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Math  /  Algebra

QuestionSolve using augmented matrix methods. 2x1−x2=−53x1+2x2=3\begin{array}{rr} 2 x_{1}-x_{2}= & -5 \\ 3 x_{1}+2 x_{2}= & 3 \end{array}

Studdy Solution

STEP 1

1. The system of equations is linear and can be represented using an augmented matrix.
2. We will use row operations to solve the system.

STEP 2

1. Write the augmented matrix for the system of equations.
2. Use row operations to achieve row-echelon form.
3. Use back substitution to find the solution.

STEP 3

Write the augmented matrix for the given system of equations:
The system of equations is: \[ \begin{align*} 2x_{1} - x_{2} &= -5 \\ 3x_{1} + 2x_{2} &= 3 \end{align*}$
The augmented matrix is: [2−1∣−532∣3]\begin{bmatrix} 2 & -1 & | & -5 \\ 3 & 2 & | & 3 \end{bmatrix}

STEP 4

Perform row operations to achieve row-echelon form.
First, we can eliminate the first element of the second row by replacing Row 2 with (Row 2−32×Row 1)(\text{Row 2} - \frac{3}{2} \times \text{Row 1}):
[2−1∣−532∣3]\begin{bmatrix} 2 & -1 & | & -5 \\ 3 & 2 & | & 3 \end{bmatrix}
Row 2 becomes: 3−32×2=0,2−32×(−1)=2+32=72,3−32×(−5)=3+152=2123 - \frac{3}{2} \times 2 = 0, \quad 2 - \frac{3}{2} \times (-1) = 2 + \frac{3}{2} = \frac{7}{2}, \quad 3 - \frac{3}{2} \times (-5) = 3 + \frac{15}{2} = \frac{21}{2}
The new matrix is: [2−1∣−5072∣212]\begin{bmatrix} 2 & -1 & | & -5 \\ 0 & \frac{7}{2} & | & \frac{21}{2} \end{bmatrix}

STEP 5

Use back substitution to find the solution.
From the second row of the matrix, we have: 72x2=212\frac{7}{2}x_{2} = \frac{21}{2}
Solve for x2 x_{2} : x2=212×27=3x_{2} = \frac{21}{2} \times \frac{2}{7} = 3
Substitute x2=3 x_{2} = 3 into the first equation: 2x1−x2=−52x_{1} - x_{2} = -5 2x1−3=−52x_{1} - 3 = -5 2x1=−5+32x_{1} = -5 + 3 2x1=−22x_{1} = -2 x1=−1x_{1} = -1
The solution to the system of equations is: x1=−1,x2=3x_{1} = -1, \quad x_{2} = 3

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