Question

Question: Use Cramer’s rule to compute the solutions of the systems in Exercises1-6.

5. \(\begin{array}{c}{x_1} + {x_2} = 3\\ - 3{x_1} + 2{x_3} = 0\\{x_2} - 2{x_3} = 2\end{array}\)

Short answer

The solutions of the systems are \({x_1} = \frac{1}{4},{x_2} = \frac{{11}}{4},{x_3} = \frac{3}{8}\).

Step-by-step solution

State the matrices \({A_1}\left( b \right)\) and \({A_2}\left( b \right)\)

For any\(n \times n\)matrix A and any b in \({\mathbb{R}^n}\), let \({A_i}\left( b \right)\) be the matrix obtained from A by replacing the column \(i\)by vector\({\mathop{\rm b}\nolimits} \).

\({A_i}\left( {\mathop{\rm b}\nolimits} \right) = \left( {\begin{array}{*{20}{c}}{{a_1}}& \cdots &{\mathop{\rm b}\nolimits} & \cdots &{{a_n}}\end{array}} \right)\)

The system of equations is equivalent to\(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \), where\({\mathop{\rm A}\nolimits} = \left( {\begin{array}{*{20}{c}}1&1&0\\{ - 3}&0&2\\0&1&{ - 2}\end{array}} \right)\)and\({\mathop{\rm b}\nolimits} = \left( {\begin{array}{*{20}{c}}3\\0\\2\end{array}} \right)\).

Matrices\({A_1}\left( b \right),{A_2}\left( b \right)\), and\({A_3}\left( b \right)\)are shown below:

\({A_1}\left( b \right) = \left( {\begin{array}{*{20}{c}}3&1&0\\0&0&2\\2&1&{ - 2}\end{array}} \right),{A_2}\left( b \right) = \left( {\begin{array}{*{20}{c}}1&3&0\\{ - 3}&0&2\\0&2&{ - 2}\end{array}} \right),{A_3}\left( b \right) = \left( {\begin{array}{*{20}{c}}1&1&3\\{ - 3}&0&0\\0&1&2\end{array}} \right)\)

Compute the determinants of the matrices

The determinant of matrix\(A\)is shown below:

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}1&1&0\\{ - 3}&0&2\\0&1&{ - 2}\end{array}} \right|\\ = 1\left( {0 - 2} \right) - 1\left( {6 - 0} \right) + 0\\ = - 2 - 6 + 0\\ = - 8\end{array}\)

The determinant of matrix\({A_1}\left( b \right)\)is shown below:

\(\begin{array}{c}\det {A_1}\left( b \right) = \left| {\begin{array}{*{20}{c}}3&1&0\\0&0&2\\2&1&{ - 2}\end{array}} \right|\\ = 3\left( {0 - 2} \right) - 1\left( {0 - 4} \right) + 0\\ = - 6 + 4\\ = - 2\end{array}\)

The determinant of matrix\({A_2}\left( b \right)\)is shown below:

\(\begin{array}{c}\det {A_2}\left( b \right) = \left| {\begin{array}{*{20}{c}}1&3&0\\{ - 3}&0&2\\0&2&{ - 2}\end{array}} \right|\\ = 1\left( {0 - 4} \right) - 3\left( {6 - 0} \right) + 0\\ = - 4 - 18\\ = - 22\end{array}\)

The determinant of matrix\({A_3}\left( b \right)\)is shown below:

\(\begin{array}{c}\det {A_3}\left( b \right) = \left| {\begin{array}{*{20}{c}}1&1&3\\{ - 3}&0&0\\0&1&2\end{array}} \right|\\ = 1\left( {0 - 0} \right) - 1\left( { - 6 - 0} \right) + 3\left( { - 3 - 0} \right)\\ = 6 - 9\\ = - 3\end{array}\)

Since \(\det A = - 8\), the system has a unique solution.

Compute the solution of the system

Let\(A\)be aninvertible\(n \times n\) matrix. Based on Cramer’s rule,for any b in \({\mathbb{R}^n}\), the unique solution \({\mathop{\rm x}\nolimits} \) of \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) has entries given by

\({x_i} = \frac{{\det {A_i}\left( b \right)}}{{\det A}},\,\,\,\,i = 1,2,...,n\).

Use Cramer’s rule to compute the solution of the system as shown below:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = \frac{{\det {A_1}\left( b \right)}}{{\det A}}\\ = \frac{1}{4}\\{{\mathop{\rm x}\nolimits} _2} = \frac{{\det {A_2}\left( b \right)}}{{\det A}}\\ = \frac{{11}}{4}\end{array}\)

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

Thus, the solutions of the systems are \({x_1} = \frac{1}{4},{x_2} = \frac{{11}}{4},{x_3} = \frac{3}{8}\).