exercises-17-20-refer-to-the-matrices-a-and-b-below-make-appropriate-calculations-that-justify-you-2

Question

Exercises $17-20$ refer to the matrices $A$ and $B$ below. Make appropriate calculations that justify your answers and mention an appropriate theorem.
$$
A=\left[\begin{array}{rrrr}{1} & {3} & {0} & {3} \ {-1} & {-1} & {-1} & {1} \ {0} & {-4} & {2} & {-8} \ {2} & {0} & {3} & {-1}\end{array}ight] \quad B=\left[\begin{array}{rrrr}{1} & {3} & {-2} & {2} \ {0} & {1} & {1} & {-5} \ {1} & {2} & {-3} & {7} \ {-2} & {-8} & {2} & {-1}\end{array}ight]
$$
Do the columns of $B$ span $\mathbb{R}^{4} ?$ Does the equation $B \mathbf{x}=\mathbf{y}$ have a solution for each $\mathbf{y}$ in $\mathbb{R}^{4} ?$

Michael Jacobsen · Accepted Answer

in this video, we&#x27;re start out with the Matrix B that&#x27;s provided here as a four by four matrix. And our goal is to determine first where the pivot positions would be. To that end, let&#x27;s begin row reduction on this matrix. We have our first pivot here and we need to eliminate the entries one and negative to found below. So first, let&#x27;s copy down the 1st 2 rows, they&#x27;re 13 negative two and two, then for row to it. 01 one and negative five Now to eliminate this century multiply row one by negative one added to row three will obtain zero negative one negative one and a five. Next. To eliminate this entry, we have to this time multiply row one by two and add it to row. For this time, we obtained zero negative Too negative too. And three. So our new pivot position to consider is here, and we&#x27;re going for just echelon form so we can analyze the pivots. That means we need Teoh. Just eliminate thes two entries below, but none above. So let me once again start by copying the 1st 2 rows which are 13 negative, two and two Then row two is 01 one and negative five. Now to eliminate this take row to and added to row three, the result is going to be zero 00 and zero. So we have a zero doubt row. Then, to eliminate this entry here, our role operation will be to multiply row two by two and added to row for we&#x27;ll obtain 00 zero and then finally we have negative 10 plus three for a negative seven in our last row operation. Let&#x27;s interchange growths three and four. So I&#x27;m gonna first start by copying 13 negative two and two. Then the next non zero is 011 negative five. And here&#x27;s where we make our interchange. The next non zero is 000 negative seven and place the zeroed out to row down below. Now we&#x27;re in echelon form and we find that there is a pivot here, here and here. Now, with the results that we have obtained above, we know that be is of size four by four. I&#x27;m going to make the number of rows special by highlighting as blue and there are we found three pivots. So there is not for pivots to correspond, to pivot in every row. This gives us to immediate conclusions. We can say that the columns of be do not span are four and the reason being would be since we require four pivots in order span a four dimensional space and we&#x27;re short on pivots. So the columns do not spend the target space are four. And this also means that if we pick in a equation to consider be X equals y then this system is not consistent for all vectors. Why in are for again in order for this equation to always be consistent we required four pivots, but we&#x27;re short on pivots and this is a big theme in linear algebra that were studying at this level, where really knowing where the pivots are can tell us a lot of fundamental information about the Matrix as well as matrix equations like these

Michael Jacobsen · Answer

in this video, we&#x27;re considering the Matrix a once again where the first thing we did with this matrix say that&#x27;s provided is we reduced it to just echelon form. We found that rose 12 and three contained pivots. So we have that three of the four rows of a our pivot rose. But in other words, we could fray this phrase this by saying, Hey does not have a pivot in every row Because of this deficiency in pivots, we now have a couple statements that we can obtain immediately as direct consequences. It falls first that there exists vectors be in are for that are not leaner combinations of the columns of a If we had four pits in this matrix say that we would say all vectors in our four are linear combinations. But being deficient pivots means there are some vectors that are not such a linear combination. Well, this language here is about leaner combinations and the set of Alden your combinations is also the span of the columns of a So from the statement here, we can also say the following the columns off a do not span are for and again. That&#x27;s because we required for pivots. But we found just three

Michael Jacobsen · Answer

in this video. We&#x27;re considering the Matrix be further. So be was given to be this four by four matrix And what we&#x27;ve done already is we have read role reduced it into its echelon form. We found that row one, row two and row three in its role reduced form our pivot columns So we can say for the following that be in particular does not have a pivot in every row since the fourth row is a zeroed out row. Now this not having a pivot in every row causes certain problems. For example, we do also know as soon as there is not a pivot Every row that there exists vectors be in our four Where are four comes from? The four rows of this matrix p So there exists Exists Victor&#x27;s being are for that are not linear combinations of the columns of be This is an important thing to consider because it means that in a vector equation such as B X equals, say why would not have a solution for every vector. Why? Let&#x27;s consider a different situation. We know the set evolving your combinations is also the span of the columns of B and we also know that we have three pivots altogether. So let&#x27;s consider the claim. The columns of be spans are three. This is an interesting claim because it really feels quite right. Sure, we don&#x27;t have a pivot and every row, so the comes of B cannot spend our four. But if there are three pivots, could the calm spend our three? Well, Sadly, this claim is false. The columns of be do not span or three and this claim is false irrespective of the number of pivots. So if we had no idea what the number of pivots were, if we were just back to the original matrix, it would still be faults. And the reason is since B is in are four by four. In other words, this four here is the most important thing to consider and it causes a mismatch with our three. So the columns of beak on lee have a chance of spanning are for not are three, not are two and so on

Michael Jacobsen · Answer

in this video, we have a four by four matrix a provided. And our goal here is to analyze the pivot positions. This can be done by first placing this matrix into its echelon form. Let&#x27;s make that our first steps. So I&#x27;ll copy of the first row 13 03 And our goal is to go to this position here and make all entries below zero. So we can add Rwanda Row two and record the result. And that will give us a zero then too negative. One and four. Next. We have a zero here, so I&#x27;ll just copy this road, then it zero negative for two and negative eight. Next. We need to eliminate this too. So multiply Row one by negative too. At it to row for And we have zero negative. Six three and negative seven. Now, since we&#x27;re going for echelon form, I&#x27;m going to pivot off this, too. We see here and go for an elimination of the entries. Negative four and negative six. But let&#x27;s ignore the three. So our next step will say that this is row equivalent to first. Let me copy the 1st 2 rows. 13 03 and zero to negative 14 Now let&#x27;s multiply row two by two, adding it to row three so we can eliminate that entry will have 000 and zero. So this is working out nicely next to eliminate the negative six. Multiply road to buy three and add it to row for we&#x27;ll obtain zero zero. Let&#x27;s see we have zero again and three by four is 12. My seven gives us a positive five. So at this stage are matrix is not quite initial on form. Let&#x27;s take that one extra step to place it there. I&#x27;ll have 13 03 in row, one row twos just fine, so it&#x27;s copies your to negative 14 The problem is Row three. We would prefer to have the zeroed out rose towards the bottom. So it&#x27;s interchange Row three with Roe for then our matrix. Let me try better with the zero that are matrix is now in echelon form. So we&#x27;re going to be able to analyze the pivots by saying here, here and here are our pivot position. So by this work it follows that three rows of a contain a pivot position and our work is complete

Thomas Emment · Answer

Okay, We&#x27;ve been given these two. Major sees A and B. We want to find a matrix. A given that four acres, three big is equal to minus two X. So it will be full of very fast, sir. Just times each and view alone it by the constant from which is forced out of my 12 minus 28 eight minus 36 on 20. And then office zero tons. Four Still be zero eyes. Gonna plus well, three lots of our base. We have minus 15 minus three offices. In times, anything is zero, and then three times three is nine minus four times three minus 12. Okay, let&#x27;s have these together, and then we just need Teoh divided by minus two minus 12. Minus 15 was going this minus 27 by 38 minus three times minus. State one. Uh, was there isn&#x27;t this once? It&#x27;s just gonna be ate my face. Six. 21st 1929 100 0 plus minus 12 B minus 12. Okay, divide the sore by minus two. We&#x27;ll get 27 to one of its A minus four. Ah, Positive was half a face. Six being on 50 18. Yeah, Yeah. So positive. 18 29. Minus 22 on, then finally, minus 12 is gonna be a positive six on this should be our next matrix.

Thomas Emment · Answer

okay, We&#x27;ll be giving these. Major sees A and B and we want to find a matrix X given, um, given that b minus X, it was for a so we can turn this around so that X is equal to B minus for a just not exposed sides. Minds for a from both sides. On that we could get cracking. So we&#x27;re gonna be taking each element of R B on minus ing four lots off its corresponding element in XY. Got minus five minus four. Lots of my street to B minus five plus 12. So that will just leave us with seven, right? Yeah. Let me got mice for a SER minus one minus four minus seven. Should have minus one plus 28. Said I will be 20. Okay, um zero minus four lots to B minus. Eights on zero minus four. Lots of minus nine would be 36 three minus four. Lots of fire will be three months. 20 which relieve us active. Minus 17 on. Then we just have minus 40 times. Anything is zero

Problem

Exercises $17-20$ refer to the matrices $A$ and $…

Problem 18 Hard Difficulty

Answer

only three rows contain pivot positions
Columns of $\mathrm{B}$ do not $\operatorname{span} \mathbb{R}^{4}$
the equation $\mathrm{Bx}=y$ does not have a solution for each $\mathrm{y}$ in $\mathbb{R}^{4}$
Check theorem 4

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Alayna H.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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only three rows contain pivot positionsColumns of $\mathrm{B}$ do not $\operatorname{span} \mathbb{R}^{4}$the equation $\mathrm{Bx}=y$ does not have a solution for each $\mathrm{y}$ in $\mathbb{R}^{4}$Check theorem 4

Linear Algebra and Its Applications

Grace H.

Anna Marie V.

Alayna H.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Anna Marie V.

Alayna H.

Caleb E.

only three rows contain pivot positions
Columns of $\mathrm{B}$ do not $\operatorname{span} \mathbb{R}^{4}$
the equation $\mathrm{Bx}=y$ does not have a solution for each $\mathrm{y}$ in $\mathbb{R}^{4}$
Check theorem 4

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications