# Math Help - Augmented Matrix with constants

1. ## Augmented Matrix with constants

How would you solve a matrix with constants inside of it (such as h,k). An example I was given was
-7x + 3y + 4z = 3
-4x - 4y - 4z = -5
29x -y + h = k
and was told to find "This system has a unique solution whenever "

What I did was bring the matrix to a form as close as I could to row echelon form, but I have no idea what the conditions are to find what k isnt equal to for a unique solution to be found.

The near row echelon form I found was

x + 11y + 12z = 13
0x + 1y + 11/10z = 47/40
0x + 0y + (9/5 + h)z = k + 67/20

and from there I'm at a complete loss as to how to solve the problem

Edit: It'd also be appreciated if someone could let me know the conditions for a similar problem, but with infinite solutions and with no solutions

Edit 2: The question was actually when , not k, sorry for the inconvenience!

Thanks!

2. ## Re: Augmented Matrix with constants

Originally Posted by virus1019
How would you solve a matrix with constants inside of it (such as h,k). An example I was given was
-7x + 3y + 4z = 3
-4x - 4y - 4z = -5
29x -y + h = k
and was told to find "This system has a unique solution whenever "

What I did was bring the matrix to a form as close as I could to row echelon form, but I have no idea what the conditions are to find what k isnt equal to for a unique solution to be found.

The near row echelon form I found was

x + 11y + 12z = 13
0x + 1y + 11/10z = 47/40
0x + 0y + (9/5 + h)z = k + 67/20

and from there I'm at a complete loss as to how to solve the problem
Whenever $\displaystyle k \neq$ what?

Have you tried evaluating the determinant of this matrix and setting it equal to 0? That will tell you what h can't be at least...

3. ## Re: Augmented Matrix with constants

oh sorry, i posted the wrong thing :S it was supposed to be h, not k

I don't think I've learned determinants in class yet though, so I'm not quite sure what you mean

4. ## Re: Augmented Matrix with constants

Have you learnt about the identity matrix or the inverse matrix?

5. ## Re: Augmented Matrix with constants

yes, ive learnt the identity matrix and the inverse matrix

6. ## Re: Augmented Matrix with constants

Originally Posted by virus1019
yes, ive learnt the identity matrix and the inverse matrix
OK, well if you think about a 2x2 matrix $\displaystyle \mathbf{A} = \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]$, the inverse matrix is $\displaystyle \mathbf{A}^{-1} = \frac{1}{ad - bc}\left[\begin{matrix}\phantom{-}d & -b \\ -c & \phantom{-}a\end{matrix}\right]$. Clearly, this inverse matrix only exists if $\displaystyle ad-bc \neq 0$, so it determines if there is a solution to the matrix equation $\displaystyle \mathbf{A}\mathbf{x} = \mathbf{b}$. This is why that quantity is called the DETERMINANT, and is given the notation $\displaystyle |\mathbf{A}|$, or $\displaystyle \left|\begin{matrix}a&b\\c&d\end{matrix}\right|$.

Square matrices of higher dimension can also have inverses, but again, only as long as its determinant is nonzero. Higher determinants can be evaluated by reducing it to smaller determinants, e.g. for a 3x3 matrix

$\displaystyle \left|\begin{matrix}a&b&c\\d&e&f\\g&h&j\end{matrix }\right| = a\left|\begin{matrix}e&f\\h&j\end{matrix}\right| - b\left|\begin{matrix}d&f\\g&j\end{matrix}\right| + c\left|\begin{matrix}d&e\\g&h\end{matrix}\right|$

The easy way to reduce a large determinant to the smaller determinants is to choose a row (like the top row), start with the first element, visualise the row and column that it lies in being erased, and the remaining elements becoming the smaller determinant. Move to the next element in the top row, mentally erase the row and column it lies in, and the remaining elements become the smaller determinant, etc. As you move along the row, the signs alternate (+,-,+,-,...)

In summary, a matrix equation only has solution when the inverse matrix exists, and the inverse matrix exists only when its determinant is nonzero.

So to answer your original question, if you evaluate the determinant and set it equal to 0, you'll find the value that h can not take for the matrix equation to have solution.

7. ## Re: Augmented Matrix with constants

But then what happens to the answers (to the individual equations) in the matrices? Do they just disappear while finding the determinant? Or do they play a role in finding the determinant as well?

8. ## Re: Augmented Matrix with constants

Originally Posted by virus1019
But then what happens to the answers (to the individual equations) in the matrices? Do they just disappear while finding the determinant? Or do they play a role in finding the determinant as well?
The determinant tells you IF there will be a solution. The determinant is also a part of the inverse matrix that you can premultiply both sides by to solve the matrix equation.

I suggest you Google for some information about determinants.

9. ## Re: Augmented Matrix with constants

You reduced to
$x+ 11y+ 12z= 13$
$0x+ y+ \frac{11}{10}z= \frac{47}{40}$
$0x+ 0y+ \left(\frac{9}{5}+ h\right)z= k+ \frac{67}{20}$

Look at that last equation. Obviously you solve for z by dividing both sides by $\frac{9}{5}+ h$.

For what value of h are you not allowed to do that?