Chapter 3.3, Problem 5E

Use Cramer’s rule to compute the solutions of the systems in Exercises 1−6.

5. $\begin{array}{ccccccc}{\text{x}}_1& +& {\text{x}}_2& & & =& 3\\ -\text{3}{\text{x}}_1& & & +& 2{\text{x}}_3& =& 0\\ & & {\text{x}}_2& -& 2{\text{x}}_3& =& 2\end{array}$

To determine

To compute: The solutions of the systems using Cramer’s rule.

Answer to Problem 5E

The solutions of the systems using Cramer’s rule is $x=\left[\begin{array}{c}\frac{1}{4}\\ \frac{11}{4}\\ \frac{3}{8}\end{array}\right]$.

Explanation of Solution

Given:

The equations are,

$\begin{array}{c}{x}_1+x_2=3\\ -3x_1+2x_3=0\\ x_2-2x_3=2\end{array}$

Rule used:

Cramer’s Rule:

Let A be an invertible $n\times n$ matrix. For any b in ${ℝ}^n$, the unique solution x of $Ax=\text{b}$ has entries given by $x_i=\frac{\det A_i\left(b\right)}{\det A}$, $i=1,2,\mathrm{...},n$.

Calculation:

The given system is of the form $Ax=b$ where $A=\left[\begin{array}{ccc}1& 1& 0\\ -3& 0& 2\\ 0& 1& -2\end{array}\right]$ and $b=\left[\begin{array}{c}3\\ 0\\ 2\end{array}\right]$.

Check the matrix $A=\left[\begin{array}{ccc}1& 1& 0\\ -3& 0& 2\\ 0& 1& -2\end{array}\right]$ is invertible or not.

$\begin{array}{c}\det A=1\times \left[\left(0\times -2\right)-\left(1\times 2\right)\right]-1\times \left[\left(-3\times -2\right)-\left(0\times 2\right)\right]+0\\ =-2-6\\ =-8\end{array}$

Here, determinant of the matrix is non-zero, the matrix is invertible.

The matrix $A_i\left(b\right)$ is obtained from A by replacing column i by the vector $b$.

Thus, the matrix $A_1\left(b\right)$ is obtained from A by replacing first column by the vector $b$ and the matrix $A_2\left(b\right)$ is obtained from A by replacing second column by the vector $b$.

That is, $A_1\left(b\right)=\left[\begin{array}{ccc}3& 1& 0\\ 0& 0& 2\\ 2& 1& -2\end{array}\right]$, $A_2\left(b\right)=\left[\begin{array}{ccc}1& 3& 0\\ -3& 0& 2\\ 0& 2& -2\end{array}\right]$ and $A_3\left(b\right)=\left[\begin{array}{ccc}1& 1& 3\\ -3& 0& 0\\ 0& 1& 2\end{array}\right]$.

From Cramer rule, the unique solution x of $Ax=\text{b}$ has entries given by the equation $x_i=\frac{\det A_i\left(b\right)}{\det A}$, $i=1,2,\mathrm{...},n$.

Obtain the determinant of the matrix $A_1\left(b\right)$.

$\begin{array}{c}\det A_1\left(b\right)=3\times \left[\left(0\times -2\right)-\left(1\times 2\right)\right]-1\times \left[\left(0\times -2\right)-\left(2\times 2\right)\right]+0\\ =-6-\left(-4\right)\\ =-2\end{array}$

Obtain the determinant of the matrix $A_2\left(b\right)$.

$\begin{array}{c}\det A_2\left(b\right)=1\times \left[\left(0\times -2\right)-\left(2\times 2\right)\right]-3\times \left[\left(-3\times -2\right)-\left(0\times 2\right)\right]+0\\ =-4-18\\ =-22\end{array}$

Obtain the determinant of the matrix $A_3\left(b\right)$.

$\begin{array}{c}\det A_3\left(b\right)=1\times \left[\left(0\times 2\right)-\left(1\times 0\right)\right]-1\times \left[\left(-3\times 2\right)-0\right]+3\times \left[\left(-3\times 1\right)-0\right]\\ =0-\left(-6\right)+3\left(-3\right)\\ =-3\end{array}$

Thus, $\det A_1\left(b\right)=-2$, $\det A_2\left(b\right)=-22$ and $\det A_2\left(b\right)=-3$.

The first entry of the solution obtained by substituting 1 for i in the equation $x_i=\frac{\det A_i\left(b\right)}{\det A}$.

$\begin{array}{c}{x}_1=\frac{\det A_1\left(b\right)}{\det A}\\ =\frac{-2}{-8}\\ =\frac{1}{4}\end{array}$

Similarly, the second entry of the solution obtained by substituting 2 for i in the equation $x_i=\frac{\det A_i\left(b\right)}{\det A}$.

$\begin{array}{c}{x}_2=\frac{\det A_2\left(b\right)}{\det A}\\ =\frac{-22}{-8}\\ =\frac{11}{4}\end{array}$

Similarly, the third entry of the solution obtained by substituting 3 for i in the equation $x_i=\frac{\det A_i\left(b\right)}{\det A}$.

$\begin{array}{c}{x}_3=\frac{\det A_3\left(b\right)}{\det A}\\ =\frac{-3}{-8}\\ =\frac{3}{8}\end{array}$

Hence, the solutions of the systems using Cramer’s rule is $x=\left[\begin{array}{c}\frac{1}{4}\\ \frac{11}{4}\\ \frac{3}{8}\end{array}\right]$.