Chapter 11.4, Problem 46AYU

To determine

To solve: System of equation using matrix inverse.

Answer to Problem 46AYU

Solution:

  $x=-7$ and $y=-28$

Explanation of Solution

Given:

  $-4x+y=0$

  $6x-2y=14$

Calculation:

The original system of equations can be written compactly as a matrix equation.

  $AX=B$

Where,

  $A=\left[\begin{array}{cc}-4& 1\\ 6& -2\end{array}\right],X=\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]$ and $B=\left[\begin{array}{c}0\\ 14\end{array}\right]$

To solve the for $x$ and $y$ , the solution of $X$ has to be found as given below:

  $AX=B$

Multiply both sides by $A^{-1}$

  $A^{-1}(AX)=A^{-1}B$

By association property of matrix multiplication, the above equation can be rewritten as,

  $\left(A^{-1}A\right)X=A^{-1}B$

By the definition of inverse matrix $\left(A^{-1}A\right)=I_n$ .

  $I_{n}X=A^{-1}B$

By the property of Identity matrix $,I_{n}X=X$ .

Hence,

  $X=A^{-1}B$

i.e. $\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]=A^{-1}B$

Step- $1_j$ Construction of $\left[A\mid I_n\right]$ .

  

Step- 2: Transform the matrix $\left[A\mid I_n\right]$ into reduced row echelon form.

  

Step-3 : Separate the matrix on the right of the vertical bar which is the inverse of $A$ .

  $A^{-1}=\left[\begin{array}{ll}-1\hfill & -\frac{1}{2}\hfill \\ -3\hfill & -2\hfill \end{array}\right]$

Solving $x$ and $y$ ,

  $\left[\begin{array}{l}x\hfill \\ y\hfill \end{array}\right]=A^{-1}B$

  $=\left[\begin{array}{ll}-1\hfill & -\frac{1}{2}\hfill \\ -3\hfill & -2\hfill \end{array}\right]\left[\begin{array}{l}0\hfill \\ 14\hfill \end{array}\right]$

  $=\left[\begin{array}{c}0-7\\ 0-28\end{array}\right]$

  $=\left[\begin{array}{c}-7\\ -28\end{array}\right]$

The solution for the given system of equations is $x=-7$ and $y=-28$ .