Question

Cbegin%7Barray%7D%7B%20c%20c%20c%7D%20%5C%3A%20%5C%3A%20%5C%3A%20%5C%3A%202%20%26%20-3%20%26%20-%205%20%5C%5C%20-%201%20%26%20%5C%3A%20%5C%3A%20%5C%3A%204%20%26%20%5C%3A%20%5C%3A%20%5C%3A%20%5C%3A%205%20%5C%5C%20%5C%3A%20%5C%3A%20%5C%3A%20%5C%3A%201%26%20-%203%20%26%20-%204%5Cend%7Barray%7D%20%5C%3A%20%5C%3A%20%5C%3A%20%5Cright%5D%20%5C%3A%20and%20%5C%3A%20B%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%20c%20c%7D%20%5C%3A%20%5C%3A%20%5C%3A%20%5C%3A%202%20%26%20-%202%20%26%20-%204%20%5C%5C%20-%201%20%26%20%5C%3A%20%5C%3A%20%5C%3A%20%5C%3A%203%20%26%20%5C%3A%20%5C%3A%20%5C%3A%20%5C%3A%204%20%5C%5C%20%5C%3A%20%5C%3A%20%5C%3A%20%5C%3A%201%20%26%20-%202%20%26%20-%203%5Cend%7Barray%7D%20%5Cright%5D%7D%7D%7D%2C%20%7B%20%5Crm%7Bshown%20%20%5C%3A%20that%20%5C%3A%20%20AB%20%3D%20A%20%20%5C%3A%20and%20%20%5C%3A%20BA%20%3D%20B%7D%7D" id="TexFormula1" title=" \\ { \rm{If \: }{\tt{A = \left [ \: \begin{array}{ c c c} \: \: \: \: 2 & -3 & - 5 \\ - 1 & \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: 1& - 3 & - 4\end{array} \: \: \: \right] \: and \: B = \left[\begin{array}{c c c} \: \: \: \: 2 & - 2 & - 4 \\ - 1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & - 2 & - 3\end{array} \right]}}}, { \rm{shown \: that \: AB = A \: and \: BA = B}}" alt=" \\ { \rm{If \: }{\tt{A = \left [ \: \begin{array}{ c c c} \: \: \: \: 2 & -3 & - 5 \\ - 1 & \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: 1& - 3 & - 4\end{array} \: \: \: \right] \: and \: B = \left[\begin{array}{c c c} \: \: \: \: 2 & - 2 & - 4 \\ - 1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & - 2 & - 3\end{array} \right]}}}, { \rm{shown \: that \: AB = A \: and \: BA = B}}" align="absmiddle" class="latex-formula">
$\:$
Don't Spam
Explain well ​

Answer 1

Answer:

Given:

$\\ {\tt{A = \left [ \: \begin{array}{ c c c} \: \: \: \: 2 & -3 & - 5 \\ - 1 & \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: 1& - 3 & - 4\end{array} \: \: \: \right] \: and \: B = \left[\begin{array}{c c c} \: \: \: \: 2 & - 2 & - 4 \\ - 1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & - 2 & - 3\end{array} \right]}}$

Matrix A is of order 3 × 3 and Matrix B is of order 3 × 3

To Show:

• Matrix AB = A , BA = B

Formula Used:

$\\ { \rm{ \left[ \begin{array}{c c c} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array} \right] \times \left[ \begin{array}{c c c} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{array} \right] = \: \left[ \begin{array}{c c c} \: \: a_{11}b_{11} + a_{12}b_{21} + a_{13}b_{31} & a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32} & a_{11}b_{13} + a_{12}b_{23} + a_{13}b_{33} \\ \: \: a_{21}b_{11} + a_{22}b_{21} + a_{23}b_{31} & a_{21}b_{12} + a_{22}b_{22} + a_{23}b_{32}&a_{21}b_{13} + a_{22}b_{23} + a_{23}b_{33} \\ \: \: a_{31}b_{11} + a_{32}b_{21} + a_{33}b_{31}&a_{31}b_{12} + a_{32}b_{22} + a_{33}b_{32} & a_{31}b_{13} + a_{32}b_{23} + a_{33}b_{33}\end{array} \right]}}$

If A is a matrix of order a×b and B is a matrix of order c×d , then matrix AB exits and is of order a×d , if and only if b = c.

$\: \:$

If A is a matrix of order a×b and B is a matrix of order c×d , then matrix BA exits and is of order c×b , if and only if d = a.

$\: \:$

____________________________________________________________________

For Matrix AB, a = 3, b = c = 3, d = 3 , thus Matrix AB is of order 3×3

$\\ { \rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right] \times \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] = \left[\begin{array}{c c c} \: \: \: \: 2(2)-3(-1)-5(1) & \: \: \: \: 2(-2)-3(3)-5(-2) & \: \: \: \: 2(-4)-3(4)-5(-3) \\ -1(2)+4(-1)+5(1) & -1(-2)+4(3)+5(-2) & -1(-4)+4(4)+5(-3)\\ \: \: \: \: 1(2)-3(-1)-4(1) & \: \: \: \: 1(-2)-3(3)-4(-2) & \: \: \: \: 1(-4)-3(4)-4(-3) \end{array} \right]}}$

$\\ {\rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 4 + 3 - 5 & \: \: \: -4 -9 + 10 & \: \: -8 - 12 + 15 \\ -2 - 4 + 5 & \: \: \: \: \: \: + 2 + 12 - 10 & \: \: \: \: \: 4 + 16 - 15 \\ \: \: \: \: \: 2 + 3 - 4 & \: \: -2 - 9 + 8 & -4 - 12 + 12 \end{array} \right] = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right]}}$

$\\ {\rm{Matrix \: AB = \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: 5 \\ \: \: \: \: 1 & -3 & -4 \end{array} \right] = Matrix \: A}}$

Matrix AB = Matrix A

____________________________________________________________________

For Matrix BA, a = 3, b = c = 3, d = 3 , thus Matrix BA is of order 3 × 3

$\\ {\rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] \times \left[ \begin{array}{c c c} \: \: \: \: 2 & -3 & -5 \\ -1 & \: \: \: \: 4 & \: \: \: \: 5 \\ \: \: \: \: \: 1 & -3 & -4 \end{array} \right]}}= \left[ \begin{array}{c c c} \: \: \: \: \: 2(2) -2(-1) -4(1) & \: \: \: \: 2(-3) -2(4) -4(-3) & \: \: \: \: 2(-5) -2(5) -4(-4) \\ \: -1(2) + 3(-1) + 4(1) & -1(-3) + 3(4) +4(-3) & -1(-5) + 3(5) + 4(-4) \\ \: \: \: \: \: 1(2) -2(-1) -3(1) & \: \: \: \: 1(-3) -2(4) -3(-3) & \: \: \: \: 1(-5) -2(5) -3(-4) \end{array} \right]$

$\\{ \rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 4 + 2 - 4 & \: \: \: -6 -8 + 12 & \: \: -10 - 10 + 16 \\ -2 - 3 + 4 & \: \: \: \: \: \: +3 + 12 - 12 & \: \: \: \: \: + 5 + 15 - 16 \\ \: \: \: \: \: 2 + 2 - 3 & \: -3 -8 +9 & \: \: \: \: \: -5 -10 + 12 \end{array} \right] = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right]}}$

$\\{ \rm{Matrix \: BA = \left[ \begin{array}{c c c} \: \: \: \: 2 & -2 & -4 \\ -1 & \: \: \: \: 3 & \: \: \: \: 4 \\ \: \: \: \: 1 & -2 & -3 \end{array} \right] = Matrix \: B}}$

Matrix BA = Matrix B

____________________________________________________________________