Question

Use an inverse matrix to solve (if possible) the system of linear equations. $$ \left\\{\begin{array}{r} -\frac{1}{4} x+\frac{3}{8} y=-2 \\ \frac{3}{2} x+\frac{3}{4} y=-12 \end{array}\right. $$

Short answer

The solution to the system of equation is \(x = 3.5\) and \(y = 5\).

Step-by-step solution

Rewrite the system in matrix form

The given system of equations can be written in the form \(Ax = b\), where \(A\) is the coefficient matrix, \(x\) is the vector of unknowns, and \(b\) is the constant vector. So we get, \(A = \left[\begin{array}{cc}-\frac{1}{4} & \frac{3}{8} \\ \frac{3}{2} & \frac{3}{4}\end{array}\right]\), \(x = \left[\begin{array}{c}x \\ y\end{array}\right]\), and \(b = \left[\begin{array}{c}-2 \\ -12\end{array}\right]\).

Calculate the inverse of the matrix A

The inverse of matrix \(A\) is computed using the formula, \[A^{-1} = \frac{1}{ad-bc} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]\], substituting \(a=-\frac{1}{4}\), \(b=\frac{3}{8}\), \(c=\frac{3}{2}\), and \(d= \frac{3}{4}\), we get \(A^{-1} = \left[\begin{array}{cc}-\frac{3}{4} & -1 \\ -2 & -1\end{array}\right]\).

Multiply the inverse of A with b

Now, to find the solution to the system of equations, multiply the inverse of \(A\) that we found in Step 2 with vector \(b\). That is, \(x = A^{-1}b\). So we get \(x = \left[\begin{array}{c}x \\ y\end{array}\right] = \left[\begin{array}{cc}-\frac{3}{4} & -1 \\ -2 & -1\end{array}\right] \cdot \left[\begin{array}{c}-2 \\ -12\end{array}\right] = \left[\begin{array}{c}\frac{14}{4} \\ \frac{20}{4}\end{array}\right] = \left[\begin{array}{c}3.5 \\ 5\end{array}\right] \).