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If A=\begin{pmatrix} 2 &2 \\ 9 &4 \end{pmatrix} and I=\begin{pmatrix} 1 &0 \\ 0 &1 \end{pmatrix} , then 10A^{-1} is equal to :
Option: 1 6I-A
Option: 5 A-6I
Option: 9 4I-A
Option: 13 A-4I
 

Answers (1)

best_answer

 

 

Inverse of a Matrix -

A non-singular square matrix “A” is said to be invertible if there exists a non-singular square matrix B such that AB = I = BA, (all matrix are of the same order, they must be for this), then B is called inverse of matrix A.

\\\mathrm{A^{-1} = B \Leftrightarrow AB = \mathbb{I}_n = BA} \\\mathrm{we \; have, } \\ \\\mathrm{A(adj A)=A\mathbb{I}_n} \\\mathrm{\Rightarrow A^{-1}A (adj A)=A^{-1}\mathbb{I}_n|A|} \\\mathrm{\mathbb{I}_n(adj A)=A^{-1}|A|\mathbb{I}_n} \\\mathrm{A^{-1}=\frac{adj A}{\left | A \right |}}

 

Inverse of 2 x 2 matrix

\\\mathrm{Let\;A\;is\;a\;square\;matrix\;of\;order\;2}\\\mathrm{A=\begin{bmatrix} a &b \\ c & d \end{bmatrix}}\\\mathrm{Then,}\\\mathrm{A^{-1}=\begin{bmatrix} a &b \\ c & d \end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix} d &-b \\- c & a \end{bmatrix}}

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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}}

 

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A=\left[\begin{array}{ll}{2} & {2} \\ {9} & {4}\end{array}\right] \text { and } A^{-1}=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]

AA^{-1}=\left[\begin{array}{ll}{2a+2c} & {2b+2d} \\ {9a+4c} & {9b+4d}\end{array}\right] =\left[\begin{array}{ll}{1} & {0} \\ {0} & {1}\end{array}\right]

On comparing we get

a=-\frac{2}{5}, \;\; b=\frac{1}{5},\;\;c=\frac{9}{10}\;\&\;d=-\frac{1}{5}

\Rightarrow \quad 10 A^{-1}=A-61

Correct Option (2)

Posted by

vishal kumar

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