Calculating Determinants: Using Laplace Expansion or Echelon Form?

  • Context: MHB 
  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    Determinants
Click For Summary

Discussion Overview

The discussion revolves around calculating the determinants of two specific matrices using different methods: the Laplace expansion theorem and transforming the matrices into echelon form. Participants explore the implications of each method and the rules governing determinant calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest that both Laplace expansion and echelon form can be used to calculate determinants.
  • It is noted that when using echelon form, certain rules apply: adding a multiple of a row to another does not change the determinant, while swapping rows changes the sign of the determinant.
  • One participant proposes starting with echelon form and then applying Laplace expansion when it becomes easier.
  • There is a discussion about the correctness of specific determinant calculations for matrix $b$, with some participants confirming the calculations presented.
  • Concerns are raised about the accuracy of intermediate steps in the determinant calculation for matrix $a$, leading to a re-evaluation of those steps.

Areas of Agreement / Disagreement

Participants generally agree that both methods can be used, but there is no consensus on which method is superior or more efficient for the specific matrices discussed. Some participants express uncertainty about specific calculations, leading to further clarification and debate.

Contextual Notes

Participants highlight the importance of following the rules for determinants when transforming matrices, but some steps in the calculations remain unresolved or are questioned, indicating potential errors or misunderstandings in the process.

mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

I want to calculate the determinants of the matrices $a=\begin{pmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{pmatrix}$ and $b=\begin{pmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 3 & 4 & 1 & 9 & -1 & 2 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{pmatrix}$

For that do we use the Laplace expansion theorem or can we transform these matrices firstly in echelon form and calculate the determinant then? But is the determinant of the initial matrix equal to the determinant of the matrix in echelon form? (Wondering)
 
Physics news on Phys.org
Hey mathmari!

We can do both. (Nod)

If we choose the echelon form, there are a couple of special rules for determinants though.
1. We can add a multiple of a row to another.
2. If we swap 2 rows, then the sign of the determinant swaps as well.
3. If we multiply a row by a factor, then the determinant is multiplied by that factor as well.
4. We can do the same with columns instead of rows.
(Thinking)
 
Klaas van Aarsen said:
We can do both. (Nod)

If we choose the echelon form, there are a couple of special rules for determinants though.
1. We can add a multiple of a row to another.
2. If we swap 2 rows, then the sign of the determinant swaps as well.
3. If we multiply a row by a factor, then the determinant is multiplied by that factor as well.
4. We can do the same with columns instead of rows.
(Thinking)

Ok! Which of the two ways do you suggest to use in this case? (Wondering)
 
mathmari said:
Ok! Which of the two ways do you suggest to use in this case? (Wondering)

A combination! (Happy)

Start with working towards echelon form or some such until Laplace expansion becomes easy enough.

If you reach echelon form, the Laplace expansion is just the product of the elements on the diagonal. (Thinking)
 
Klaas van Aarsen said:
If we choose the echelon form, there are a couple of special rules for determinants though.
1. We can add a multiple of a row to another.

How does the determinant change in this case? Do we multiply also the determinant by this factor? Or doesn't the determinant change by using this rule? (Wondering)
 
mathmari said:
How does the determinant change in this case? Do we multiply also the determinant by this factor? Or doesn't the determinant change by using this rule?

The determinant does not change if we add a multiple of a row to a different row. (Emo)
 
Klaas van Aarsen said:
The determinant does not change if we add a multiple of a row to a different row. (Emo)

Ok! So, as for the matrix $b$ we have the following:
\begin{align*}\det b=&\begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 3 & 4 & 1 & 9 & -1 & 2 \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix} \ \overset{R_2:R_2-\frac{3}{2}\cdot R_1}{=} \ \begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 0 & \frac{5}{2} & 4 & \frac{15}{2} & -\frac{23}{2} & -\frac{5}{2} \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 1 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix} \\ & \ \overset{R_4:R_4-R_3}{=} \ \begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 0 & \frac{5}{2} & 4 & \frac{15}{2} & -\frac{23}{2} & -\frac{5}{2} \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 1 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix} \ \overset{R_5:R_5-R_4}{=} \ \begin{vmatrix}2 & 1 & -2 & 1 & 7 & 3 \\ 0 & \frac{5}{2} & 4 & \frac{15}{2} & -\frac{23}{2} & -\frac{5}{2} \\ 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 6 \\ 0 & 0 & 0 & 0 & 1 & -7 \\ 0 & 0 & 0 & 0 & 0 & 3\end{vmatrix} \\ & =2\cdot \frac{5}{2}\cdot 1\cdot 1\cdot 1\cdot 3=15
\end{align*}
Right? (Wondering)
 
Yep. (Nod)
 
Klaas van Aarsen said:
Yep. (Nod)

And as for matrix $a$ we have:
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}\\ & \ \overset{R_3:R_3-3\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & -2 & -4 & -11 & -13 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_4:R_4-4\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \\ & \ \overset{R_5:R_5-5\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_3:R_3-2\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \\ & \ \overset{R_4:R_4-3\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_5:R_5-9\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & 0 & 5 & 10 & 15\end{vmatrix} \\ & \ \overset{R_5:R_5+\frac{5}{2}\cdot R_4}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & 0 & 0 & -\frac{5}{2} & \frac{5}{2}\end{vmatrix} \ \overset{R_5:R_5-\frac{1}{2}\cdot R_3}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & 0 & 0 & 0 & 5\end{vmatrix} \\ & \ \overset{R_3\leftrightarrow R_4}{=} \ -\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 0 & 0 & -2 & -5 & -5 \\ 0 & 0 & 0 & -5 & -5 \\ 0 & 0 & 0 & 0 & 5\end{vmatrix}=-1\cdot (-1)\cdot (-2)\cdot (-5)\cdot 5=50
\end{align*}

Is this correct? And is this the way you meant or is there a better/faster way if we use also the Laplace expansion? (Wondering)
 
  • #10
mathmari said:
And as for matrix $a$ we have:
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -4 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}
\end{align*}

Shouldn't that be
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & {\color{red}\mathbf{-\enclose{circle}{9}}} \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}
\end{align*}
(Worried)

mathmari said:
And is this the way you meant or is there a better/faster way if we use also the Laplace expansion?

This is the way yes. I think it is the fastest way to calculate this determinant. (Thinking)
 
  • #11
Klaas van Aarsen said:
Shouldn't that be
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & {\color{red}\mathbf{-\enclose{circle}{9}}} \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}
\end{align*}
(Worried)

Now I get
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}\\ & \ \overset{R_3:R_3-3\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_4:R_4-4\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \\ & \ \overset{R_5:R_5-5\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_3:R_3-2\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \\ & \ \overset{R_4:R_4-3\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_5:R_5-9\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & 0 & 5 & 10 & 60\end{vmatrix} \\ & \ \overset{R_5:R_5+\frac{5}{2}\cdot R_4}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & 0 & 0 & -\frac{5}{2} & 85\end{vmatrix} \ \overset{R_5:R_5-\frac{1}{2}\cdot R_3}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & 0 & 0 & 0 & \frac{165}{2}\end{vmatrix} \\ & \ \overset{R_3\leftrightarrow R_4}{=} \ -\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & 0 & 0 & \frac{165}{2}\end{vmatrix}=-1\cdot (-1)\cdot (-2)\cdot (-5)\cdot \frac{165}{2}=825
\end{align*}

But according to Wolfram the result is $1875$ but I don't see where I could have done something wrong. Do you maybe have an idea? (Wondering)
 
  • #12
mathmari said:
\begin{align*}\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} & \ \overset{R_4:R_4-3\cdot R_2}{=} \
\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -2 & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix}\end{align*}

But according to Wolfram the result is $1875$ but I don't see where I could have done something wrong. Do you maybe have an idea? (Wondering)

Shouldn't it be:
\begin{align*}\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} & \ \overset{R_4:R_4-3\cdot R_2}{=} \
\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & {\color{red}-\enclose{circle}{5}} & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix}\end{align*}
(Worried)
 
  • #13
Klaas van Aarsen said:
Shouldn't it be:
\begin{align*}\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} & \ \overset{R_4:R_4-3\cdot R_2}{=} \
\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & {\color{red}-\enclose{circle}{5}} & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix}\end{align*}
(Worried)

Oh yes! So it must be as follows:
\begin{align*}\det a&=\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 2 & 3 & 4 & 5 & 1 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_2:R_2-2\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 3 & 4 & 5 & 1 & 2 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix}\\ & \ \overset{R_3:R_3-3\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 4 & 5 & 1 & 2 & 3 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \ \overset{R_4:R_4-4\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 5 & 1 & 2 & 3 & 4\end{vmatrix} \\ & \ \overset{R_5:R_5-5\cdot R_1}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & -2 & -4 & -11 & -13 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_3:R_3-2\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & -3 & -11 & -14 & -17 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \\ & \ \overset{R_4:R_4-3\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & -9 & -13 & -17 & -21\end{vmatrix} \ \overset{R_5:R_5-9\cdot R_2}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & 0 & 5 & 10 & 60\end{vmatrix} \\ & \ \overset{R_5:R_5+R_4}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & 0 & 0 & 5 & 70\end{vmatrix} \ \overset{R_5:R_5+ R_3}{=} \ \begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & 0 & 0 & 0 & 75\end{vmatrix} \\ & \ \overset{R_3\leftrightarrow R_4}{=} \ -\begin{vmatrix}1 & 2 & 3 & 4 & 5 \\ 0 & -1 & -2 & -3 & -9 \\ 0 & 0 & -5 & -5 & 10 \\ 0 & 0 & 0 & -5 & 5 \\ 0 & 0 & 0 & 0 & 75\end{vmatrix}=-1\cdot (-1)\cdot (-5)\cdot (-5)\cdot 75=1875
\end{align*}

Now the result is correct (Malthe)
 
  • #14
mathmari said:
Now the result is correct (Malthe)

Nice! (Cake)
 

Similar threads

MHB Determine the solution set of the system using the echelon form
  • · Replies 2 ·
Replies
2
Views
1K
MHB Row echelon form : Can the first column contain only zeros?
  • · Replies 14 ·
Replies
14
Views
3K
Undergrad Solve the system of linear equations
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad Row space, Column space, Null space & Left null space
  • · Replies 8 ·
Replies
8
Views
3K
MHB Determine a matrix C such that T = CA has echelon form
  • · Replies 4 ·
Replies
4
Views
2K
MHB Linear system of equations: Echelon form/Solutions
  • · Replies 3 ·
Replies
3
Views
1K
MHB Define matrix to get a row operation of type 1
  • · Replies 2 ·
Replies
2
Views
2K
MHB Householder method : Which α do we take?
  • · Replies 4 ·
Replies
4
Views
1K
MHB System no/infinitely many solution(s)
  • · Replies 15 ·
Replies
15
Views
2K
MHB How to determine the bases B and C
  • · Replies 8 ·
Replies
8
Views
2K