Let Upper A equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 4 2nd Row 1st Column 1 2nd Column 3 EndMatrix right bracketA=−2 41 3, and Upper B equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 1 2nd Row 1st Column 3 2nd Column 7 EndMatrix right bracketB=−2 13 7.a. FindABAB,if possible. b. FindBABA,if possible.c. Are the answers in parts a and b the same?d. In general, for matrices A and B such that AB and BA both exist, does AB always equal BA?a. FindABAB,if possible. Select the correct choice below and, ifnecessary, fill in the answer boxes to complete your choice.

Question

rainbowsadie4653 rainbowsadie4653

07-10-2019
Mathematics

contestada

Let Upper A equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 4 2nd Row 1st Column 1 2nd Column 3 EndMatrix right bracketA=−2 41 3, and Upper B equals left bracket Start 2 By 2 Matrix 1st Row 1st Column negative 2 2nd Column 1 2nd Row 1st Column 3 2nd Column 7 EndMatrix right bracketB=−2 13 7.a. FindABAB,if possible. b. FindBABA,if possible.c. Are the answers in parts a and b the same?d. In general, for matrices A and B such that AB and BA both exist, does AB always equal BA?a. FindABAB,if possible. Select the correct choice below and, ifnecessary, fill in the answer boxes to complete your choice.

Respuesta :

Otras preguntas

explain fajan's rule

frika frika · Answer 1 · 2019-10-09T22:47:21+02:00

Answer:

See explanation

Step-by-step explanation:

Given:

[tex]A=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right][/tex]

[tex]B=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right][/tex]

A. Find AB:

[tex]AB=\left[\begin{array}{cc}-2&4\\1&3\end{array}\right]\cdot \left[\begin{array}{cc}-2&1\\3&7\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+4\cdot 3&-2\cdot 1+4\cdot 7\\1\cdot (-2)+3\cdot 3&1\cdot 1+3\cdot 7\end{array}\right]=\left[\begin{array}{cc}16&26\\7&22\end{array}\right][/tex]

B. Find BA:

[tex]BA=\left[\begin{array}{cc}-2&1\\3&7\end{array}\right]\cdot \left[\begin{array}{cc}-2&4\\1&3\end{array}\right]=\left[\begin{array}{cc}-2\cdot (-2)+1\cdot 1&-2\cdot 4+1\cdot 3\\3\cdot (-2)+7\cdot 1&3\cdot 4+7\cdot 3\end{array}\right]=\left[\begin{array}{cc}5&-5\\1&33\end{array}\right][/tex]

C. Answers are not the same

D. Matrices multiplication is not commutastive in general, so

[tex]AB\neq BA[/tex]

MrRoyal MrRoyal · Answer 2 · 2022-04-19T19:13:00+02:00

For matrices A and B such that AB and BA both exist, AB is not always equal to BA

The matrix AB

The matrices are given as:

[tex]A = \left[\begin{array}{cc}-2&4\\1&3\\\end{array}\right][/tex] and [tex]B = \left[\begin{array}{cc}-2&1\\3&7\\\end{array}\right][/tex]

The product of the matrices A and B is calculated as follows:

[tex]AB = \left[\begin{array}{cc}-2*-2 + 4 * 3&-2 * 1+ 4 * 7\\1 * -2 + 3 * 3&1 * 1 + 3 * 7\\\end{array}\right][/tex]

Evaluate the sum

[tex]AB = \left[\begin{array}{cc}16&26\\7&22\\\end{array}\right][/tex]

The matrix BA

The matrices are given as:

[tex]A = \left[\begin{array}{cc}-2&4\\1&3\\\end{array}\right][/tex] and [tex]B = \left[\begin{array}{cc}-2&1\\3&7\\\end{array}\right][/tex]

The product of the matrices B and A is calculated as follows:

[tex]BA = \left[\begin{array}{cc}-2*-2 + 1 * 1&-2 * 4+ 1 * 3\\3 * -2 + 7 * 1&3 * 4 + 7 * 3\\\end{array}\right][/tex]

Evaluate the sum

[tex]BA = \left[\begin{array}{cc}5&-5\\1&33\\\end{array}\right][/tex]

Are the answers the same?

In (a) and (b), we have:

[tex]AB = \left[\begin{array}{cc}16&26\\7&22\\\end{array}\right][/tex] and [tex]BA = \left[\begin{array}{cc}5&-5\\1&33\\\end{array}\right][/tex]

By comparison, both answers are not the same

The conclusion

In general, for matrices A and B such that AB and BA both exist, AB is not always equal to BA