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8 Here a,b are real numbers Let2 bA =3Given are the following statements: det( A) depends on the value of a but not on the value of b (II) A is invertible as long...

Question

8 Here a,b are real numbers Let2 bA =3Given are the following statements: det( A) depends on the value of a but not on the value of b (II) A is invertible as long as 6 is different from 2 and 3 (III) The row reduced echelon form of A has exactly two leading 1's for all real &,b Of these statements; indicate which are correct: a ) Land III IL only II and III I only [and II

8 Here a,b are real numbers Let 2 b A = 3 Given are the following statements: det( A) depends on the value of a but not on the value of b (II) A is invertible as long as 6 is different from 2 and 3 (III) The row reduced echelon form of A has exactly two leading 1's for all real &,b Of these statements; indicate which are correct: a ) Land III IL only II and III I only [and II



Answers

If $a$ and $b$ are real numbers such that $a b=0,$ then $a=0$ or $b=0 .$ However, if $A$ and $B$ are matrices such that $A B=O$ , it is not necessarily true that $A=O$ or $B=O .$ Illustrate this using the following matrices.
$A=\left[ \begin{array}{ll}{3} & {3} \\ {4} & {4}\end{array}\right], \quad B=\left[ \begin{array}{rr}{1} & {-1} \\ {-1} & {1}\end{array}\right]$

Hello guys to solve this problem we need to multiply too much receives the multiplied by B. And as we know that the multiplication of mattresses differ from the multiplication of two real numbers we have here. 33 44. This is matrix A. Multiplied by matrix B. Which is one negative one negative one one. And as we know a fee multiplied two x 2 matrices by each other. We shall get two by two matrix the result of multiplication. And if we need to calculate the first element which is the in the first true and first column. So we need to multiply the fairest grew by the first code on here. So if we multiply both to each other we got zero. And if we need to calculate the second element of the first truth when we need to multiply by the first True by the second column here. And if we multiply it also we get zero. And if we need to get the first element of the second group. So we need to multiply the second group by the first column which is zero and here the second element and the second rule. We need to multiply the second rule by the second column here which is also zero and that's it.

They were given with three metrics A, B and C, and we to prove a plus. B is B plus A that is using community law. Oh, sorry that we have to prove it. This community. Okay, so let's start with left hand side. That is a plus B and we can right here a +11 a 1 to +8218 to to write on adding B, which is B +11 b +12 b 21 b two. Okay, credit. Now we know the addition in metrics is done by element wise. Right means first row first element is added to first throw first element. So we get here a 11 plus b 11 that is our first, then similarly, a 12 plus B 12 a 21 plus B 21 and last twenties they to plus B two. Now we know as you mean these aerial numbers or any, uh, numbers in complex domain as well their communicative in addition. Okay, so any number is commuted. In addition, so we can write. This is we won one plus a 11 Similarly, here be 12 plus a 12 b 21 plus 8 to 1 B two plus eight or two. Okay, so we're done. So now we just have to do is write it into metrics. Right. So this will become be 11 B one to B 21 be to plus the second my Twix A +11 a 1 to 8 to 18 or two. Yeah, on this is B plus A correct. So we approach. Thank you.

When a equal to metrics, A B C, b C and c A B. Now we if we are finding the determinant okay determinant of equal to a hindu. B c minus is square minus being do or B squared minus A C plus Sindhu A B minus C square. Then by expanding we get a B C. Mine is a cube minus B Q. Let's say B C plus cbc minor CQ. Then we get taking a minus outside. We get a cube plus B cube plus C Q minus three baby. We can abc we can write like this. Okay, determinate date. No we have a transport sequel to same as a when we look the metric so you know metrics is symmetric. Matics. Therefore a transports equal to itself. The four a into a transports will be is clear equal to X. Sorry not a transport into a hey into equal to esquire and when we find the esquire we get it is equal to I. Okay then determinant of a square will be determinant of I that is equal to one. For determinant of a will be plus or -1 and we have found the value A Q plus B Q plus CQ minus three. Abc. This is the determinant day that we got us plus or -1 and we also have A B C. Equal to one before A Q plus B Q plus c cube equal to Plus or -1 plus three. And he decided two or four. What determinant of faith. Okay the for the correct options are B and B. Thank you

Okay, so we want to show it at thes two are equal, So let's multiply ntd into our matrix. A. So here we have CD a 11 CD, a 12 and so on. You need to to and then on our right hand side, we first wanted more supply and RG to have d a 11 d a want to e a one? Actually, that's not one of the one that's too in de eight. You too. And I was multiply inner city. So we get TD anyone? One CD a 12 CD, a 21 win in TV 8 to 2 Now will notice here that this on the right hand side is equal to our left outside.


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