If A- - aij -2- 2 such that aij-i-j-3 then find A

The correct option is D [ 3 amp; 2; 4 amp; 3 ] Given A is a 2 × 2 matrix with the rule of the elements a_ij = i j + 3 So, ...

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Inverse of a Matrix

If A= [ aij ...

Question

If A= ${[a_{i j}]}_{2 \times 2}$ such that $a_{i j} = i - j + 3$ then find $A$

[\begin{matrix} 2 & 3 4 & 2 \end{matrix}]

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[\begin{matrix} 3 & 4 2 & 3 \end{matrix}]

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[\begin{matrix} 4 & 2 2 & 3 \end{matrix}]

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[\begin{matrix} 3 & 2 4 & 3 \end{matrix}]

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Solution

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Similar questions

A = [a_{i j}]_{2 \times 2}

a_{i j} = i + j

A = [a_{i j}]_{2 \times 2}

a_{i j} = i + j

A

A = | a_{i j} |_{2 \times 2}

a_{i j} = {\begin{matrix} i + j, & i f & i \neq j i^{2} - 2 j, & i f & i = j \end{matrix}

A^{- 1} = ?

A = {[a_{i j}]}_{3 \times 3}

B = {[a_{i j}]}_{3 \times 3}

b_{i j} = 2^{i + j} a_{i j}

i, j

i, j .

| A | = 2

| B | =

\frac{(i + j)^{2}}{2}

a_{i j} = \frac{(i - j)^{2}}{2}

a_{i j} = \frac{(i - 2 j)^{2}}{2}

a_{i j} = \frac{(2 i + j)^{2}}{2}

a_{i j} = \frac{|2 i - 3 i|}{2}

a_{i j} = \frac{|- 3 i + j|}{2}

a_{i j} = e^{2 i x} \sin (x j)

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