in Linear Algebra
1,712 views
1 vote
1 vote

Choose a coefficient b that makes this system singular. Then choose a right-hand side g that makes it solvable. Find two solutions in that singular case.

             $2x +by = 16$

            $4x + 8y = g$

in Linear Algebra
1.7k views

1 Answer

4 votes
4 votes
Best answer

$A = \begin{bmatrix}
2 & b\\ 
4 & 8
\end{bmatrix}$

and let the augmented matrix be 

$A|B = \begin{bmatrix}
2 & b & 16\\ 
4 & 8 & g
\end{bmatrix}$


For singularity, $|A| = 0$ but this says that the system of linear equations can have either no solution or infinite solution .

We get $b=4$

Now, since to make it solvable i.e. having infinite solutions, we need to make the last row of augmented matrix = $0$.

Now, this is only possible if we take $g=32$, then last row completely gets eliminated and hence, we have infinite solutions .

selected by
by

3 Comments

Thank you. Few silly doubts:

1) Why have you taken b = 4? so that second column is a multiple of first and hence leading us to get no solution or infinite solutions?

2) Solvable here means infinitely many solutions?

So what I've understood is whether the system has any solution depends on the left hand side of the equations and whether it has no solution or infinitely many solutions depends on the right hand side, Am I correct?
0
0

Taking $b=4$ ensures that the matrix $A$ is singular, i.e $det(A)=0$.


Showing a system of linear equations is not solvable (has no solutions) is, by definition, the same thing as showing that the system of linear equations is "inconsistent".

Here, we need to show solvable (consistent solution) and by luck here we got infinite solutions .


So what I've understood is whether the system has any solution depends on the left hand side of the equations

Only, in case of $det(A) = 0$, we can say either no or infinite in case of non homogenous equations

and

only and only infinite in case of homogenous equations .


An ex :- 

Take the system of linear equations (Non homogenous)

$$\begin{cases}x+3y-z=4\\4x-y+2z=8\\2x-7y+4z=0,\end{cases}$$ 

Here, $det(A) =0$, and solving the complete system we have infinite solutions .

and now 

Take the same system of linear equations

$$\begin{cases}x+3y-z=4\\4x-y+2z=8\\2x-7y+4z=-3,\end{cases}$$ 

Here, also  $det(A) =0$, and solving the complete system we have no solutions .


Note :- I just changed $0$ to $-3$ in last equation RHS .

1
1
Wonderful. Thank you (Y)
1
1

Related questions

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true