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Which of the following matrices does not satisfy the following statement:Ax 0 has only the trivial solution:0 a1 ~-1b.7] 0 c0 d.e. None of the above...

Question

Which of the following matrices does not satisfy the following statement:Ax 0 has only the trivial solution:0 a1 ~-1b.7] 0 c0 d.e. None of the above

Which of the following matrices does not satisfy the following statement: Ax 0 has only the trivial solution: 0 a 1 ~ -1 b. 7] 0 c 0 d. e. None of the above



Answers

Which of the following matrices is in reduced row echelon form? (a) $\left[\begin{array}{rr|r}{1} & {2} & {9} \\ {3} & {-1} & {-1}\end{array}\right] \quad\left(\text { b) }\left[\begin{array}{ll|l}{1} & {0} & {1} \\ {0} & {1} & {4}\end{array}\right]\right.$ $\left.\begin{array}{ll}{\text { (c) }\left[\begin{array}{ll|l}{1} & {2} & {1} \\ {0} & {0}\end{array} |\right.} & {28}\end{array}\right] \quad$ (d) $\left[\begin{array}{ll|l}{1} & {2} & {1} \\ {0} & {1} & {4}\end{array}\right]$

So in the given question we have four matrices that are given in the question. The first metrics metrics as 1, 1, 1, 2. And the question we are asked. Which of the four matrices that are given in the question has non real characteristic values. Right? So we can take the first matrix over here and find the determinant of those metrics. So what we do to find the character, characteristic value of the determinant is the characteristic value is lambda, right? And what we do is we can find lambda by taking, say, let's say this determinant is a the characteristic equation is made by taking the determinant off a minus lambda I, so a minus lambda equal to zero, is the character characteristic equation. And using the characteristic equation, we find the characteristic values of the given metrics. Right? So the Landau here is called the different values of London would be called the characteristic values of this uh metrics. So first, let's form the equation a minus lambda I. So a minus lambda, I would be the metrics one minus lambda one, one, 2- Lambda. Right? This would be the metrics. And when we take the determinant of this matrix, what we would get is We would have 1- learned at times two minus lambda -1 which is equal to lambda square -3 λ Plus two. So this is what we get Plus 2 -1. Right? So we would have lambda square Ministry lambda plus two minus one to minus one is one, so one and this is equal to zero. So this is the first equation. And from this we can see mhm. That the values of lander are the solutions of this quadratic equation. Right. And to know whether the solution of this quadratic equation is real or not. We can just find B squared minus four A. C. Which is the discriminate where B is the question of lambda, that is minus three squared minus four times A C. Where is the coefficient of lambda square? And see is the constant in the equation. So this is the equation that we're using right. The quadratic equation. So what we'll have here is -3 square is Niall. 9 -4 is equal to five which is greater than zero. Which means if The the Square -4 is is greater than zero, we will have real roots, which means lambda is real. Lambda is real. For the first metrics. Right next up we have the second metrics that is given in the question. So the second metrics that is given in the question is zero minus four, one minus seven. So this is the metrics. And now let's uh find the characteristic equation. So the determinant that we are taking that is a minus lambda. I would be of the form minus lambda minus 41 minus seven minus lambda. Right? And when we take the determinant, what we would have is lambda minus lambda times minus seven minus lambda, which would be equal to land at times lambda plus seven or seven plus lambda. Since the uh minus one times minus one would give us a positive one. So lambda times lambda seven plus lambda plus four. Right, So this is the determinant which we are equating to zero to get the characteristic equation. So what we will have is named uh squared plus 7 λ Plus four equal to zero. So let's find the discriminate over here which is b squared minus four A C. If it is greater than zero we can say the values of lambda are real, right? So be is seven, so seven square minus four times A. S. The coefficient of Lander square which is one CS the constant In the equation which is forward, so what we have is 49 -16, which is definitely greater than zero, which means λ is real over here. Also, next we have the third matrix that is given in the question. The 3rd matrix is given us 1, 1, So The determine that we need to find is 1 -10, 3, 5- Lambda. And this would be equal to one minus nanda times five minus lambda, which we can write a slammed a minus one times lambda minus five minus street. And this gives us the quadratic equation named uh squared minus six. Nanda plus two is equal to zero. So in the same way, let's take the discriminate of the quadratic equation over here. So b squared minus four A C equal to B. S. The collection of lambda, which is minus six minus four times A. Is the coefficient of lambda square which is one and sees the constant in the equation which is two minus six square is 36 36 minus eight is definitely greater than zero, which means base square minus four, A. C. Is greater than zero. So the roots of the situation, which are the values of the values of lambda would definitely be real values solando Israel in the third matrix that is given in the question. Also, next up, we have the fourth metrics, that is the last metrics that is given in the question. So the matrix is 3 -1, 70 And the determinant that we need to find us three minus lambda minus 170 minus lambda is minus lambda, which would give us the determinant, lambda times lambda Ministry, Lander times lambda minus three minus seven times minus one Which would be plus seven. So what we have is Lander times lambda minus three plus seven, which would give us the quadratic equation. Lambda square minus three. λ Plus seven Which is equal to zero, is the characteristic equation. And the discriminative of the situation can be found as B squared minus four. Is is equal to B. S minus three, which is the coefficient of lambda minus four times A. Is the question of lambda square and sees the constant in the equation. So this is nine minus 28 right? Which is less than zero. So since nine minus 28 is less than zero, we can say that the discriminative or re squared minus four A. C. Is negative, which means the roots of the situation would be complex rules and the rules of the situation at values of lambda, which means the metrics that has non real characteristic value, non real characteristic values. Yes, the metrics deep. So the answer, the answer is metrics and the metrics given an option deep. So I hope you understood the method. Thank you.

Hello caution is taken for matrices and the ocean is if the metric says it will do 0 -6 ι And then Zio and Matrix B is equal to 10 geo. Also 100 miners from these are metrics A N B, then A B plus B S. So that's pretty big. The value of A B must be. Let me evaluate the value of A B. A B is equal to is 0 -6 ι And then zero and B is equal to 10 to-. Let us use multiplication zero in 200 plus zero minus. I am trying to zero is for a +00 and +200 minus. I am trying to minus one And I obtained 1 ι. Are you going to 00? I obtained 2000 to -10. Okay, so that will be equal to B is equal to z O iota iota zero. In the similar violators evaluate B A. Is equal to. So if B A. Is equal to 100 -1 and 80 -6 ι zero. And that is the one in 20000 and into minus to minus 0 to 00. She went to the u minus. I obtained too minus went into iota is minus io zero into minus iota zero minus went into 00. So that will be equal to the F is equal to zero minus Toyota minus iota and then zero adding a B plus B. A B plus B. A may be equal to zero plus zero iota minus two to minus iota iota, and then zero plus zero, which is equal to which is a national metrics. Okay, so the correct terms of discussion is in al metrics option is the correct answer. Hope this clears your doubt and thank you.

We're going to work on finding the inverse of this special matrix, which is which has only the variables A, B, C and D surrounded by zeros, and you'll notice the A, B, C and D are going in a diagonal. So whenever we're working with a matrix, that's more than two by two. We're going to start by making it into a for by eight matrix with the original matrix on the left and the identity matrix on the right. So now that we have this matrix set up as a four by eight matrix, we're going to see how we could change this first road to be equal to 10 serious zero. Well, since we have a in this first position, if we take one times one over A, that would be equal to grow one. Right. So our solution will have 1000 and one over a 000 Next for road to we want 0100 So you have be in this second position where we want a one. So if we divide this whole row by B, we'll get road, too. So we'll have 0100 if we divide Bi Bi Bi. So now we'll have 01 divided by B is one Overbey Syria, Syria. For this third room, we're going to have we're going to 10010 So the way to get a one in this third position is to divide by sea. So now we have 00 c, divided by sea is one and 001 divided by sea is one oversee and finally, we're going to have row for is going to be divided by D so that we can get a one in that final position. So have 000 de divided by de is one and 0001 divided by D is one over D. So our inverse matrix is going to be as such as long as a, B, C and D are not equal to zero because in that case, these fractions will not exist.

It's for this question. They're looking for you slow so t e n reduced. Or then we're gonna see that we've ones in the diagonals and everything else Hostile zeros. And when we've got our chances. So when you're looking at your place is here, your first place doesn't have it has ones in your diagram. But rest is a is out. He has ones of your deal. Every else that one's good to go see Has a to in there that work. And he has to Well, so the one that we're looking for here would be me that once in a reduced form, has its got that diagonal said once everything else is zero and then it just has are that's it.


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