1. ## Matrices

here I put my answer as a picture

2. Originally Posted by r-soy
here I put my answer as a picture

Dear r-soy,

1) $A^{-1}=\left(\begin{array}{ccc}-1&1&1\\-2&3&2\\0.5&-0.25&-0.25\end{array}\right)$

2) A) $x_{1}=-1~and~x_{2}=3$

B) $x_{1}=1~and~x_{2}=2$

3. Originally Posted by Sudharaka
Dear r-soy,

1) $A^{-1}=\left(\begin{array}{ccc}-1&1&1\\-2&3&2\\0.5&-0.25&-0.25\end{array}\right)$

2) A) $x_{1}=-1~and~x_{2}=3$

B) $x_{1}=1~and~x_{2}=2$

But I wrote same answer which you wrote here

?????????????????????

where my mistake

4. Originally Posted by r-soy
But I wrote same answer which you wrote here

?????????????????????

where my mistake
Dear r-soy,

I solved these problems myself and wrote down only the answers. If they tally with your answers that means you have done these problems correctly. Congratulations!!

5. Originally Posted by Sudharaka
Dear r-soy,

I solved these problems myself and wrote down only the answers. If they tally with your answers that means you have done these problems correctly. Congratulations!!

see In Q1 Is the check is true or not ?

and also any mistake in Q1

6. Originally Posted by r-soy
see In Q1 Is the check is true or not ?

and also any mistake in Q1
Dear r-soy,

I really don't understand what you mean?? If you think that there is a mistake in Q1, my answer is "NO".

7. but some teacher said You have obviously made a mistake somewhere, either in the inverse or in your check. When you multiplied A-1 by A, you should have gotten the identity matrix I, but you didn't. That should be a clue that you've done something wrong.

8. Originally Posted by r-soy
but some teacher said You have obviously made a mistake somewhere, either in the inverse or in your check. When you multiplied A-1 by A, you should have gotten the identity matrix I, but you didn't. That should be a clue that you've done something wrong.

Dear r-soy,

Yes. You have done something wrong in getting the inverse of A. Your answer is not exactly mine.

My answer: $A^{-1}=\left(\begin{array}{ccc}-1&1&1\\-2&3&2\\0.5&-0.25&-0.25\end{array}\right)$

Your answer: $A^{-1}=\left(\begin{array}{ccc}-1&1&1\\-2&3&2\\0.5&{\color{red}0.25}&-0.25\end{array}\right)$

9. that means there is a mistake on the my checking ? ? ? ? ? ? ? ?

10. Originally Posted by r-soy
that means there is a mistake on the my checking ? ? ? ? ? ? ? ?
Dear r-soy,

co-factor matrix= $\left(\begin{array}{ccc}(4-0)&-(-8-0)&(2-4)\\-(0+4)&(4-16)&{\color{red}-(0+1)}\\(0-4)&-(0+8)&(1-0)\end{array}\right)$

11. I try to correct my answer

help me if true now or not >>

12. Originally Posted by r-soy
I try to correct my answer

help me if true now or not >>

Dear r-soy,

Your "check" is wrong. You should get the identitiy matrix. Go through your matrix multiplication again.

13. my dear i try but i get same the answer

$\left(\begin{array}{ccc}-1&1&1\\-2&3&2\\{\frac{2}{4}}&{-\frac{1}{4}}&{-\frac{1}{4}}\end{array}\right)\left(\begin{array}{ ccc}1&0&4\\-2&1&0\\4&-1&4\end{array}\right)=\left(\begin{array}{ccc}1&0& 0\\0&1&0\\{\color{red}-1}&{\color{red}-\frac{2}{4}}&1\end{array}\right)$