5 a find the numbers a and b such that a beginbmatrix 1 a 1...

Name: 5 a find the numbers a and b such that a beginbmatrix 1 a 1 2 endbmatrix and b beginbmatrix 2 2 1 b endbmatrix are similar matrices b find the numbers c d and e such that c beginbmatrix c 0 80433
Uploaded: 2022-02-24T03:21:54-08:00
Duration: 5 min 30 s
Channel: Nick Johnson
Description: 5 a find the numbers a and b such that a beginbmatrix 1 a 1 2 endbmatrix and b beginbmatrix 2 2 1 b endbmatrix are similar matrices b find the numbers c d and e such that c beginbmatrix c 0 80433

5. (a) Find the numbers $a$ and $b$ such that $A = \begin{bmatrix} 1 & a \\ 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 2 \\ 1 & b \end{bmatrix}$ are similar matrices. (b) Find the numbers $c, d,$ and $e$ such that $C = \begin{bmatrix} c & 0 & 0 \\ 3 & 2 & 0 \\ 3 & 2 & 1 \end{bmatrix}$ and $D = \begin{bmatrix} d & e & 0 \\ 1 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ are similar matrices.

Transcript

00:01 Okay, so here we consider the matrices a, which is matrix 101, and b, 0101, and b, 0101, and now we assume that m is equal to the matrix a, b, c, d, and then we consider m -b is equal to a -m, so we set a -b -c -d times 0 -1 -1 -1, equal to 1 -0 -1 times a -b -c -d.

00:25 That's going to give us the matrices.

00:27 We have 0, and then a -b -c -d.

00:31 Plus b and then we have zero and then c plus d and that's going to be equal to ab ab which then is going to imply that a is equal to zero a plus b is equal to b and c plus d is equal to b so therefore for m b equal to m a is going to require that either a is equal to zero or b is equal to c plus d so we take the values for b c and d so it's the matrix should be invertible.

01:04 So since a is equal to zero, so one of the possible values for b, c, and d is b, is 1, c is d, and d is 0, and then the matrices a and b are similar, and one of the possible matrices here is going to be m, which is equal to the matrix 0 -1 -1 -0, so that m -b is equal to am.

01:29 And then for part b, we consider the matrices, so now we have b or a is equal to the matrix with all -1s.

01:41 And then b is equal to 1, negative 1, negative 1, 1, and m, again, it's equal to a, b, c, d.

01:51 So, doing the multiplication, again, mb equal to am, this is now going to give us the matrix c is a minus a minus b, and then minus a plus b, and then c minus a plus b, and then c plus b, that implies that a minus d and negative c plus b is equal to a plus c, and then b plus d.

02:11 B is equal to a plus c, negative a plus b is equal to b plus d, c minus d is equal to a plus c, and negative c and negative c plus d, and negative c, so solving the equations we're going to get for a, b, and c, we get negative b is equal to c, we get negative a is equal to d, we get negative d is equal to a, and we get the negative c is equal to b.

02:39 So therefore, for mb, equals a, m is going to require that a be equal to negative d and b be equal to negative c.

02:52 So taking up for one of the values a, b, c, and d so that m should be revertible.

02:55 The possible values of a, b, c, and d are a is 1, b is 2, c is negative 2, and d is negative 1.

03:02 So the matrices a and b are similar, and one of the possible matrices m is going to be 1, 2, negative 2, negative 1...

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