91Ó°ÊÓ

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\({\mathop{\rm A}\nolimits} = \left( {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 4}&6\end{array}} \right)\)and\({\mathop{\rm b}\nolimits} = \left( {\begin{array}{*{20}{c}}3\\{ - 5}\end{array}} \right)\).

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

\({A_1}\left( b \right) = \left( {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 5}&6\end{array}} \right),{A_2}\left( b \right) = \left( {\begin{array}{*{20}{c}}3&3\\{ - 4}&{ - 5}\end{array}} \right)\)

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

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 4}&6\end{array}} \right|\\ = 18 - 8\\ = 10\end{array}\)

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

\(\begin{array}{c}\det {A_1}\left( b \right) = \left| {\begin{array}{*{20}{c}}3&{ - 2}\\{ - 5}&6\end{array}} \right|\\ = 18 - 10\\ = 8\end{array}\)

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

\(\begin{array}{c}\det {A_2}\left( b \right) = \left| {\begin{array}{*{20}{c}}3&3\\{ - 4}&{ - 5}\end{array}} \right|\\ = - 15 + 12\\ = - 3\end{array}\)

Since \(\det A = 10\), the system has a unique solution.

Let\(A\)be aninvertible\(n \times n\) matrix. Based on Cramer’s rule,for any b in\({\mathbb{R}^n}\), theunique 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:

\(\begin{array}{c}{{\mathop{\rm x}\nolimits} _1} = \frac{{\det {A_1}\left( b \right)}}{{\det A}}\\ = \frac{4}{5}\\{{\mathop{\rm x}\nolimits} _2} = \frac{{\det {A_2}\left( b \right)}}{{\det A}}\\ = - \frac{3}{{10}}\end{array}\)

Thus, the solutions of the systems are \({x_1} = \frac{4}{5},{x_2} = - \frac{3}{{10}}\).