5
$\begingroup$

I'm preparing for an exam currently, and I came across this question: The particular question I found

I have noticed that A can be constructed from the matrix on the left by a series of row operations, so I had the idea maybe to express A as a product of elementary matrices as well as the matrix on the left and, maybe there was some fact about the determinants of elementary matrices? So then I could just use some properties of determinants, as well as knowing that the determinant of the matrix on the left is 2 to figure out $det(A)$

Entirely confused though because I do not know what the determinant of elementary matrices are, and if I should be multiplying all the elementary matrices together etc etc?

Any help would be hugely appreciated.

$\endgroup$
6
  • $\begingroup$ Well get the factor $4$ out (i.e. replace first row by $g,h,i$ and consider $\det(A)/4$). Then you van transform the determinant to the one on the left by only row additions subtraction. How is the determinant affected by such operations ? $\endgroup$
    – zwim
    May 26 at 21:32
  • $\begingroup$ Why should it be $\frac{det(A)}{4^3}$ ? This is where im getting confused because im not sure I know the rules for what you can and cant do $\endgroup$
    – HarveyR
    May 26 at 21:34
  • $\begingroup$ sorry typo, only $4$ because $\det(A)=4g |\cdots|-4h|\cdots|+4i|\cdots|=4\times(g |\cdots|-h|\cdots|+i|\cdots|)$ this is just developping the determinant along terms of first row. $\endgroup$
    – zwim
    May 26 at 21:36
  • 1
    $\begingroup$ Because row $g,h,i$ is in first position, while originally it is in last position. Each time you swap rows, you multiply by $-1$ (here swap row1 and row3 one time). $\endgroup$
    – zwim
    May 26 at 21:41
  • 1
    $\begingroup$ @HarveyR if an answer solved your question, don't forget to accept it by clicking on the check mark below the voting arrows ;) $\endgroup$
    – StratosFair
    May 27 at 7:19

3 Answers 3

Reset to default
9
$\begingroup$

Let $B = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$.

If you take the operations that transform $B$ into $A$, and apply them to the identity matrix, you get:

$$C = \begin{pmatrix} 0 & 0 & 4 \\ 0 & 1 & -1 \\ 1 & -2 & 0 \end{pmatrix}$$

So $A = CB$, which means that $\det(A) = \det(C) \det(B)$. You should be able to calculate that $\det(C) = -4$, and you are given that $\det(B) = 2$. Therefore, $\det(A) = -8$.

Since this is nonzero, then yes, $A$ is invertible.

$\endgroup$
1
  • 2
    $\begingroup$ Thank you! I didn't realise that you could combine the elementary matrices into one matrix like this. From that point I agree, finding $det(C)$ is straightforward. Thank you so much! $\endgroup$
    – HarveyR
    May 26 at 22:13
5
$\begingroup$

If we are uncertain about the row / col transform we could represent it in a matrix multiplication fashion. $$ \begin{bmatrix} 4g & 4h & 4i \\ d-g & e-h & f-i \\ a-2d & b-2e & c-2f\\ \end{bmatrix}= \begin{bmatrix} 4 & 0 & 0\\ -1 & 1 & 0\\ 0 & -2 & 1\\ \end{bmatrix} \begin{bmatrix} g & h & i\\ d & e & f\\ a & b & c\\ \end{bmatrix} $$

$$ \begin{bmatrix} g & h & i\\ d & e & f\\ a & b & c\\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\\ \end{bmatrix} \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\\ \end{bmatrix} $$ Easy to see $$ \det\begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0\\ 1 & 0 & 0\\ \end{bmatrix}=-1\\ \det\begin{bmatrix} 4 & 0 & 0\\ -1 & 1 & 0\\ 0 & -2 & 1\\ \end{bmatrix}=4 $$

Since the determinant $\det AB=\det A \det B$

We get $\det A=-4\det \begin{bmatrix} a & b & c\\ d & e & f\\ g & h & i\\ \end{bmatrix}=-8 $

$\det A\neq 0$ so it's invertible

$\endgroup$
3
$\begingroup$

Entirely confused though because I do not know what the determinant of elementary matrices are...

You don't even need to know those elementary matrices. You just need to know how elementary row operations affect the determinant. In this case, we need all three types of operations, and I write the effect in the parentheses behind.

  1. Multiply a row by a non-zero number. (determinant multiplied by this number)
  2. Interchange two rows. (determinant multiplied by $-1$)
  3. Add a scalar multiple of one row to another row. (determinant unchanged)

It's obvious that we can multiply the first row of $A$ by $\frac{1}{4}$ and add to the second row (operation type 3) to get $(d\ e\ f)$. Then multiplying this row by $\frac{1}{2}$ and adding to the second row (type 3) gives us a clean row $(a\ b\ c)$. Now we have a matrix that can be related to $A$ by elementary row operations and have the same determinant as $A$: $$ \begin{pmatrix} 4g & 4h & 4i\\ d & e & f\\ a & b & c \end{pmatrix} $$

Divide the first row by 4 (type 1) and interchange the first and the second last row (type 2), we get the original matrix whose determinant is known to be $2$. Since we know consequences of three types of operation, it's easy to conclude that $$ \det (A) = -4\times 2=-8 $$

P.S.

I personally don't remember those matrices performing elementary row operations as matrix multiplications. And I don't understand the necessity of working out those operation matrces (or the overall matrix $C$ in @Dan's answer). Doing matrix multiplication is always tedious, after all. As I have said, observing that $A$ can be related to the original matrix by row operations suffice to give the answer.

$\endgroup$
1
  • $\begingroup$ +1 Thanks for nicely detailing what I hurriedly said in comments, this is effectively the most efficient way to deal with determinants. I would just add that it works similarly with columns operations since $\det(M)=\det(^tM)$ $\endgroup$
    – zwim
    May 27 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.