Chapter 10, Problem 10T

(a)

To determine

The matrix equation equivalent to the system of equation $\begin{array}{c}4x-3y=10\\ 3x-2y=30\end{array}$.

Answer to Problem 10T

The matrix equation is $\left[\begin{array}{cc}4& -3\\ 3& -2\end{array}\right]\times \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}10\\ 30\end{array}\right]$.

Explanation of Solution

Given:

The system of equations is $\begin{array}{c}4x-3y=10\\ 3x-2y=30\end{array}$.

The equivalent matrix equation is,

$AX=B$

Where, $A$ is coefficient matrix, $X$ is a variable matrix and $B$ is a known matrix.

Calculation:

Write the system of equation in equivalent matrix form.

$\left[\begin{array}{cc}4& -3\\ 3& -2\end{array}\right]\times \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}10\\ 30\end{array}\right]$

Thus, the matrix equation is $\left[\begin{array}{cc}4& -3\\ 3& -2\end{array}\right]\times \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}10\\ 30\end{array}\right]$.

(b)

To determine

The inverse of the coefficient matrix and solve the system.

Answer to Problem 10T

The inverse of matrix $A$ is $\left[\begin{array}{cc}-2& 3\\ -3& 4\end{array}\right]$ and solution of the system of equation is $\left(70,90\right)$.

Explanation of Solution

Given:

The system of equations is $\begin{array}{c}4x-3y=10\\ 3x-2y=30\end{array}$.

Calculation:

The coefficient matrix is $\left[\begin{array}{cc}4& -3\\ 3& -2\end{array}\right]$.

The matrix $2\times 2$ is $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

Compare the matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ with $A=\left[\begin{array}{cc}4& -3\\ 3& -2\end{array}\right]$, the value of a is 4, b is $-3$, c is 3 and d is $-2$.

Formula to calculate the inverse of $2\times 2$ matrix is,

$A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$

Substitute $4$ for $a$, $-3$ for $b$, $3$ for $c$ and $-2$ for $d$ in above expression to find $A^{-1}$.

$\begin{array}{c}{A}^{-1}=\frac{1}{-2\times 4+3\times 3}\left[\begin{array}{cc}-2& 3\\ -3& 4\end{array}\right]\\ =\left[\begin{array}{cc}-2& 3\\ -3& 4\end{array}\right]\end{array}$

Thus, the inverse of matrix $A$ is $\left[\begin{array}{cc}-2& 3\\ -3& 4\end{array}\right]$.

The equivalent matrix equation is,

$\begin{array}{c}AX=B\\ X=A^{-1}B\end{array}$ (1)

Substitute $\left[\begin{array}{c}x\\ y\end{array}\right]$ for $X$, $\left[\begin{array}{cc}-2& 3\\ -3& 4\end{array}\right]$ for $A^{-1}$ and $\left[\begin{array}{c}10\\ 30\end{array}\right]$ for $B$ in above equation.

$\begin{array}{c}\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{cc}-2& 3\\ -3& 4\end{array}\right]\times \left[\begin{array}{c}10\\ 30\end{array}\right]\\ =\left[\begin{array}{c}-20+90\\ -30+120\end{array}\right]\\ =\left[\begin{array}{c}70\\ 90\end{array}\right]\end{array}$

Thus, the solution of the given system of equations is $\left(70,90\right)$.