show-that-the-block-upper-triangular-matrix-a-in-example-5-is-invertible-if-and-only-if-both-a_11-an

Question

Show that the block upper triangular matrix $A$ in Example 5 is invertible if and only if both $A_{11}$ and $A_{22}$ are invertible. IHint: If $A_{11}$ and $A_{22}$ are invertible, the formula for $A^{-1}$ given in Example 5 actually works as the inverse of $A . ]$ This fact about $A$ is an important part of several computer algorithms that estimate eigenvalues of matrices. Eigenvalues are discussed in Chapter $5 .$

Runpeng Li · Accepted Answer

Okay, so four problem. 14. I want to show that block block up a triangular matrix. Same example. Five convertible even. Only if Jose Won 1 82 aren&#x27;t vertical. So here&#x27;s the given matrix in, in example five. I have already set it down and by our example five. The result from our example. Five. It says, uh, the members of a ISS equal to such a matrix. It&#x27;s a one y in purse and detective, a 11 in verse and a 12 a lot of times a 22 in verse zero a 22 members. So by observe destructor off the of the avian bird is keeping by our example we find out that only we only require we only require a 11 and a 22 our Ian Birbal and it is okay for a one to to be in an uber go because we don&#x27;t we don&#x27;t know what we don&#x27;t need to calculate the checkmate, the inverse of a 12 As we use our eyes, we apply. It is formula. So that means if that means even if Metro&#x27;s ese invert both, then they won&#x27;t want a one a 22 are both in Birbal and vice versa. Because he&#x27;s obviously this is a equivalent, uh, process as we as we did, or 44 as we as what we did in the example.

Michael Jacobsen · Answer

for this exercise, we&#x27;re starting off with three. And by n matrices we&#x27;re told that the first matrix A is in vertebral. And given the equation that a Times B equals eight time, see, our goal is to show that the Matrix B must be equal to see our strategy here is going to be to start with the equation and subtract a C from both sides. So we have first that a B minus A C is equal to zero. Then we can left factor A from this left side to get a Times B minus C equals zero. Next, let&#x27;s set up some variables here for what be my A. C is going to be equal to let B minus c be equal to this matrix where the first vector I&#x27;m going to denote it by X one. The second column vector is X two and so on. Up until xn. The last column vector. Now that we have a Times B equals zero in this equation, we can use the following idea from matrix multiplication will say, then eight times the first factor X one must be equal to zero vector That&#x27;s becomes B minus. C has the first column, X one next eight times the second, called Vector of B minus E, is equal to the zero vector and so on. We also must insist that eight times the last column vector must be equal to the zero vector as well, since the right hand side is equal to zero matrix. Now we&#x27;re able to invoke the convertible matrix Theorem convertible matrix Syrup says, Let&#x27;s write first by the by m t standing for a veritable matrix serum, we must have that x one. Is this your a vector? X two must be equal to the zero factor and so on up until the last vector, X N must also be equal to zero vector. And the reason for this is since a is convertible, and that was given here from our hypotheses. Now notice that as soon as we said that X ones is euro vector all the way through except Ben. That means this vector is the zero factor. But that corresponds to B minus c. So now we can conclude the following. Let&#x27;s say this shows B minus C equals zero metrics, and so b must be equal to see this is what we want to show. And so this is the end of the proof

Ankit Gupta · Answer

given pushing on subsidies thing. Well then one minus C minus play and e equals. But you know well, you know, in divorce a inverse is equal fun You might need by and when you have one finest minus you like I minus fight. But you have a night drinks one flying and on living it triggered in Waas is equal one divided waas You applied by the metrics one I then thins Oh, one day or did was zero is undefined. Is it? Does no and and in worse.

Mauricio Araiza Canizales · Answer

So in the following problem, we want to show that a is the embers matrix of me. So in other words, we want to show that a physical to be members. One of the properties of this matrices, if that is a is the university. Then a times B must equal I subdue, which is the identity matrix off actuator metrics, which I said too. It&#x27;s just equal to 1100 So then we were gonna prove that. Mr Multiply a times B, we have a matrix 457 Sirrah multiplied by 01 over seven won over five and negative for the vote of a 35. So we&#x27;re gonna move to play this first row by this first column, so we&#x27;ll have four times zero and then five times one of her five. Next, we&#x27;re gonna multiply this first throw by the second column so we&#x27;ll have four attempts one over seven plus five times native for regrettably 35. Now we&#x27;re gonna multiplied this second row by the first columns will have seven time zero plus syrup terms. One of her five. Then we&#x27;re going to multiply the second row by the second column so we&#x27;ll have seven times one over seven plus zero attempts to negative four provided by 35. This time right here is zero is when it&#x27;s also zero when he stole the serum and this is also zero. So we&#x27;re gonna simplify this. So five divided by one of her five is just one four times. One over seven is just four divided by seven. And if we take a five over here, we have that. This is just equal. Your negative four divided by seven. Then this term is just your own. And then seven times one over seven it&#x27;s just one. And so distant right here is also zero. So we have shown that eight times be physical to the identity matrix. Refrigerate your column. Which means that a a sequel to be embers and that be physical to a embers. And this is the solution to the question

Mauricio Araiza Canizales · Answer

So we&#x27;re given two matrices A and B and we want to show that a is the number seven that is one to show that a is being arrested the negative one. So if a is the embers matrix of me, then we know that eight times being must equal the identity matrix free three way three, which is just represented by the year. They only lives once zeros and this side and zeros on this one. So this state into two matrix for a three way three. So if we multiply a times B and we obtained these matrix right here, we have proven that a is the inverse matrix of the so to perform this multiplication. We&#x27;re gonna take this first row off, eh? And multiply it times the first column of B to obtain the first term of the first column of the of the first row. Sorry, which is this one right here being we&#x27;re gonna multiply this frisk row over here times a second column to obtain the second term of the first row under new matrix. Finally, we&#x27;re gonna repeat the process where Third column to find 1/3 element in her first row and we&#x27;re gonna repeat this process with the remaining rose in Matrix A. So this multiplication will look like three times. Negative. Six is just negative. 18. Plus, we&#x27;re gonna at all of these terms eight times seven, which is just 56 then true times. Negative one. Which is just negative too. I&#x27;m gonna draw a line over here. You&#x27;re so we can have enough space to determine which terms belong Toe which wrote. And Colin. Now we&#x27;re gonna multiply three times 84 which is just 252. No, we&#x27;re gonna want to play eight times. Native 26 which is just negative. 208. And finally, two temps native 22 which is just maybe 44. Now, if we do that with the third column, we&#x27;re gonna obtain three times. Native six. It&#x27;s just negative. 18 eight times one is just eight, and then two times five is just fine. You just stand. Sorry. Now for the second for the second row. Sorry. We&#x27;re have one times negative. Six. So have negative. Six one plus seven. Sorry. Want them? 77 in one. Temp&#x27;s 81 Just needed one now. One times 84 is just 84. One times negative. 26 just need 26 1 times. Think 22. Just native training too. And so for the less. For the last time, we were just gonna happen in six. Close one in plus five for our third row. We&#x27;re gonna multiplying this row right here by all of this column shipped in three different, three different terms. So for the 1st 1 we&#x27;ll have five times negative six, which is just native 30 six times seven, which is just 42 12 times 91 which is negative growth. The second term is gonna be composed of five gems 84. Sorry. So have 5 10 city for which is just 400 training six times negative 26 which is just negative. 1 56 and then 12th times negative. 22. It&#x27;s just equal to negative. 2 64 finally, will have negative 30 was six plus 60. And remember that all of this term, this matrix is multiplied by one over 36. So h James b gonna be equal to one over 36 terms of matrix. So this time right here, this one the first term just simplifies to 36. This second term simplifies to zero as well. Of the third, their first term on second column is your zero. These one right here is 36 again, The last German The second column zero as one of the first and second from the third row. And finally this term right here just tickle to 36. And so, if we multiply this matrix by one over 36 we have found that this is indeed the product of the Matrix. A Thames matrix B isn&#x27;t debate into two Matrix, so we can see that a is the embers matrix of B. And so he is also inverse matrix a paid, and this is the answer to the question.

Ankit Gupta · Answer

it is given that the metrics A is equal to be too gross to metrics with entries a B c. D. Now that the dominant off metrics, it is equal to be a minus PC, which is given Toby known zero. So this implies that in was off metrics it exists. Now we know that in worse off, my taxi is equal to bid on at doing Toaff a divided by the determinant of it. Now the eminent is equal to be a T minus Basie. So it becomes one upon a D minus specie. Now that a joint off is a too gross to metrics comes out by using the concept off defectors. Now the co factors off a cool factor if a is equal to B D and the co factors off cool Factor of B is equal to B minus C and the cool factor F C is equal to B minus. P on the cool factor of T is equal to be a nobody taking transports. This will become one upon a T minus pc, but deployed with a metrics with n please d minus B minus c be Now, this is the required inverse off the metrics a

Problem

Suppose the block matrix $A$ on the left side of …

Problem 14 Easy Difficulty

Answer

block upper triangular matrix $A$ is invertible if and only if both the block matrices $A _ { 11 }$ and $A _ { 22 }$ are invertible.

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Alayna H.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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block upper triangular matrix $A$ is invertible if and only if both the block matrices $A _ { 11 }$ and $A _ { 22 }$ are invertible.

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Alayna H.

Michael J.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Anna Marie V.

Alayna H.

Michael J.

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

Linear Algebra and Its Applications