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Question

Let $\mathbf{b}_{1}=\left[\begin{array}{l}{3} \ {0}\end{array}ight], \mathbf{b}_{2}=\left[\begin{array}{r}{-1} \ {2}\end{array}ight], \mathbf{w}=\left[\begin{array}{r}{7} \ {-2}\end{array}ight], \mathbf{x}=\left[\begin{array}{l}{4} \ {1}\end{array}ight], \quad$ and $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}ight\} .$ Use the figure to estimate $[\mathbf{w}]_{\mathcal{B}}$ and $[\mathbf{x}]_{\mathcal{B}}$ .
Confirm your estimate of $[\mathbf{x}]_{\mathcal{B}}$ by using it and $\left\{\mathbf{b}_{1}, \mathbf{b}_{2}ight\}$ to compute $\mathbf{x} .$

James Chok · Accepted Answer

I get in this question, we want to estimate xB wybie that being using this graph so it&#x27;s too many X B. First, this is equal to well, we have to go positive, too, in the B one direction and negative one in the B two directions. So two and 91 why be well in this case will have to go 1.5 in the B one direction and one in the basic direction at one point by one. That be is equal to negative 045 in the negative, one in the B one direction and negative 0.5 in two directions. Native one and native 0.5. So to confirm our estimates will have for why we have one point. Why would be one? So is there too plus one title be choosers to one just equal to 0.51 point five times. Syracuse is one times two gives you two plus two 1.5 times to give you three. Just one time. One gives you one. This is your four, which is a good buy. Now Ford is dead. We have negative one times like be one. So it is right minus 0.5 ton of feature, which is 21 So this is equal to negative. One time zero gives us no negative. 0.5 times like to give your minus one negative one times are too did in 92 negative 0.5 times by one getting next zero point just equal to make to one negative 2.5, which is equal to that.

Linh Vu · Answer

it was. But the I ego Ju ju one month do that to be you Could you 130 and from here first want to find Ah, I am. Just be. And if I don&#x27;t get any code you which and acclaimed wise here and then we should get big Ojito 34 I&#x27;m honest. You on. I&#x27;m so going to find ah, a ministry Be so I&#x27;ve gotten better gloom guys to minus three temps one so it could U minus one and one month three times. Streak attuned minus eight. Here we have the minus two minus three times there also you Could you banished you? I&#x27;m so what you find in? No, I&#x27;m the far I am blessed to be Cost a four minute you consider the phone I just to be first. So for times to blessed two times one so it could be 10 and far. Attempts one blessed two times three echo to tennis Well and fun times, Months to encourage minus eight times and plus zero equal to minus eight. And now they know me Go to the square root off the 10 square breast 10 square breast when it&#x27;s a square therefore again, they&#x27;re square it on the 100 plus 100 plus 64. Then we get a could discredit it to 64. So we have credible to 64. It was another nice number, so we should give it like that.

Sneha Ravi · Answer

So in this question, we were asked to determine if A and B are in verses off each other. So to do that, let&#x27;s first calculate a B. We have ABC equals 2 to 1 32 times to one minus three to let&#x27;s see if that&#x27;s equal to the identity matrix. So we have in our first point four minus three, which is one second point, we have to plus two, which is four third point. We have six minus six. That 0/4 point. We have three plus four, just seven. So we know that 1407 is not equal to the identity matrix. And so therefore, since a B isnot equals to the identity matrix A and we are not in verses, no, and the answer is no.

Joe Lesueur · Answer

first step in this problem is to solve for the Matrix eight B, I noticed that Matrix A is a two by three matrix and Matrix B is a three by two matrix so multiplying these together will result in a two by two matrix. Do you multiply these two matrices Together? We will first take the first comma B and multiply by the first road off a three times one is three, two times negative. Four is eight and three times A is gonna be three a. Next people take the second column of Matrix B and multiply by the first row of Matrix A for our next value in the two by two matrix. The result be times one is B plus two times a is to a plus three times B is three beats. Next we will multiply the again first calm and measured to be by the second row in matrix a day three times to six negative four times five is negative 20 and seven times a is seven. A next move Multiply the second column in Matrix B by the second row in matrix, a two times be is to be plus five times A It&#x27;s five a plus seven times b is seven beating and this is our matrix. Now, this matrix could use slight simplification before you move forward. So combining like terms, we have the matrix A B, eagle 23 a minus five seven a minus 14 to a plus four b and finally five A plus nine b. The next step in this problem is to solve for a and B lower case and be in order for eight B to equal the identity matrix and a two by two grid. To do this, we know that this matrix is going to look like the following but the diagonal being ones and the outsides being zeros, which is too too identity matrix. To do this, we can set the value three a minus five equal to one. Since we know that this most equal this for this to be an identity matrix doing this, we find that a most equal to not to solve for B, we can plug in this value instead of equal toe one as well. So five and now we know a is too, so we can put two in for a and add mine be which we know should equal want. And this gives us that B equals negative one. And if you plug in those values into these equations, if you plug in the A and B values into this matrix, you will end up getting the identity matrix of 1001 The next stop is part in this problem is to solve for the Matrix B A with the values of lower case and be in mind. So be a remember multiplication is important. We were multiple matrix B, but adding in the values of Lower case and be by matrix. A Sorry, please disregard the line drawn through a problem. And keep in mind that this multiplication is a three by two matrix multiplied by a two by three matrix. The resultant will be a three by three matrix following the same steps as we did before with multiplication. We&#x27;ll start with the first column and major today and multiplied by the first row in Matrix B. And then we will move on to Cinco Matrix B times. The first row, followed by the third column, multiplied by the first row. And then we will continue all three of these for the next rose short In this video, the final answer of Matrix B A. After doing all these multiplication is correctly is the phone.

Sneha Ravi · Answer

So in this question, we are asked to determine whether A and B are in verses off each other. So to do that, let&#x27;s calculate a being and be a and see if it&#x27;s equal to the identity matrix. So a B is 2311 times minus 131 minus two. So here we get minus two plus three, which is one. And then the next point we get six minus six, which is zero. The third point we get minus one plus one, which is zero. In the fourth point, we get three minus two, which is one, so ABC holds to the identity matrix. Now let&#x27;s look at B A. So be here. We have minus 131 minus two times 231 and one. So in our first row first column, we get minus two last three, which is one. And then here we have minus three plus three zero to minus two, which is zero, then three minus two, which is one which is equal to the identity matrix. And so, since baby and be a are equal to the identity matrix, yes, they are. Inverse is off each other. And that&#x27;s our answer

Problem

Let $\mathbf{b}_{1}=\left[\begin{array}{l}{0} \\ …

Problem 7 Easy Difficulty

Answer

$[\mathbf{w}]_{\mathrm{p}}=\left[\begin{array}{l}{2} \\ {-1}\end{array}\right]$ and $[\mathbf{x}]_{\mathrm{p}}=\left[\begin{array}{l}{1.5} \\ {0.5}\end{array}\right]$
$\left[\begin{array}{l}{4} \\ {1}\end{array}\right]$

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

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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$[\mathbf{w}]_{\mathrm{p}}=\left[\begin{array}{l}{2} \\ {-1}\end{array}\right]$ and $[\mathbf{x}]_{\mathrm{p}}=\left[\begin{array}{l}{1.5} \\ {0.5}\end{array}\right]$$\left[\begin{array}{l}{4} \\ {1}\end{array}\right]$

Linear Algebra and Its Applications

Catherine R.

Alayna H.

Caleb E.

Kristen K.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Alayna H.

Caleb E.

Kristen K.

$[\mathbf{w}]_{\mathrm{p}}=\left[\begin{array}{l}{2} \\ {-1}\end{array}\right]$ and $[\mathbf{x}]_{\mathrm{p}}=\left[\begin{array}{l}{1.5} \\ {0.5}\end{array}\right]$
$\left[\begin{array}{l}{4} \\ {1}\end{array}\right]$

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

Linear Algebra and Its Applications