Question

aij=1÷2(3i-2j) and A=[aij]2cross 2is​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{a_{ij} = \frac{1}{2} (3i - 2j)}$

TO DETERMINE

$\sf{A = \big[ a_{ij} \big] _{2 \times 2}}$

EVALUATION

Here it is given that

$\displaystyle \sf{a_{ij} = \frac{1}{2} (3i - 2j)}$

$\displaystyle \sf{ a_{11} = \frac{1}{2} (3 - 2) = \frac{1}{2} }$

$\displaystyle \sf{a_{12} = \frac{1}{2} (3 - 4) = - \frac{1}{2} }$

$\displaystyle \sf{a_{21} = \frac{1}{2} (6 - 2) = 2}$

$\displaystyle \sf{a_{22} = \frac{1}{2} (6 - 4) = 1}$

Now

$\sf{A = \big[ a_{ij} \big] _{2 \times 2}}$

$= \displaystyle\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}$

$= \displaystyle\begin{pmatrix} \frac{1}{2} & - \frac{1}{2} \\ 2 & 1 \end{pmatrix}$

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