Question

Use inverse matrix to solve the linear system. Solve #19

Answer 1

19)

The given system of equations is,

$\begin{gathered} 4x-3y=11 \\ 5x-2y=12 \end{gathered}$

The above system of equations can be written in matrix form as,

$\begin{bmatrix}{4} & {-3} & {} \\ {5} & {-2} & {} \\ {} & {} & \end{bmatrix}\begin{bmatrix}{x} & {} & {} \\ {y} & & {} \\ {} & {} & \end{bmatrix}=\begin{bmatrix}{11} & {} & \\ {12} & {} & \\ {} & {} & \end{bmatrix}\text{ -----(1)}$

Here,

$A=\begin{bmatrix}{4} & {-3} & {} \\ {5} & {-2} & {} \\ {} & {} & \end{bmatrix},\text{ X=}\begin{bmatrix}{x} & {} & {} \\ {y} & & {} \\ {} & {} & \end{bmatrix}\text{ },\text{ B=}\begin{bmatrix}{11} & {} & \\ {12} & {} & \\ {} & {} & \end{bmatrix}$

Therefore, equation (1) can be written as,

$AX=B$

Therefore,

$X=A^{-1}B\text{ -------(2)}$

Now, we need to calculate the inverse of A.

(Note:

Let a 2x2 matrix P is of the form given below.

$P=\begin{bmatrix}{a} & {b} & {} \\ {c} & {d} & {} \\ {} & {} & {}\end{bmatrix}$

The inverse of the matrix P is,

$\begin{gathered} P^{-1}=\frac{1}{|P|}\begin{bmatrix}{d} & {-b} & {} \\ {-c} & {a} & {} \\ {} & {} & {}\end{bmatrix} \\ =\frac{1}{ad-bc}\begin{bmatrix}{d} & {-b} & {} \\ {-c} & {a} & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}$

)

Similar to the inverse matrix of 2x2 matrix P, the inverse matrix of A can be written as,

$\begin{gathered} A^{-1}=\frac{1}{4\times(-2)-(-3)\times5}\begin{bmatrix}{-2} & {3} & {} \\ {-5} & {4} & {} \\ {} & {} & {}\end{bmatrix} \\ =\frac{1}{-8+15}\begin{bmatrix}{-2} & {3} & {} \\ {-5} & {4} & {} \\ {} & {} & {}\end{bmatrix} \\ =\frac{1}{7}\begin{bmatrix}{-2} & {3} & {} \\ {-5} & {4} & {} \\ {} & {} & {}\end{bmatrix} \\ =\begin{bmatrix}{\frac{-2}{7}} & {\frac{3}{7}} & {} \\ {\frac{-5}{7}} & {\frac{4}{7}} & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}$

Now, put the values in equation (2) to find the solution to the system of equations.

$\begin{gathered} X=A^{-1^{}}B \\ \begin{bmatrix}{x} & {} & {} \\ {y} & & {} \\ {} & {} & \end{bmatrix}=\begin{bmatrix}{\frac{-2}{7}} & {\frac{3}{7}} & {} \\ {\frac{-5}{7}} & {\frac{4}{7}} & {} \\ {} & {} & {}\end{bmatrix}\begin{bmatrix}{11} & {} & \\ {12} & {} & \\ {} & {} & \end{bmatrix} \\ =\begin{bmatrix}{\frac{-2}{7}\times11+\frac{3}{7}\times12} & {} & {} \\ {\frac{-5}{7}\times11+\frac{4}{7}\times12} & & {} \\ {} & {} & {}\end{bmatrix} \\ =\begin{bmatrix}{\frac{-22}{7}+\frac{36}{7}} & {} & {} \\ {\frac{-55}{7}+\frac{48}{7}} & & {} \\ {} & {} & {}\end{bmatrix} \\ =\begin{bmatrix}{\frac{-22+36}{7}} & {} & {} \\ {\frac{-55+48}{7}} & & {} \\ {} & {} & {}\end{bmatrix} \\ =\begin{bmatrix}{\frac{14}{7}} & {} & {} \\ {\frac{-7}{7}} & & {} \\ {} & {} & {}\end{bmatrix} \\ \begin{bmatrix}{x} & {} & {} \\ {y} & & {} \\ {} & {} & \end{bmatrix}=\begin{bmatrix}{2} & {} & {} \\ {-1} & & {} \\ {} & {} & {}\end{bmatrix} \end{gathered}$

Therefore, the solution to the system of equations using inverse matrix is x=2 and y=-1.