Chapter 11, Problem 5CRT

a.

To determine

To Write: the Matrix equation of $AX=B$

Answer to Problem 5CRT

  $\begin{array}{l}AX=B\\ \left[\begin{array}{cc}5& -3\\ 6& -4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 0\end{array}\right]\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}5x-3y=5\\ 6x-4y=0\end{array}$

Concept Used:

On converting the given equations to matrix form we can obtain the equation.

Calculation:

  $\begin{array}{l}\left[\begin{array}{cc}5& -3\\ 6& -4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 0\end{array}\right]\\ AX=B\end{array}$

Conclusion:

The required matrix equation is $\begin{array}{l}\left[\begin{array}{cc}5& -3\\ 6& -4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 0\end{array}\right]\\ AX=B\end{array}$

b.

To determine

To Find: $A^{-1}$ as the inverse of coefficient matrix.

Answer to Problem 5CRT

  $A^{-1}=\frac{1}{-2}\left[\begin{array}{cc}-4& 3\\ -6& 5\end{array}\right]$

Explanation of Solution

Given information:

  $\begin{array}{l}5x-3y=5\\ 6x-4y=0\end{array}$

Concept Used:

  $A^{-1}=\frac{adjA}{\left|A\right|}\text{ (or) }\frac{adjA}{\det A}$

Now, adjoint matrix is found by reversing (or) replacing the values in principal diagonal matrix among then also, by changing the sign of other diagonal elements.

Also, determinant of matrix id found by cross multiplying the elements of principal diagonal and other diagonal and subtracting them.

Calculation:

  $adjA=\left[\begin{array}{cc}-4& 3\\ -6& 5\end{array}\right]$

And $\det A=\left(-4\right)\left(5\right)-\left(3\right)\left(-6\right)$

  $\begin{array}{l}=-20+18\\ \Rightarrow \det A\text{ or }\left|A\right|=-2\end{array}$

So, $A^{-1}=\frac{adjA}{\left|A\right|}=\frac{1}{-2}\left[\begin{array}{cc}-4& 3\\ -6& 5\end{array}\right]$

Conclusion:

Hence, $A^{-1}=\frac{adjA}{\left|A\right|}=\frac{1}{-2}\left[\begin{array}{cc}-4& 3\\ -6& 5\end{array}\right]$ is the inverse of coefficient matrix.

c.

To determine

To Solve: Matrix equation by multiplying $A^{-1}$ on both sides.

Answer to Problem 5CRT

  $x=10$

  $y=15$

Explanation of Solution

Given information:

  $\begin{array}{l}5x-3y=5\\ 6x-4y=0\end{array}$

Concept Used:

By matrix inversion method

  $X=A^{-1}B$

Calculation:

Let find solution of given equations using multiplying $A^{-1}$ on both sides of $AX=B$

  $\begin{array}{l}\Rightarrow X=A^{-1}B\\ X=\frac{1}{-2}\left[\begin{array}{cc}-4& 3\\ -6& 5\end{array}\right]\left[\begin{array}{c}5\\ 0\end{array}\right]\left(\text{from b}\right)\\ \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{-2}\left[\begin{array}{c}-20+0\\ -30+0\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\frac{-1}{2}\left[\begin{array}{c}-20\\ -30\end{array}\right]\\ \Rightarrow \left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}-10\\ -15\end{array}\right]\end{array}$

Conclusion:

So, solutions are $x=10,y=15$

d.

To determine

To Solve: Matrix equation using Cramer’s rule.

Answer to Problem 5CRT

  $\begin{array}{l}x=10\\ y=15\end{array}$

Explanation of Solution

Given information:

  $5x-3y=5\text{ \& }6x-4y=0$

Solve them using crammer’s method

Concept Used:

  $x=\frac{Dx}{D};y=\frac{Dy}{D}$

Calculation:

From b we get $AX=B$

  $\left[\begin{array}{cc}5& -3\\ 6& -4\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}5\\ 0\end{array}\right]$

Now, let $D=\det A\text{ or }\left|A\right|$

Also; $D_x=$ determinant of matrix formed by replacing first column elements of A with constant matrix elements

  $D_y=$ determination of matrix is formed by replacing second column elements of A with constant matrix elements.

  $D=\left|A\right|=-2$

(from b)

  $\begin{array}{l}{D}_x=\left[\begin{array}{cc}5& -3\\ 0& -4\end{array}\right]=5\left(-4\right)-\left(-3\right)\left(0\right)\\ =-20\\ D_x=-20\\ D_y=\left[\begin{array}{cc}5& 5\\ 6& 0\end{array}\right]=5\left(0\right)-\left(5\right)\left(6\right)\\ =-30\\ D_y=-30\end{array}$

Now, solutions; $x=\frac{Dx}{D}=\frac{-20}{-2}=10$

  $y=\frac{Dy}{D}=\frac{-30}{-2}=15$

Conclusion:

Hence, solutions from (b) and (d) are same

i.e. $x=10;y=15$

e.

To determine

To estimate: The value of the function.

Answer to Problem 5CRT

The value of function is $4$

Explanation of Solution

Given information:

Graph of $f(x)$

Calculation:

  $\underset{x\to 5}{{\displaystyle \lim }}f(x)=\underset{x\to 5^{+}}{{\displaystyle \lim }}f(x)=\underset{x\to 5}{{\displaystyle \lim }}f(x)=4$

So, $f(5)=4$

Conclusion:

From graph it can be observed at $x=5$ ,hence the graph is continuous