 0 & \frac{3}{2} &\bigm |& \frac{1}{14}\end{bmatrix}

Multiply row 33 by scalar 23\frac{2}{3} to make the leading term 11.

[10−12∣91401−1∣107001∣121]\begin{bmatrix}1 & 0 & - \frac{1}{2} &\bigm |& \frac{9}{14}\\0 & 1 & -1 &\bigm |& \frac{10}{7}\\0 & 0 & 1 &\bigm |& \frac{1}{21}\end{bmatrix}

Multiply row 33 by scalar 12\frac{1}{2} and add it to row 11.

[100∣2301−1∣107001∣121]\begin{bmatrix}1 & 0 & 0 &\bigm |& \frac{2}{3}\\0 & 1 & -1 &\bigm |& \frac{10}{7}\\0 & 0 & 1 &\bigm |& \frac{1}{21}\end{bmatrix}

Multiply row 33 by scalar 11 and add it to row 22.

[100∣23010∣3121001∣121]\begin{bmatrix}1 & 0 & 0 &\bigm |& \frac{2}{3}\\0 & 1 & 0 &\bigm |& \frac{31}{21}\\0 & 0 & 1 &\bigm |& \frac{1}{21}\end{bmatrix}

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=23x=23\begin{aligned}1 \cdot x+ 0 \cdot y+ 0 \cdot z = \frac{2}{3} \\ x = \frac{2}{3}\end{aligned}0⋅x+1⋅y+0⋅z=3121y=3121\begin{aligned}0 \cdot x+ 1 \cdot y+ 0 \cdot z = \frac{31}{21} \\ y = \frac{31}{21}\end{aligned}0⋅x+0⋅y+1⋅z=121z=121\begin{aligned}0 \cdot x+ 0 \cdot y+ 1 \cdot z = \frac{1}{21} \\ z = \frac{1}{21}\end{aligned}

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=BA \times X = B where AA is the coefficient matrix, XX is the matrix of unknowns, and BB is the constant matrix.[2−32−32141−3]×[xyz]=[−314]\left[\begin{matrix}2 & -3 & 2\\-3 & 2 & 1\\4 & 1 & -3\end{matrix}\right] \times \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}-3\\1\\4\end{matrix}\right]

The product of AA and its inverse A−1A^{-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=BA−1×A×X=A−1×BI×X=A−1×BX=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}

Using a computer algebra system, calculate A−1A^{-1}. [13131352123821112123521]\left[\begin{matrix}\frac{1}{3} & \frac{1}{3} & \frac{1}{3}\\\frac{5}{21} & \frac{2}{3} & \frac{8}{21}\\\frac{11}{21} & \frac{2}{3} & \frac{5}{21}\end{matrix}\right]

Multiply both sides of the equation by the inverse. [13131352123821112123521]×[2−32−32141−3]×[xyz]=[13131352123821112123521]×[−314]\left[\begin{matrix}\frac{1}{3} & \frac{1}{3} & \frac{1}{3}\\\frac{5}{21} & \frac{2}{3} & \frac{8}{21}\\\frac{11}{21} & \frac{2}{3} & \frac{5}{21}\end{matrix}\right] \times \left[\begin{matrix}2 & -3 & 2\\-3 & 2 & 1\\4 & 1 & -3\end{matrix}\right] \times \left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}\frac{1}{3} & \frac{1}{3} & \frac{1}{3}\\\frac{5}{21} & \frac{2}{3} & \frac{8}{21}\\\frac{11}{21} & \frac{2}{3} & \frac{5}{21}\end{matrix}\right] \times \left[\begin{matrix}-3\\1\\4\end{matrix}\right] [xyz]=[233121121]\left[\begin{matrix}x\\y\\z\end{matrix}\right] = \left[\begin{matrix}\frac{2}{3}\\\frac{31}{21}\\\frac{1}{21}\end{matrix}\right]

x=23x = \frac{2}{3}y=3121y = \frac{31}{21}z=121z = \frac{1}{21}

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