Matrices trouble, making a,b,c no sol, a unique, and infin solutions

[mr_coffee]

Hello everyone. I'm having troubles understanding if I'm doing this right.
I have the matrix
[tex]\left( {\begin{array}{*{20}c} 2 & {-3} & {-3} & a \\ {-1} & 1 & 2 & b \\ 1 & {-3} & 0 & c \\ \end{array} } \right)[/tex]

I row reduced it to:
[tex]\left( {\begin{array}{*{20}c} 2 & {-3} & {-3} & a \\ 0 & {-1} & 1 & 2b+a \\ 0 & {0} & 0 & {-3b-2a+c} \\ \end{array} } \right)[/tex]

I'm suppose to find, In each case find if possible conditions on the numbers a, b, and c that the given syhstem has no solution, a unique solution, or infintnety many s9lutions. So would i let b = 0, a = 0 and c equal 0 to make it have an infinit many solutions. and then let a, b and c be any number so the expression: -3b-2a+c wil not equal 0, so 0 = 4, no solution. How would i find a unique solution? if the last row is already 0 = -3b-2a+c? thanks!

 

[TD]

In order to have at least one solution, the last row cannot by all 0 and a non-zero final element. So in order to have solutions, you have to let [itex]-3b-2a+c = 0[/itex].

Can you take it from here?

 

[quicknote]

Hello, it seems like you are struggling with solving systems of equations using matrices. Don't worry, it can be a bit confusing at first. Let me explain a few things to help you understand better.

First, let's talk about what a matrix is. A matrix is a rectangular array of numbers or symbols arranged in rows and columns. It is used to represent equations or systems of equations in a more organized way.

Now, let's look at the matrix you provided. This is a 3x4 matrix, meaning it has 3 rows and 4 columns. Each row represents an equation and each column represents a variable. The last column is called the "augmented column" and it includes the constants in the equations.

When we perform row operations on a matrix, we are essentially manipulating the equations to solve for the variables. The goal is to get the matrix into a specific form called "row echelon form" or "reduced row echelon form." In this form, the matrix will tell us if the system of equations has no solution, a unique solution, or infinitely many solutions.

In your example, you have already reduced the matrix to row echelon form. Let's look at the last row: 0 = -3b-2a+c. This means that -3b-2a+c must equal 0 for the system to have a solution. Therefore, the condition for a unique solution is -3b-2a+c = 0. This means that there is only one combination of values for a, b, and c that will satisfy all three equations.

For infinitely many solutions, we need to have a case where the last row is all zeros, like this: 0 = 0. This means that the equation is always true, regardless of the values of a, b, and c. In your example, if we let b = 0, a = 0, and c = 0, the last row will be all zeros. This means that any values for a, b, and c will satisfy the system of equations.

Finally, for no solution, we need to have a case where the last row is in the form of 0 = k, where k is a non-zero constant. This means that the equation is always false, regardless of the values of a, b, and c. In your example, if we let a = 1, b = 2, and c = 3