Easy Determinant Help: Solving Row Operations for Matrix | -9 is Wrong?

[Augustine Duran]

Homework Statement

\begin{vmatrix}
1 & -4 & 3 & 4 \\
0 & -9 & 6 & 8 \\
0 & -6 & 5 & 5 \\
0 & 0 & -3 & 2
\end{vmatrix}

so the determinant of this matrix is -9, apparently I am doing something illegal in my row operations.

I want to get -6 in row 3 to be 0 so...

2R₂ - 3R₃ = 0 0 -3 1

\begin{vmatrix}
1 & -4 & 3 & 4 \\
0 & -9 & 6 & 8 \\
0 & 0 & -3 & 1 \\
0 & 0 & -3 & 2
\end{vmatrix}

i want to get -3 in row 4 to be 0 so..

R₄ - R₃ = 0 0 0 1

\begin{vmatrix}
1 & -4 & 3 & 4 \\
0 & -9 & 6 & 8 \\
0 & 0 & -3 & 1 \\
0 & 0 & 0 & 1
\end{vmatrix}

here i multiply the main diagonal and get 27 which is wrong. Can someone point out what I am doing wrong

 

[fresh_42]

Which determinant do you get, if you apply the same operations on the identity matrix?

 

[Augustine Duran]

-3?

 

[fresh_42]

So the determinant changed from $1$ to $-3$. Now why should the determinant of your given matrix remain unchanged, when the determinant of another matrix changes by the same operations?

Do you see, why a correction by this factor yields the correct result?

 

[Augustine Duran]

to be honest i don't know. maybe you can rephrase your question so i can better understand?

all i know is the theorems in listed in the book

-If a multiple of one row of A is added to another row to produce a matrix B,
then detB = detA.

-If two rows of A are interchanged to produce B, then detB = detA

-If one row of A is multiplied by k to produce B, then detB = k ⋅ detA.

 

[Augustine Duran]

unless your trying to point out that the determinant of my given matrix shouldn't change if the determinant of the identity matrix doesn't change?

 

[fresh_42]

Did you already had matrix multiplication and the property of determinants $\det (A\cdot B) = \det(A)\cdot \det(B)$?

These rules cannot be true. E.g. if I swap rows of $\begin{bmatrix}1&0 \\ 0 &1 \end{bmatrix}$ then I get $\begin{bmatrix}0 &1 \\ 1 & 0 \end{bmatrix}$, but the first one has determinate $1$ and the second one $-1$.

unless your trying to point out that the determinant of my given matrix shouldn't change if the determinant of the identity matrix doesn't change?

Either both will change or neither, if we apply the same operations.

 

[Augustine Duran]

ok so the Multiplicative Property u gave me is 2 pages ahead from where I am at, I am trying to understand an example using those theorems given to me.

im aware that if you swap rows the detB = - detA

 

[fresh_42]

ok so the Multiplicative Property u gave me is 2 pages ahead from where I am at, I am trying to understand an example using those theorems given to me.

im aware that if you swap rows the detB = - detA

For short: only operations are allowed which don't change the determinant or you have to keep track of them, e.g. by applying the same operations to the identity matrix, and at the end adjust your result accordingly. In our case this would mean, divide the $27$ by $-3$.
I wanted to tell you why this is possible. As it is ahead of you, I'll say it and you can come back on it, when you are a few pages further.

Row operations applied on a matrix $A$ the way you did it, means you transformed $A$ into a matrix $R\cdot A$. Now you calculated $\det (R\cdot A)$ which is equal to $\det R \cdot \det A$. So we still have to divide by $\det R$ in order to get $\det A$.
If we apply the same operations on the identity matrix, then we change $I$ with $\det I =1$ to $R\cdot I = R$ with $\det R = \det R \cdot \det I = \det (R\cdot I)$ which gave in our case the value $\det R = -3$ as you said in post #3.

 

[Augustine Duran]

so dividing det(R⋅A) by det R gives me det A. what's bugging me is that, do i have apply the same operations to the identity matrix to get det R, and then divide det( R⋅A) by det R every time to find the determinant? the example in the book just did some row operations then multiplied the main diagonal to get the determinant
(i did different row operations than the book so I am thinking that's where my error is but i can't see where i went wrong)

 

[fresh_42]

so dividing det(R⋅A) by det R gives me det A. what's bugging me is that, do i have apply the same operations to the identity matrix to get det R, and then divide det( R⋅A) by det R every time to find the determinant? the example in the book just did some row operations then multiplied the main diagonal to get the determinant
(i did different row operations than the book so I am thinking that's where my error is but i can't see where i went wrong)

The alternative would be: only apply operations, which do not change the determinant. Addition of rows is o.k., multiplications change it (see last rule in post #5) as well as interchanging two neighboring rows does it by a factor $-1$. Your first step $2R_2 - 3R_3$ changed the determinant, because you performed multiplications (see last rule in post #5):

• factor $2$ from row two $R_2$
• factor $-3$ from row three $R_3$
• addition for free
• reversing the operation on row two by dividing it by $2$

So you changed the determinant by a factor $2\cdot (-3) \cdot \dfrac{1}{2} = -3$

 

[Augustine Duran]

so...
-If one row of A is multiplied by k to produce B, then detB = k ⋅ detA

so correct me if I am wrong, i will still end up with the correct determinant if i multiply that determinant by k since detB = k ⋅ detA

 

[fresh_42]

Yes. But don't forget, that you multiplied several times to prepare the the rows for addition. All three have to be put into consideration.

 

[Augustine Duran]

ok final question. overall it seems i multiplied it by -3, which would seem to be my k. But multiplying my determinant (27) by -3 obviously gives the wrong answer. So shouldn't it be defined as 1/k instead of just k?

 

[fresh_42]

ok final question. overall it seems i multiplied it by -3, which would seem to be my k. But multiplying my determinant (27) by -3 obviously gives the wrong answer. So shouldn't it be defined as 1/k instead of just k?

You increased your determinant by a factor $-3$ during the process. So to reverse it, you have to divide it again. But in the named rule, there has to be k.

 

[Augustine Duran]

ok thanks for all the help and being patient, i really appreciate it!