Use Cramer’s rule to solve the system of linear equations
\begin{align*}
3x_1-2x_2&=5\\
7x_1+4x_2&=-1.
\end{align*}
 

Solution.

Let
\[A=[A_1, A_2]=\begin{bmatrix}
3 & -2\\
7& 4
\end{bmatrix},\] be the coefficient matrix of the system, where $A_1, A_2$ are column vectors of $A$.
Let $\mathbf{b}=\begin{bmatrix}
5 \\
-1
\end{bmatrix}$ be the constant term vector. Then the system can be written as
\[A\mathbf{x}=\mathbf{b},\] where $\mathbf{x}=\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}$.

We form
\[B_1=[\mathbf{b}, A_2]=\begin{bmatrix}
5 & -2\\
-1& 4
\end{bmatrix}\] and
\[B_2=[A_1, \mathbf{b}]=\begin{bmatrix}
3 & 5\\
7& -1
\end{bmatrix}.\]

Then Cramer’s rule gives the formula for solutions
\[x_1=\frac{\det(B_1)}{\det(A)} \text{ and } x_2=\frac{\det(B_2)}{\det(A)}. \tag{*}\] Thus, it remains to compute the determinants.
We have
\begin{align*}
\det(A)=\begin{vmatrix}
3 & -2\\
7& 4
\end{vmatrix}=3\cdot 4 -(-2)\cdot 7 =26.
\end{align*}
Similarly, a calculation shows that
\[\det(B_1)=18 \text{ and } \det(B_2)=-38.\]

Therefore by Cramer’s rule (*), we obtain
\[x_1=\frac{18}{26}=\frac{9}{13} \text{ and } x_2=\frac{-38}{26}=-\frac{19}{13}.\] Click here if solved 63

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