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  • Let <img alt="A=\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}" src="https://learn.careers360.com/latex-image/?A%3D%5Cbegin%7Bbmatrix%7D%201%20%26%202%5C%5C%203%20%26%204%20%5Cend%7Bbmatr

Let A=\begin{bmatrix} 1 & 2\\ 3 & 4 \end{bmatrix}  and B=\begin{bmatrix} a & b\\ c & d \end{bmatrix} are two matrices such that AB = BA and c\neq 0, then value of \frac{a-d}{3b-c} is:

Option: 1 0

Option: 2 2

Option: 3 -2

Option: 4 -1

Answers (1)

best_answer

 

Multiplication of matrices -

 

\huge \large\bigl(\begin{smallmatrix} a_{11} &a_{12} &a_{13} \\ a_{21}& a_{22} &a_{23} \\ a_{31}&a_{32} &a_{33} \end{smallmatrix}\bigr)\times \bigl(\begin{smallmatrix} b_{11} &b_{12} &b_{13} \\ b_{21}& b_{22} &b_{23} \\ b_{31}&b_{32} &b_{33} \end{smallmatrix}\bigr)=

\huge \large\bigl(\begin{smallmatrix} 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{smallmatrix}\bigr)

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AB=\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}\begin{bmatrix} a &b \\ c&d \end{bmatrix}=\begin{bmatrix} a+2c &b+2d \\ 3a+4c&3b+4d \end{bmatrix}

BA=\begin{bmatrix} a &b \\ c&d \end{bmatrix}\begin{bmatrix} 1 &2 \\ 3&4 \end{bmatrix}=\begin{bmatrix} a+3b &2a+4b \\ c+3d &2c+4d \end{bmatrix}

if     AB = BA,        then    a + 2c = a + 3b

\Rightarrow 2c=3b\Rightarrow b\neq 0

b+2d=2a+4b

\Rightarrow 2a-2d=-3b

\frac{a-d}{3b-c}=\frac{-\frac{3}{2}b}{3b-\frac{3}{2}b}=-1

 

Posted by

Pankaj

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