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Question

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Answers

LetA be the given matrix. Find $A^{-1}$. $$ \left[\begin{array}{llll} 3 & 1 & 0 & 0 \\ 1 & 3 & 1 & 0 \\ 0 & 1 & 3 & 1 \\ 0 & 0 & 1 & 3 \end{array}\right] $$

We have to find the rank of metrics equal to metrics one minus one to minus 341020314 And 0102. So in order to find the rank of these metrics, we need to do the column operations and roll operations. First time doing some column operations, that C two changes to C two plus. Even then C three changes to see three minus two C one. And seafood changes to C four plus three C one. Then C two becomes four plus 15, sorry, one plus 104 plus 150 plus 33 and zero plus 11. So we can write the metrics after doing all this column transformations, that is equal to metrics 1000 05 -8 14 0314 0102. Okay now we need to uh we need one in place of this. So here we have one. We can interchange these €2 are two in the change with our four next to a patient. Then we get the metrics 1000 0100 0314 and 05 -84. Now we need to make this to #0 for that. We can do the column operation. Yeah. Mhm. Before that we can do one column operation. Sea food changes to C 4 -2 C2. Thank you. That becomes metrics 1000 01 00. All right then C 4 -2 C. to that is 4 -6 -2 and 4 -10. That is minus six. For -10. uh 10- who that is. Yeah wow Still. So we get 031 -205 -84. Okay regardless metrics now making these two numbers zero. We do the operation uh Three changes to our three minus three or two. Let me get the metrics last weekend. Make one element zero. Okay. 10 05 -84. No. Yeah let's do a column operation C. Four changes to C. Four plus two C. Three. And it becomes the metrics become on 000100001 and minus two plus to that is zero. Then 05 minus eight fourth plus minus 84 plus minus hopeless menacing. It's become Uh huh. Yeah minus two. It no we need to make this element to want the four. We use the C. Four with minus one by 12 C. Phone. Okay? Yeah let me get 100001000010 and 0001. This is the identity matrix of order four. So by uh the definition we can say rank of A is equal to four. That is option B. Thank you

Alright in this problem again, we're wanting to figure out what is ta of X. And so we need to remember this key idea in the green box here. That that notation just means we're going to take the matrix A times the vector X. That were given. All right. So in the first problem were given A. And X. Like this. So what we need to do is just take the matrix A and multiply it by that kind of generic vector X. We're not going to get numbers for answers. We're going to have um variables in our answer this time. So X one X two X three. Right. Remember our matrix multiplication? We're gonna do negative two times X one plus one times X two plus four times X three. And that's gonna go in my first slot. So I'm going to get negative two x one plus one x two. So just X two plus four X three. Okay, for the second part in my vector I'm going to do the three X 15 times the X two and seven times the X three. So three X one plus five X two plus seven X three again just doing matrix multiplication. And in the last spot we've got six times the X 10 times the X two. And negative one times the X three. So that will leave me with six X one. No X choose. So I don't need to write that down and then minus X. Three already. Great. And then let's take a look at the second part of the problem. It's the same idea except this time eh is the matrix negative 11 24 78 And my ex this time is a two by one vector X. One next to. Okay, so again they ask us to find T. Sub A. Of X. And that just means take the matrix A. And multiply it by X. So when I do that this time just going to take that back that matrix negative 11 2478 And just like I did in the previous part multiply that by X. One X. Two. And so I will get for my answer I'll get negative X. One plus X two. Doing my matrix multiplication two X one plus four X two. Yeah. And lastly seven X one plus eight X two people. Okay once again in the problem they tell us to leave our answer in matrix form. So they just want this and this for our answer already. Great. Have a good rest of your assignment.


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