find-all-mathbfx-in-mathbbr4-that-are-mapped-into-the-zero-vector-by-the-transformation-mathbfx-ma-2

Question

Find all $\mathbf{x}$ in $\mathbb{R}^{4}$ that are mapped into the zero vector by the transformation $\mathbf{x} \mapsto A \mathbf{x}$ for the given matrix $A .$
$A=\left[\begin{array}{rrrr}{1} & {3} & {9} & {2} \ {1} & {0} & {3} & {-4} \ {0} & {1} & {2} & {3} \ {-2} & {3} & {0} & {5}\end{array}ight]$

Ashley Boni · Accepted Answer

So we want to find all vectors x such that a X maps to the zero vector. So we want to reproduce. This augmented me tricks of a and zero vector. So first we know we want to get rid of these two. So first and third row are going to remain the same. So let&#x27;s get rid of the one in the second equation. So I&#x27;m gonna take negative one times the first equation and add it to the second. So we get a zero here. Negative three. Ah, negative. Nine plus three is negative. Six, we get another negative six and a zero. Now, I&#x27;m gonna take twice the first equation and add it to the fourth to get a zero here. Two times three is six plus three is nine. Two times nine is 18 plus 0 18 Ah, two times two is four plus five is nine. And, uh, anything would do a zero. Here is zero. Ah, So next I want a pivot in the second row. Second call a position. So in this position here. So in order to do that, I&#x27;m going to swap the second and third equations. So 139 to 0. Second becomes 0123 Then we have zero minus three, minus six months. Six and zero, nine 18 9 So now we have one where we want it. And we can use this one to get rid of, um, everything below it. So let&#x27;s do that. So first column remains the same, and so does the second, their first and second row. So now I&#x27;m gonna take three times the second equation and add it to the third. So we get a 00 another zero, three and zero. Now, I&#x27;m gonna take negative nine times the first equation at it to the third. So we get a zero Where the nine waas Negative. 18 plus 18. A zero. Um, we get a negative. 27 plus nine is negative. 18 and zero. So now, um, we want to go to the next pivot position, which would be right here, but we can&#x27;t make this a one, so we have to move on to the next s. So we have this negative, Uh, this three here. So let&#x27;s just take the third equation and divided by three so we can get that our next pit. So everything else remains the same. So dividing everything By three, we still have zeros and our three becomes a one, which is what we want. We want our pivot in the third row and last Rover means the same. So our last step, we need this guy to be a zero. So I&#x27;m just gonna take third equation multiplied by 18 and add So all that&#x27;s gonna do is switch that to a zero. So let&#x27;s try to squeeze. Listen here if we can. So Last Room&#x27;s always zeros. So now it&#x27;s right out on the equations that we have. So we have x one plus three x two plus nine x three plus two x four is equal to zero. Second equation gives us x two plus two x three plus three x four equal zero third equation gives us x four equal zero and the last equation tells us nothing. So we can see that since we didn&#x27;t have that pivot in the third column, we have x three is ah, free variable, so x Ford&#x27;s equal to zero. We know that. So this first equation becomes next. One plus three x two plus nine x three Because for getting rid of these terms. Second equation becomes x two plus two x three So we know X three is free so everything&#x27;s gonna be written in terms of X 30. So the second equation um, we could get an expression for X two, so x two is equal to negative thio x three and let&#x27;s plug this into the first equations. So we have x one plus three x two plus nine x three equal zero. So now we can get an expression for X one. Uh so first let&#x27;s simplify x one minus six x three plus nine x three Equal zero negative six plus nine is three so x one is equal to negative three x three So let&#x27;s see what we have next one next to x three x four is equal to x one is equal to negative three x three x two is negative two x three Ah x three is just x three because that&#x27;s what we&#x27;re ready in terms of and x 40 So we can just rewrite this as a vector times, uh, X three to be a little bit more clean. So we have negatives. Very negative, too. One and zero. So this is our final answer

Michael Jacobsen · Answer

in this example, we have the transformation where X is mapped to the matrix, a times X where a is defined as follows. Here. The goal is to determine which vectors x arm empt to zero vector, which in this case would be the three by 10 Vector. Well, given the transformation X maps to a X. What we need to do is determine which vectors x result in the mapping being sent to the zero Vector. So really, we&#x27;re just considering the matrix equation eight times X equals zero. This could be solved by forming the augmented matrix, which will have columns 102 negative 41 negative six seven negative 46 negative 53 negative four augmented from here with zero vector. Now for our next steps. In order to solve for X, we&#x27;re going to roll reduced into reduced echelon form. So for our first pivot located here, we already have a zero below. But we need to eliminate this entry. So let&#x27;s copy row one, then to eliminate the century Well, maybe first copy wrote to as well, then, to eliminate this entry to here, we can multiply row one by negative too and add it to Row three. We obtained zero to negative eight, six and zero. Now we&#x27;re in good shape at this pivot position. We need to eliminate the entry here and here. Let&#x27;s make that our next set of role operations, a copy Row two, which is 01 negative for three and zero. Then let&#x27;s start with eliminating this entry if we multiply row one by negative to at that result to Row three, we obtained 00 then 00 and zero, so row three is eliminated. Next, multiply row two by four added to Row one, and we will obtain altogether 10 negative nine, seven and zero. And don&#x27;t forget that this is augmented matrix. Now. At this stage, Columns one and two are pivot columns and recall that column one corresponds to X one. Then we have x two x three x four and we&#x27;re setting the equal to the values on the right hand side in this way From column one. If we saw for the X one variable, we&#x27;ll see The X one is equal to positive nine x three minus seven x four, then for X two, take row to solve for X two and we have four x three minus three times x four. Now the last two variables here are free variables and so will simply record that X three equals x three, meaning there is no restriction and x four equals x four. At this stage, we ready to describe X the set of all vectors X That map to the zero vector under this given transformation is going to be first. I&#x27;ll put in a coif or a parameter x three tens of constant vector plus x four times Another constant vector x three Next four came from the set of free variables here, as well as from here and here. Then we obtained these vectors by just pulling coefficients. So coefficients of X three are 941 and zero, then coefficients for X four Read Negative seven negative 30 and one. So this is the set of all vectors X, where X three and X four are in our So that could be any real numbers that are mapped to the zero factor

Ankit Gupta · Answer

we have metrics. A is equal to 204 three one life minus one one minus. We have to find a inverse if it exists without using calculator. So first we have to find determinant off A That is you in tow minus two minus five minus zero. Indo minus six plus five plus four into three. Plus one. By solving it, we get determinant off is do as determinant Off a is not a close to zero which means hey in verse exist No formula to find in verse is one upon determinant off a Indo and joint for a joint A. We have to find co factors that is a is equal to even one even to a 13 a new one it will do 8 to 3 a 31 832 a three three 1st 1 indicated through and 2nd 1 indicates column as one place managed to That is even number even number indicate positive sign One person who is three threes are number one or number indicate negative sign. So we have plus minus plus minus blows minus place minus plus Now in a 11 we have to hide first room and first column. So we get minus seven in Avon. Do we have to hide? First row and second column. We have minus one in e 13 We have to hide. First drove and third column. So we have. For as it is, we have minus four Zito cool. Minus four, minus two. And to no, according to the signs of even and odd. We have a is equal to minus seven one four 40 minus two minus four to school. No, we have to find transport off a transport off a is convert throws into columns, so we get transpose off A That is minus seven one. Food 40 minus two. Minus four. Do cool. No. In verse. Off it is. Once upon determinant off it is two in tow. A joint A That is minus seven four minus four 10 Do food minus two and do no in verse. Off is minus seven upon Do tu minus two one upon 20 one tu minus one one. This is the answer

Ankit Gupta · Answer

we have my drinks. A is equal to minus one minus one minus one 450 01 minus 30. We have to find a inverse if it exists without using calculator. So first we have to find determinant off A That is minus one into 15 minus minus 15 minus zero plus one in tow. Minus 12 miners zero minus one into four, minus zero. Now determinant off is minus one as determinant off is not equal to zero, which means in verse exist. So for Miller to find a inverse is one upon determinant off a Indo joint. Hey, for a joint A. We have to find co factors. That is a is equal to even one even to a 13 8 to 1, it will do a new tree a 31 832 a tree three 1st 1 indicates through and 2nd 1 indicates column As one person is true, that is even number even number indicates positive signs and our number indicates negative sign one blessed with old, which means negative sign as it is plus minus, plus minus, plus minus plus no to find even one. We have to hide first row and first column. So we get minus 15 and in even, do we have a wide first row and second column? So we get minus 12 on here for 43 minus one 54 minus one according to the signs off. Positive and negative as even and order. We have mine of 15 grad food minus four three one five minus four minus one. No transport off a is. We have to convert Rose into columns. So we get transport off A That is minus 15 grand four minus four three. One for you. Minus four minus one. No, in verse is equal to one upon determinant. Over That is minus one into a joint A That is minus 15 minus 45 grand three minus four 41 minus one No. A inverse is equal to 15 four, minus five, minus 12 minus three. Food minus four minus 11 This is answer

Ankit Gupta · Answer

we have metrics. It is equal to one, three, three, 14 three and 134 We have to find a inverse if it exists without using calculator. So first we have to find determinant off A That is one in tow. 16 minus nine minus three into four, minus three plus three. Indo three minus four. So we get determinant off A is equal to two as determinant off a is not a question zero which means in verse exist now formula to find a inverse is one upon determinant off a Indo at joint okay for their joint A. We have to find co factors that is a is equal to a 11 even toe a 13 a new one it dodo a to three a 31 832 833 1st 1 in the gates through and 2nd 1 indicates column as one plus one is 22 is even number. So even number indicates a positive sign. One place to his three that is old number or number indicates negative sign. So we have plus minus plus minus plus minus plus minus place No, in fact, even one We have to hide. First group and first column. So we have seven in Avon. Do we have to hide? First row and second column. So we have one in a 13 We have minus one as it is. Three 10 minus 301 No, according to the signs off even. And Lord, we have a is equal to seven minus one minus one minus 310 minus 301 No transport off a in transport off A. We have to convert Rose into columns. So we get seven minus one minus one minus 310 minus 301 No. A inverse is equal to one upon the dominant off is to into a joint a 37 minus three minus three minus 110 minus 101 No. A inverse is seven upon to minus three upon to minus three upon to minus one Upon Do Vanna born do zero minus one upon 201 upon. This is the on. So

Ankit Gupta · Answer

we have determinant. A is equal to five minus three Do minus five three minus two one Zito one. We have to find a inverse if it exists without using calculator. So determinant A is equal to five into three into 13 minus zero into minus two Zito plus three into minus five into one minus five. Minus two into one plus two. Place go into five into zero. Zero three into 13 No, we have determinant off is equal to zero as determinant off is equals to zero Dismay and in verse does not exist This is an

Problem

Let $\mathbf{b}=\left[\begin{array}{r}{-1} \\ {1}…

Problem 10 Medium Difficulty

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

Linear Algebra and Its Applications

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Catherine R.

Alayna H.

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Absolute Value - Example 1

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Linear Algebra and Its Applications

Grace H.

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Absolute Value - Example 1

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

Grace H.

Catherine R.

Alayna H.

Maria G.

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

Linear Algebra and Its Applications