Chapter 10, Problem 1PRE

For Exercises 1-4, solve the system of equations using

1. The substitution method or the addition method (see Sections 9. 1 and 9.2).
2. Gaussian elimination (see Section 10.1)
3. Gauss-Jordan elimination (see Section 10.1).
4. The inverse of the coefficient matrix (see Section 10.4).
5. Cramer's rule (see Section 10.5).

1. $x=-3y-10$
$-3x-7y=22$

To determine

To solve: The system of equations using substitution method.

Answer to Problem 1PRE

The required solution of the system of equations by substitution method is $x\text{=2}$ , and $y\text{=- 4}$ .

Explanation of Solution

Given:

The given system of equationsis

  $x\text{=-}3y\text{-}10$  ........ (1)

  $\text{-}3x\text{-7}y\text{=22}$  ........ (2)

Method Used:

The steps used in substitution method are:

Step1: Choose one of the two equations and solve it for one of the two variables. (Make sure avoiding fractions, if possible.)

Step2: Substitute the value of variable of step 1 into the equation that is not used in step 1 and then solve resulted linear equation for one variable.

Step3: Substitute the result of step 2 into the expression obtained in step 1 to find the value of the other variable.

Calculations:

The value of variable x from equation (1) is $x\text{=-3}y\text{-10}$ . Substitute this value of x into the equation (2)

  $\text{-}3\left(\text{-3}y\text{-10}\right)\text{-7}y\text{=22}$

  $\Rightarrow \text{9}y\text{+30-7}y\text{=22}\Rightarrow \text{2}y\text{=22-30=-8}$

  $\Rightarrow \text{2}y\text{=-8}\Rightarrow y\text{=}\frac{\text{-8}}{2}\text{=-}\text{4}$

Now substitute value of y back into the first equation $x\text{=-3}\left(\text{- }4\right)\text{-10=12-10=2}$ .

Thus $x\text{=2}$ and $y\text{=- 4}$ .

Conclusion:

The solution of given system of equations by substitution method is $x\text{=2}$ , $y\text{=- 4}$ .

To determine

To solve: The system of equations using Gaussian elimination method.

Answer to Problem 1PRE

The solution of the system of equations using Gaussian elimination method is $x\text{=2}$ , $y\text{=- 4}$ .

Explanation of Solution

Given:

The system of equations in part (a)

Method used:

The steps used in Gauss elimination are:

Step 1: Write augmented matrix for the system of equations

Step 2: Using elementary operations write augmented matrix in “row echelon form”

Step 3: Using back substitution solve the resulted set of equations

Calculations:

The given system of equations can be written as: $x\text{+}3y\text{=-}10$ , $\text{-}3x\text{-7}y\text{=22}$ .

The coefficient matrix and the augmented matrix for the given system of equations are $A\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}\right]$ and $\left[A|B\right]\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}|\begin{array}{c}\text{-10}\\ \text{22}\end{array}\right]$ respectively. Now applying row operations:

  $\left[A|B\right]\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}|\begin{array}{c}\text{-10}\\ \text{22}\end{array}\right]\underrightarrow{R_2\to R_2\text{+}3R_1}\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{0}& \text{2}\end{array}|\begin{array}{c}\text{-10}\\ \text{-}8\end{array}\right]$

  $\underrightarrow{R_2\to \frac{1}{2}{R}_2}\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{0}& \text{1}\end{array}|\begin{array}{c}\text{-10}\\ \text{-}\text{4}\end{array}\right]\begin{array}{c}\to \\ \to \end{array}\begin{array}{c}x\text{+}3y\text{=-}10\\ 0x\text{+}y\text{=-}\text{4}\text{or}\end{array}\begin{array}{c}\text{or}x\text{+}3y\text{=-}10\\ y\text{=-}\text{4}\end{array}$ .

Thus, value of $y\text{=- }4$ . Now substituting back the value of y in resulted equation $x\text{+}3y\text{=-}10$ ; $x\text{+}3y\text{=-}10\Rightarrow x\text{=-}3y\text{-}10\text{=-}3\left(\text{-4}\right)\text{-}10\text{=}12\text{-}10\text{=2}$ .

The solution of system of equations is $x\text{=2}$ , and $y\text{=- 4}$ .

Conclusion:The solution of the system of equations using Gaussian elimination method is $x\text{=2}$ , $y\text{=- 4}$ .

To determine

To solve: The system of equations using Gauss- Jordan elimination method.

Answer to Problem 1PRE

The solution of the system of equations using Gauss- Jordan elimination method is $x\text{=2}$

  $y=-4$ .

Explanation of Solution

Given:

The system of equations in part (a)

Method used:

In Gauss- Jordan elimination method a “reduced row Echelon matrix” is obtained using appropriate elementary row operations as given below:

Step 1: Choosing the leftmost nonzero column and using row operation get a 1 at the top.

Step2: Use multiples of the rows containing 1 from step 1, and get zeros in all remaining places in the column containing this 1.

Step 3: Repeat step 1 with the sub-matrixformed by deleting (in mind only) the row used in step 2 and all rows above this row.

Step 4: Repeat step 2 with the entire matrix until the entire matrix get transformed in reduced row Echelon form.

Calculations:

The given system of equations can be written as: $x\text{+}3y\text{=-}10$ , $\text{-}3x\text{-7}y\text{=22}$ .

The coefficient matrix and the augmented matrix for the given system of equations are $A\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}\right]$ and $\left[A|B\right]\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}|\begin{array}{c}\text{-10}\\ \text{22}\end{array}\right]$ respectively.

Now applying row operations:

  $\left[A|B\right]\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}|\begin{array}{c}\text{-10}\\ \text{22}\end{array}\right]\underrightarrow{R_2\to R_2\text{+}3R_1}\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{0}& \text{2}\end{array}|\begin{array}{c}\text{-10}\\ \text{-}8\end{array}\right]$

  $\underrightarrow{R_2\to \frac{1}{2}{R}_2}\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{0}& \text{1}\end{array}|\begin{array}{c}\text{-10}\\ \text{-}\text{4}\end{array}\right]\underrightarrow{R_1\to R_1\text{-3}{R}_2}\text{=}\left[\begin{array}{cc}\text{1}& 0\\ \text{0}& \text{1}\end{array}|\begin{array}{c}\text{2}\\ \text{-}\text{4}\end{array}\right]\begin{array}{c}\to \\ \to \end{array}\begin{array}{c}x\text{+0}y\text{=}2\\ 0x\text{+}y\text{=-}\text{4}\end{array}\begin{array}{c}\Rightarrow \\ \Rightarrow \end{array}\begin{array}{c}x\text{=}2\\ y\text{=-}\text{4}\end{array}$ .

The solution of system of equations is $x\text{=2}$ , and $y\text{=- 4}$ .

Conclusion: The solution of system of equations by Gauss-Jordan elimination method is $x\text{=2}$ , $y\text{=- 4}$ .

To determine

To solve: The system of equations using inverse of the coefficient matrix.

Answer to Problem 1PRE

The solution of the system of equations using inverse of coefficient matrix is $x\text{=2}$ , $y\text{=- 4}$ .

Explanation of Solution

Given:

The system of equations in part (a)

Method/ Formula used:

The system of equations can be written as $AX\text{=}B$ or $X\text{=}{A}^{\text{-1}}B$ , where A is the coefficient matrix and $A^{\text{-1}}$ its inverse. The inverse of the matrix $A\text{=}\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is $A^{\text{-}\text{1}}\text{=}\frac{1}{\left(ad-bc\right)}\left(\begin{array}{cc}d& \text{-}b\\ \text{-}c& a\end{array}\right)$ .

Calculations:

The given system of equations can be written as: $x\text{+}3y\text{=-}10$ , $\text{-}3x\text{-7}y\text{=22}$ . Also, in coefficient matrix formthis system of equations is written as $AX\text{=}B\Rightarrow \left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\text{=}\left[\begin{array}{c}\text{-10}\\ \text{22}\end{array}\right]$

  $\Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]\text{=}{\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}\right]}^{\text{-1}}\cdot \left[\begin{array}{c}\text{-10}\\ \text{22}\end{array}\right]$ .

The inverse of A using formula is $A^{\text{-}\text{1}}\text{=}{\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}\right]}^{\text{-1}}\text{=}\frac{\text{1}}{\left[\left(\text{1×-7}\right)\text{-}\left(\text{3×-3}\right)\right]}\left[\begin{array}{cc}\text{-7}& \text{-3}\\ \text{3}& \text{1}\end{array}\right]\text{=}\frac{\text{1}}{\text{2}}\left[\begin{array}{cc}\text{-7}& \text{-3}\\ \text{3}& \text{1}\end{array}\right]=\left[\begin{array}{cc}\text{-7/2}& \text{-3/2}\\ \text{3/2}& \text{1/2}\end{array}\right]$ .

Therefore, the solution of the system of equations is $\left[\begin{array}{c}x\\ y\end{array}\right]\text{=}\left[\begin{array}{cc}\text{-7/2}& \text{-3/2}\\ \text{3/2}& \text{1/2}\end{array}\right]\text{×}\left[\begin{array}{c}\text{-10}\\ \text{22}\end{array}\right]\text{=}\left[\begin{array}{c}\left(\frac{\text{-7}}{\text{2}}\right)\text{×-10+}\left(\text{-}\frac{\text{3}}{\text{2}}\right)\text{×22}\\ \frac{\text{3}}{\text{2}}\text{×-10+}\frac{\text{1}}{\text{2}}\text{×22}\end{array}\right]\text{=}\left[\begin{array}{c}\text{35-33}\\ \text{-15+11}\end{array}\right]\text{=}\left[\begin{array}{c}\text{2}\\ \text{- 4}\end{array}\right]$ .

The solution of system of equations is $x\text{=2}$ , and $y\text{=- 4}$ .

Conclusion: The solution of system of equations by inverse of coefficient matrix method is $x\text{=2}$ , $y=-4$ .

To determine

To solve: The system of equations using Cramer’s rule.

Answer to Problem 1PRE

The solution of the system of equations using Cramer’s rule is $x\text{=2}$ , $y\text{=- 4}$ .

Explanation of Solution

Given:

The system of equations in part (a)

Method/ Formula used:

For two variable system of equations $a_{1}x\text{+}{b}_{1}y\text{=}{d}_1$ , $a_{2}x+b_{2}y\text{=}{d}_2$ , the solution using Cramer’s rule is $x\text{=}\frac{D_x}{D},$

  $y\text{=}\frac{D_y}{D}$ , where $D$ , $D_x$ , and $D_y$ are the determinants defined as:

  $D\text{=}\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$ , $D_x\text{=}\left[\begin{array}{cc}{d}_1& b_1\\ d_2& b_2\end{array}\right]$ and $D_y=\left[\begin{array}{cc}{a}_1& d_1\\ a_2& d_2\end{array}\right]$ .

Calculations:

The given system of equations can be written as: $x\text{+}3y\text{=-}10$ , $\text{-}3x\text{-7}y\text{=22}$ .

For the system of equations, the determinant of coefficient matrix $D\text{=}\left[\begin{array}{cc}\text{1}& \text{3}\\ \text{-3}& \text{-7}\end{array}\right]\text{=-7+9=2}$ , and other determinants are $D_x\text{=}\left[\begin{array}{cc}\text{-10}& \text{3}\\ \text{22}& \text{-7}\end{array}\right]=70-66=4$ and $D_y\text{=}\left[\begin{array}{cc}\text{1}& \text{-10}\\ \text{-3}& \text{22}\end{array}\right]\text{=22-30=-8}$ .

Therefore, values of x and y are $x\text{=}\frac{D_x}{D}\text{=}\frac{\text{4}}{\text{2}}\text{=2}$ , $y\text{=}\frac{D_y}{D}\text{=}\frac{\text{-8}}{\text{2}}\text{=- 4}$

The solution of system of equations is $x\text{=2}$ , and $y\text{=-4}$ .

Conclusion: The solution of the system of equations usingCramer’s is $x\text{=2}$ , and $y\text{=- 4}$ .