Basic Linear Algebra Question

[burt]

I was given this question, and I'm getting stuck. It's a pretty basic question, I think.


My understanding is a pivot is the first number in that row. So, if it has a pivot in each column, then the matrix looks like an upside down staircase. I don't know what must be true for the system to have a unique solution.
Can you help?

 

[R.M.]

What does it mean to have a pivot in every row?
What does it mean to have a pivot in every column?
The answer to your question can be found by answering these 2 questions.

 

[burt]

What does it mean to have a pivot in every row?
What does it mean to have a pivot in every column?
The answer to your question can be found by answering these 2 questions.

If it has a pivot in each row, and there are four equations in four variables, that means it has a pivot in each row and column. This means it is in row-echelon form. But, I still don't see the answer to the question.

 

[R.M.]

Let's say you had a 3x5 matrix, with a pivot in each of the 3 rows. How many solutions?
Now let's say you had 5x3 matrix, with a pivot in each column. How many solutions?
Now let's use your 4x4 example. Each row and each column has a pivot. How many solutions?

 

[burt]

Let's say you had a 3x5 matrix, with a pivot in each of the 3 rows. How many solutions?
Now let's say you had 5x3 matrix, with a pivot in each column. How many solutions?
Now let's use your 4x4 example. Each row and each column has a pivot. How many solutions?

I'm not understanding. Thank you for helping me go through this.

 

[R.M.]

[MATH] \left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 5 & 0 & 0\\ 0 & 0 & 6 & -3 & 0 \\ \end{matrix}\right] [/MATH]Every row has a pivot. Does this matrix have a solution? Yes, it has at least 1. Relationship between # of rows & columns: ___________.


[MATH] \left[\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \\ 0 & 0 & 0\\ 0 & 0 & 0 \end{matrix}\right] [/MATH]Every column has a pivot, so there is either one solution or none (which does this have?). Relationship between # of rows & columns: ___________.


[MATH] \left[\begin{matrix} 1 & 0 & 0 & 0 \\ 0 & 2 & 5 & 0 \\ 0 & 0 & 6 & 0 \\ 0 & 0 & 0 & -1 \end{matrix}\right] [/MATH]Every row has a pivot (at least 1 solution) AND every column has a pivot (either 1 solution or none). Conclusion?
Relationship between # of rows & columns: ___________.

 

[HallsofIvy]

In a square matrix, in particular a 4 by 4, that is "upper triangular" the equation is of the form
\(\displaystyle \begin{bmatrix}a & b & c & d \\ 0 & e & f & g \\ 0 & 0 & h & i \\ 0 & 0 & 0 & j\end{bmatrix} \begin{bmatrix}t \\ x \\ y \\ z\end{bmatrix}\)\(\displaystyle = \begin{bmatrix}at+ bx+ cy+ dz \\ ex+ fy+ gz \\ hy+ jz \\ jz\end{bmatrix}=\)\(\displaystyle \begin{bmatrix}b_1 \\ b_2 \\ b_3\\ b_4 \end{bmatrix}\).

That is equivalent to the four equations \(\displaystyle at+ bx+ cy+ dz= b_1\), \(\displaystyle ex+ fy+ gz= b_2\), \(\displaystyle hy+ jz= b_3\), and \(\displaystyle jz= b_4\). Do you see how easy those are to solve? The last one just has the single unknown, z, and immediately gives \(\displaystyle z= b_4/z\). Now you can replace z in \(\displaystyle hy+ jz= b_3\) by that and get an equation in the single unknown, y. Solve that, and replace y and z \(\displaystyle ex+ fy+ gz= b_2\) to get an equation in x only, then use that solution to get an equation in t only. (This is called "back substitution".)