Question

In Problems 19 to 22, solve each set of equations by the method of finding the inverse of the coefficient matrix. Hint: See Example 3.

$$\begin{gathered} \{\begin{array}{l}x-2y=5 \\ 3x+y=15\\ \end{array} \end{gathered}$$

Short answer

The value of the variables are x=5 , and y=0 .

Step-by-step solution

Definition of Inverse of a Matrix:

The inverse of a matrix M (if it has one) as the matrix ${\mathit{M}}^{\mathbf{1}}$such that $\mathit{M}{\mathit{M}}^{\mathbf{-}\mathbf{1}}$ and $\mathit{M}{\mathit{M}}^{\mathbf{-}\mathbf{1}}$ are both equal to a unit matrix I .

And also, only square matrices whose determinant is non-zero and for which the inverse matrix can be calculated is called an invertible matrix.

Given parameters:

Solve the given equations by the method to find the inverse of the coefficient matrix.

$$\begin{gathered} \left\{\begin{array}{l}x-2y=5 \\ 3x+y=15\\ \end{array}\right. \end{gathered}$$

Compare with the general equation $Mr=k$, where M is the matrix coefficient, r is an unknown vector.

localid="1658984740423" $$\begin{gathered} here,M=\left(1-2 \\ 31\right),K=\left(5 \\ 15\right)andr=\left(x \\ y\right) \end{gathered}$$

Finding inverse of the matrix:

To find inverse of the matrix: Find the cofactors $C_{ij}$ of all elements, write the matrix C whose elements are $C_{ij}$, and transpose it (interchange rows and columns), and divide by $\left(M\right)$.

$$\begin{gathered} C_{11}=1 \\ {C}_{12}=-3 \\ {C}_{21}=2 \\ {C}_{22}=1 \end{gathered}$$

Therefore, cofactors of the matrix , $$\begin{gathered} M,C=\left(1-3 \\ 21\right) \end{gathered}$$.

Transpose of matrix C is $$\begin{gathered} C^T=\left(12 \\ -31\right) \end{gathered}$$.

Find role="math" localid="1658985176008" $det\left(M\right)$.

$$\begin{gathered} det\left(M\right)=\left(1\times 1\right)-3\times \left(-2\right) \\ =7 \end{gathered}$$

Substitute it into the formula.

$$\begin{gathered} M^{-1}=\frac{1}{detM}{C}^T \\ =\frac{1}{7}\left(12 \\ -31\right) \end{gathered}$$

Use $M^{-1}$ to solve the given equation and the solution is given by the column matrix $r=M^{-1}k$.

$$\begin{gathered} \left(x \\ y\right)=\frac{1}{7}\left(12 \\ -31\right)\left(5 \\ 15\right) \\ =\frac{1}{7}\left(35 \\ 0\right) \\ =\left(5 \\ 0\right) \end{gathered}$$

Therefore, x=5 and y=0 .