Question about given solution for Determinant (3×3 matrix)

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sunny123

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I have one more question for today

13061

I am not able to understand the solution given in rectangle , Can anyone expand it step by step, with explanation. ( Rest is clear)
 
Ok this is what I have tried so far, my last part is not matching the solution given in rectangle. Please Help

13062
 
Otis said:
Hello sunny. The underlined part in the prior step contains a mistake. There's supposed to be only one a2c term, not two.

The final solution is correct, and it can also be expressed as:

(b - a)(b - c)(a - c)

?
Click to expand...

Thanx alot, it fixed now, yes got the same answer. Can't thank you enough
 
Otis said:
Hello sunny. The underlined part in the prior step contains a mistake. There's supposed to be only one a2c term, not two.

The final solution is correct, and it can also be expressed as:

(b - a)(b - c)(a - c)

?
Click to expand...

Can you please also check this, answer is not matching in this question as well.

This is the Text book answer.

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My answer is exactly the same but I am getting C (Sq)

13067
 
The last part should be \(\displaystyle c^2+ d^2\). Not only should the "c" be squared but it should be "+" not "-" because \(\displaystyle i^2= -1\).

Your text book seems to have a lot of printing errors!
 
Otis said:
Hello sunny. The underlined part in the prior step contains a mistake. There's supposed to be only one a2c term, not two.

The final solution is correct, and it can also be expressed as:

(b - a)(b - c)(a - c)

?
Click to expand...
Can You tell me How did you derived (b - a)(b - c)(a - c)? Is there any general formula?
 
Dr.Peterson said:
You replaced i with -1. It's i^2 that is -1. Be careful!

Their c should be c^2, s has been said. Their -d^2 is correct.
Click to expand...

Thanx, I am making silly mistakes, you saved me from disaster. Please check if my answer is correct now.

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Last edited:
HallsofIvy said:
The last part should be \(\displaystyle c^2+ d^2\). Not only should the "c" be squared but it should be "+" not "-" because \(\displaystyle i^2= -1\).

Your text book seems to have a lot of printing errors!
Click to expand...

Can you please explain me how? I tried to solve the equation and if the answer should have c^2 then it should be "-" please see the attached image above.

PS: Tell me about it, I wish I could change the System. Text books are designed in such a way that it becomes impossible for students to understand.

My Idea of a Text Book, it should be interactive, it should speak to the student and help him on the very spot. It should show how the method was derived, what formula was used, what are the pre-requisite for learning that particular topic etc. Last but not least it should not have errors!
 
Last edited:
manasan said:
Can You tell me How did you derived (b - a)(b - c)(a - c)? Is there any general formula?
Click to expand...
Just do term by term.
\(\displaystyle \begin{align*}(b-a)(b-c)(a-c)& \\(b^2-bc-ab+ac)&(a-c)\\b^2a-bca-a^2b+& a^2c -b^2c+bc^2+abc-ac^2\\ab^2-ac^2-a^2b+a^2c & +bc^2-b^2c \end{align*}\)
SEE HERE
 
You had \(\displaystyle (c+ id)(c- id)= c(c- id)+ id(c- id)=c^2- icd+ icd- i^2d^2\). "\(\displaystyle -icd\)" and "\(\displaystyle icd\)" cancel and \(\displaystyle i^2= -1\) so \(\displaystyle (c+ id)(c- id)= c^2+ d^2\).

(I am assuming your "i" is the imaginary identity. If that is not true- if "i" is just another variable- then \(\displaystyle (c+ id)(c- id)= c^2- i^2d^2\).)
 
pka said:
Just do term by term.
\(\displaystyle \begin{align*}(b-a)(b-c)(a-c)& \\(b^2-bc-ab+ac)&(a-c)\\b^2a-bca-a^2b+& a^2c -b^2c+bc^2+abc-ac^2\\ab^2-ac^2-a^2b+a^2c & +bc^2-b^2c \end{align*}\)
SEE HERE
Click to expand...
I know how to get \(
ab^2−ac^2−a^2b+a^2c+bc^2−b^2c \text{ from }
(b−a)(b−c)(a−c) \) but How to do it in reverse. How can I factorize \(
(b−a)(b−c)(a−c) \text{ to get } ab^2−ac^2−a^2b+a^2c+bc^2−b^2c\)
 
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