Math, asked by Tanisha9717, 5 months ago

please help it's urgent ​

Attachments:

Answers

Answered by mathdude500
6

\large\underline{\sf{Solution-}}

Let the required matrix be A whose order is 2 × 3 is

\begin{gathered}\rm :\longmapsto\:\rm Let\:A=\left[\begin{array}{ccc}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{array}\right]\end{gathered}

Now,

According to statement,

\rm :\longmapsto\:a_{ij}  \: =  \: \dfrac{1}{2} \:  |i - 2j|

So,

\rm :\longmapsto\:a_{11}  \: =  \: \dfrac{1}{2} \:  |1 - 2|   =  \dfrac{1}{2}

\rm :\longmapsto\:a_{12}  \: =  \: \dfrac{1}{2} \:  |1 - 4|   =  \dfrac{3}{2}

\rm :\longmapsto\:a_{13}  \: =  \: \dfrac{1}{2} \:  |1 - 6|   =  \dfrac{5}{2}

\rm :\longmapsto\:a_{21}  \: =  \: \dfrac{1}{2} \:  |2 - 2|   =  \dfrac{0}{2} = 0

\rm :\longmapsto\:a_{22}  \: =  \: \dfrac{1}{2} \:  |2 - 4|   =  \dfrac{2}{2} = 1

\rm :\longmapsto\:a_{23}  \: =  \: \dfrac{1}{2} \:  |2 - 6|   =  \dfrac{4}{2} = 2

Hence,

\begin{gathered}\rm :\longmapsto\:\rm \:A=\left[\begin{array}{ccc}\dfrac{1}{2} &\dfrac{3}{2} &\dfrac{5}{2} \\ \\ 0&1&2\end{array}\right]\end{gathered}

Additional Information :-

1. Addition of two matrices A and B is possible only when order of A and B is same.

2. Subtraction of two matrices A and B is possible only when order of A and B is same.

3. Matrix multiplication is possible only when number of columns of pre multiplier is equals to number of rows of post multiplier.

Answered by Abhay00129
3

I dosent know my special one join the meeting aur not

Similar questions