1. Dec 21, 2007

### jubej

Im going crazy here please help me, i have tried to solve this but i dont know where to start.

1) Solve the following matrix equeations for a,b,c and d.

$$\left[\begin{array}{cc}a-b & b+c \\ 3d+c & 2a-4d\end{array}\right]$$ = $$\left[\begin{array}{cc}8 & 1 \\ 7 & 6\end{array}\right]$$

this is another separate problem:
2)
a) show that if ad-bc $$\neq$$ 0, then the reduced row- echelon form of

$$\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$$ is $$\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]$$

b) Use part a) to show that the system
ax+by = k
cx+dy = l

has exactly one solution when ad-bc $$\neq$$ 0

thnx for any help i have started elementary algebra and everyone its in chapter 5 in the book and I am in the 1 chapter, but dont want to give up. the book of the course its

Elemtary linear algebra by Howard Anton and Chris Rorres. i dont know if its a good book. but i try to understand it, do you have any other suggestion of a great book for people like me....noob :(

2. Dec 21, 2007

### Defennder

Firstly, thanks so much for writing in LaTeX. Too many users, myself included are too lazy at times to post in LaTeX and the resulting mess scares off potential homework helpers in these forums.

Ok, let's look at the first question. Note that it is written in the form of a matrix, but that it may be written equivalently as a system of 4 equations (write out all the components of the matrix of the LHS and equate it with the RHS):

a - b = 8
b + c = 1
3d + c = 7
2a - 4d = 6

Now do you know how to solve this system of equations for values of a,b,c,d?

For (2), firstly, ask yourself what is ad-bc ? I don't mean the numerical value of ad-bc for any 2x2 matrix, but what mathematical interpretation is given to ad-bc for 2x2 matrices? Once you figured out, ask yourself next what does it mean for that mathematical interpretation of ad - bc to satisfy ≠ 0 ?

b) follows from a). Once you realise what ad-bc means for 2x2 matrices, and also if that ≠0, then you would realise (if you remember the implication for a system of 2 equations and 2 unknowns to have 1 solution exactly and what entails it to possess such a property) the answer.

3. Dec 21, 2007

### allanworks

im nogod at math i dont even now algebra but i want to give this a try to a-b=8 b+c=1 that means 8-0=8 and 0+1=1 right

4. Dec 23, 2007

### jubej

hello thnx for the reply Defennder. but i still get it wrong and i dont know why.

1) $$\left[\begin{array}{cc}a-b & b+c \\ 3d+c & 2a-4d\end{array}\right]$$
I treat it like you suggested
a-b=8
b+c=1
3d+c=7
2a+4d=6

and then i did something like this:

$$\left[\begin{array}{cc}1,-1,0,0,8 \\ 0, 1,1,0,1\\0,0,1,3,7\\2,0,0,4,6\end{array}\right]$$

and then solve it like regular matrix like the first step is multiply the first row by -2 and ad it to the fourth row. and etc but then after i get to the reduced row -echelon form i get that a=3
b=-5
c=6
and d= 1/3. but in the book the answer is a=5, b=-3, c=4 and d= 1

so where do i do wrong?

and the second i dont really understand what ad-bc means, i see that its the diagonal of the system minus the diagonal of the second colum, but what does it really mean?

by the why i try to write in LaTeX to write the formulas but its quite hard, like who to get the augmented matrise to stay in the same colum? as you see the first row and the rest differ from eachother.

Last edited: Dec 23, 2007
5. Dec 23, 2007

### HallsofIvy

Staff Emeritus
Those are two equations in 3 unknowns. a= 8, b= 0, c= 1 is one possible solution but there are infinitely more. You have not used all the equations: a-b+ 8, b+ c= 1, 3d+ c= 7, and 2a-4c= 6, four equations is four unknowns which is simple.

6. Dec 24, 2007

### rohanprabhu

When using determinants to solve linear equations, you equate a side of variables to a constant and solve the matrix equation.

The matrix equation we are looking at is:

$$-\left\begin{bmatrix}1 & -1 & 0 & 0\\ 0 & 1 & 1 & 0\\0 & 0 & 1 & 3\\2 & 0 & 0 & 4\end{array}\right]\left[\begin{array}{cc}a\\ b\\ c\\ d\end{array}\right] = \left[\begin{array}{cc}8\\1\\7\\6\end{array}\right]$$

which gives:

$$-\left[\begin{array}{cc}a\\b\\c\\d\end{array}\right] = \left\begin{bmatrix}0.4 & 0.4 & -0.4 & 0.3\\-0.6 & 0.4 & -0.4 & 0.3\\0.6 & 0.6 & 0.4 & -0.3\\-0.2 & -0.2 & 0.2 & 0.1\end{array}\right]\left[\begin{array}{cc}8\\1\\7\\6\end{array}\right]$$

solving it, we get:

$$-\left[\begin{array}{cc}a\\b\\c\\d\end{array}\right] = \left[\begin{array}{cc}2.6\\ -5.4\\ 6.4\\ 0.2\end{array}\right]$$

which essentially gives:

$$a = -2.6 ; b = 5.4 ; c = -6.4 ; d = -0.2$$

7. Dec 25, 2007

### Gokul43201

Staff Emeritus
You missed a minus sign before 4.

Last edited: Dec 25, 2007
8. Dec 25, 2007

### Defennder

I trust the others have explained the first part clearly enough. For the 2nd part, have you learnt anything about determinants yet? If you have, then you should realise that ad-bc is the determinant of a 2x2 matrix. Now ask yourself what it means for det(2x2 matrix) ≠ 0. This is essentially the answer to 2a).

For 2b), you need to know what does the fact that a system of 2 linearly independent equations with 2 unknowns have exactly one solution entail. In other words, for a generalised theorem, if there's a system of n linearly independent equations with n unknowns, then that system of linear equations would have exactly one solution for all the unknowns. Remember this?

Now when you have that in mind, consider what it means if the system of n linear equations with n unknowns is to be represented in the form of a nxn matrix. If there is only one unique solution for the system of linear equations (in your case for 2 linear equations), then the nxn matrix which represents the the system would have one special property. Can you figure out what it is? And that is essentially the answer to 2b)