MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Difference of Two Sets

If A =19-8144...

Question

If $A = \frac{1}{9} [\begin{matrix} - 8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & - 8 & 4 \end{matrix}]$ , prove that $A^{- 1} = A^{3}$ .

Open in App

Solution

$A = \frac{1}{9} [- 8 1 4 4 4 7 1 - 8 4] = [\frac{- 8}{9} \frac{1}{9} \frac{4}{9} \frac{4}{9} \frac{4}{9} \frac{7}{9} \frac{1}{9} \frac{- 8}{9} \frac{4}{9}] \Rightarrow A^{T} = [\frac{- 8}{9} \frac{4}{9} \frac{1}{9} \frac{1}{9} \frac{4}{9} \frac{- 8}{9} \frac{4}{9} \frac{7}{9} \frac{4}{9}] = \frac{1}{9} [- 8 4 1 1 4 - 8 4 7 4] . . . (1) |A| = |\frac{- 8}{9} \frac{1}{9} \frac{4}{9} \frac{4}{9} \frac{4}{9} \frac{7}{9} \frac{1}{9} \frac{- 8}{9} \frac{4}{9}| = \frac{1}{9 \times 9 \times 9} |- 8 1 4 4 4 7 1 - 8 4| = \frac{1}{9 \times 9 \times 9} [(- 8 \times 72) - (1 \times 9) + \{4 \times (- 36)\}] = \frac{1}{9 \times 9 \times 9} \times 9 \times \{- 64 - 1 - 16\} = - \frac{9 \times 81}{9 \times 9 \times 9} = - 1 If C_{i j} is a cofactor of a_{i j} such that A = [a_{i j}], then we have C_{11} = \frac{8}{9} C_{12} = \frac{- 1}{9} C_{13} = \frac{- 4}{9} C_{21} = \frac{- 4}{9} C_{22} = \frac{- 4}{9} C_{23} = \frac{- 7}{9} C_{31} = \frac{- 1}{9} C_{32} = \frac{8}{9} C_{33} = \frac{- 4}{9} Now, adj A = {[\frac{8}{9} \frac{- 1}{9} \frac{- 4}{9} \frac{- 4}{9} \frac{- 4}{9} \frac{- 7}{9} \frac{- 1}{9} \frac{8}{9} \frac{- 4}{9}]}^{T} = [\frac{8}{9} \frac{- 4}{9} \frac{- 1}{9} \frac{- 1}{9} \frac{- 4}{9} \frac{8}{9} \frac{- 4}{9} \frac{- 7}{9} \frac{- 4}{9}] ∴ A^{- 1} = \frac{1}{|A|} a d j A = - 1 \times \frac{1}{9} [8 - 4 - 1 - 1 - 4 8 - 4 - 7 - 4] = \frac{1}{9} [- 8 4 1 1 4 - 8 4 7 4] = A^{T} [From (1)] \Rightarrow A^{- 1} = A^{T}$

Suggest Corrections

thumbs-up

0

Similar questions

A = ∣ ∣ ∣ \begin{matrix} 1 & - 1 2 & - 1 \end{matrix} ∣ ∣ ∣

then prove that

A^{- 1} = A^{3}

cos θ = \frac{1}{2} (a + \frac{1}{a})

, then prove that

cos 3 θ = \frac{1}{2} (a^{3} + \frac{1}{a^{3}})

csc A - sin A = a^{3}

sec A - cos A = b^{3}

a^{2} b^{2} (a^{2} + b^{2}) = 1

> 0

, then prove that

(a^{3} + b^{3} + c^{3} + \frac{1}{a^{3} b^{3} c^{3}}) \geq 4

\frac{{sin}^{4} θ}{a} + \frac{{cos}^{4} θ}{b} = \frac{1}{(a + b)}

then prove that

\frac{{sin}^{8} θ}{a^{3}} + \frac{{cos}^{8} θ}{b^{3}} = \frac{2}{(a + b)^{3}}

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Operations on Sets

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Difference of Two Sets

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon