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  • Compute AxB if <img alt="A=\left[\begin{array}{cccc}{1} & {2} & {-1} & {1} \\ {3} & {1} & {\;3} & {5}\end{array}\right], B=\left[\begin{array}{ccc}{-1} & {3} &

Compute AxB if A=\left[\begin{array}{cccc}{1} & {2} & {-1} & {1} \\ {3} & {1} & {\;3} & {5}\end{array}\right], B=\left[\begin{array}{ccc}{-1} & {3} & {2} \\ {\;\;1} & {5} & {2} \\ {-3} & {2} & {4} \\ {-2} & {5} & {6}\end{array}\right]

Option: 1

A\cdot B=\left[\begin{array}{ccc}\;\;\;2 &16 &8 \\ -21 & 45 & 50 \end{array}\right]


Option: 2

A\cdot B=\left[\begin{array}{ccc}2 &16 &8 \\ 21 & 45 & 50 \end{array}\right]


Option: 3

A\cdot B=\left[\begin{array}{ccc}\;\;\;2 &16 &18 \\ -21 & 43 & 50 \end{array}\right]


Option: 4

A\cdot B=\left[\begin{array}{ccc}-5 &16 &8 \\ -21 & 45 & 50 \end{array}\right]


Answers (1)

best_answer

 

 

Multiplication of two matrices -

Matrix multiplication: 

Two matrices  A and B are conformable for the product AB if the number of columns in A and the number of rows in B is equal. Otherwise, these two matrices will be non-conformable for matrix multiplication. So on that basis,

i) AB is defined only if col(A) = row(B)

ii) BA is defined only if col(B) = row(A)

If 

    \\\mathrm{A = \left [ a_{ij} \right ]_{m\times n}} \\\mathrm{\\B=\left [ b_{ij} \right ]_{n\times p}}

    \\\mathrm{C = AB = \left [ c_{ij} \right ]_{m\times p}} \\\mathrm{Where\;\; c_{ij} = \sum_{j=1}^{n}a_{ij}b_{jk}, 1\leq i\leq m,1\leq k\leq p} \\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=a_{i1}b_{1k} + a_{i2}b_{2k} + a_{i3}b_{3k}+ ... + a_{in}b_{nk}}

For examples

\\\mathrm{Suppose,\;two\;matrices\;are\;given}\\\mathrm{A=\begin{bmatrix} a_{11} &a_{12} &a_{13} \\ a_{21} &a_{22} & a_{33} \end{bmatrix}_{2\times3}\;\;\;and\;\;\;B=\begin{bmatrix} b_{11}& b_{12} &b_{13} \\b_{21} &b_{22} &b_{23} \\b_{31} &b_{32} &b_{33} \end{bmatrix}_{3\times3}}\\\\\mathrm{To\:obtain\:the\:entries\:in\:row\:\mathit{i}\:of\:AB,\:we\:multiply\:the\:entries\:in\:row\:\mathit{i}\:of\:A\:by\:}\\\mathrm{column\:\mathit{j}\:in\:B\:and\:add.}\\\mathrm{given\:matrices\:A\:and\:B,\:where\:the\:order\:of\:A\:are\:2\times3\:and\:the\:order\:of\:B\:are\:3\times3,}\\\mathrm{the\:product\:of\:AB\:will\:be\:a\:2\times3\:matrix.}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:1\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:first\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{11}\\b_{21} \\b_{31} \end{bmatrix}=a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}}

\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:2\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:second\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{12}\\b_{22} \\b_{32} \end{bmatrix}=a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}}\\\\\mathrm{To\:obtain\:the\:entry\:in\:row\:1,\:column\:3\:of\:AB,\:multiply\:the\:}\\\mathrm{first\:row\:in\:A\:by\:the\:thired\:column\:in\:B,and\:add.}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\begin{bmatrix} a_{11} &a_{12} &a_{13} \end{bmatrix}\begin{bmatrix} b_{13}\\b_{23} \\b_{33} \end{bmatrix}=a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33}}\\\\\mathrm{We\:proceed\:the\:same\:way\:to\:obtain\:the\:second\:row\:of\:AB.\:In\:other\:words,\:}\\\mathrm{row\:2\:of\:A\:times\:column\:1\:of\:B;}\\\mathrm{row\:2\:of\:A\:times\:column\:2\:of\:B;}\\\mathrm{row\:2\:of\:A\;times\:column\:3\:of\:B.}

\\\mathrm{When\:complete,\:the\:product\:matrix\:will\:be}\\\\\mathrm{AB=\begin{bmatrix} a_{11}\cdot b_{11}+a_{12}\cdot b_{21}+a_{13}\cdot b_{31}\;\;& a_{11}\cdot b_{12}+a_{12}\cdot b_{22}+a_{13}\cdot b_{32}\;\; &a_{11}\cdot b_{13}+a_{12}\cdot b_{23}+a_{13}\cdot b_{33} \\ a_{21}\cdot b_{11}+a_{22}\cdot b_{21}+a_{23}\cdot b_{31} \;\;& a_{21}\cdot b_{12}+a_{22}\cdot b_{22}+a_{23}\cdot b_{32} \;\;& a_{21}\cdot b_{13}+a_{22}\cdot b_{23}+a_{23}\cdot b_{33} \end{bmatrix}}

 

-

 

 

A=\left[\begin{array}{cccc}{1} & {2} & {-1} & {1} \\ {3} & {1} & {\;3} & {5}\end{array}\right], B=\left[\begin{array}{ccc}{-1} & {3} & {2} \\ {\;\;1} & {5} & {2} \\ {-3} & {2} & {4} \\ {-2} & {5} & {6}\end{array}\right]

A\cdot B=\left[\begin{array}{cccc}{1} & {2} & {-1} & {1} \\ {3} & {1} & {\;3} & {5}\end{array}\right]\left[\begin{array}{ccc}{-1} & {3} & {2} \\ {\;\;1} & {5} & {2} \\ {-3} & {2} & {4} \\ {-2} & {5} & {6}\end{array}\right]=\left[\begin{array}{ccc}C_{11} &C_ {12} &C_ {13} \\ C_{21} & C_{22} & C_{23} \end{array}\right]

C_{11} = a_{11}.b_{11} + a_{12}.b_{21} + a_{13}.b_{31} + a_{14}.b_{41} = 1(-1) + 2(1 +(-1)(-3) +1(-2) = 2 \\ \\ C_{12} = a_{11}.b_{12} + a_{12}.b_{22} + a_{13}.b_{32} + a_{14}.b_{42} = 1(3) + 2.(5)+ (-1).2 + 1(5) = 16 \\ \\ C_{13} = a_{11}.b_{13} + a_{12}.b_{23} + a_{13}.b_{33} + a_{14}.b_{43} = 1.(2) + 2. (2) +(-1).4 + 1.(6) = 8 \\ \\ C_{21} = a_{21}.b_{11} + a_{22}.b_{21} + a_{23}.b_{31} + a_{24}.b_{41} = 3.(-1) + 1.(1) + 3 . (-3) + 5.(-2) = -21 \\ \\ C_{22} = a_{21}.b_{12} + a_{22}.b_{22} + a_{23}.b_{32} + a_{24}.b_{42} = 3.(3) + 1.(5) + 3.(2)+5.(5) =45 \\ \\ C_{23} = a_{21}.b_{13} + a_{22}.b_{23} + a_{23}.b_{33} + a_{24}.b_{43} = 3.(2)+1.(2)+3.(4)+5.(6) = 50

A\cdot B=\left[\begin{array}{ccc}C_{11} &C_ {12} &C_ {13} \\ C_{21} & C_{22} & C_{23} \end{array}\right]=\left[\begin{array}{ccc}\;\;\;2 &16 &8 \\ -21 & 45 & 50 \end{array}\right]

hence option (a) is correct

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