Math, asked by Lakhan50741, 8 months ago

यदि $A-1= \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ और $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$ ,हो तो(48) क्का मान ज्ञात कीजिए।

Answers

Answered by amitnrw

Given : A⁻¹ = $\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ B = $\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$

To Find : (AB)⁻¹ का मान ज्ञात कीजिए

Solution:

(AB)⁻¹ = B⁻¹ A⁻¹

B = $\begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$

B⁻¹ = (adj B) / | B|

$A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33} \end{bmatrix}$

मान | A | = a₁₁ ( a₂₂ * a₃₃ - a₃₂ * a₂₃) - a₁₂ (a₂₁ * a₃₃ - a₃₁ * a₂₃) + a₁₃(a₂₁*a₃₂ - a₃₁ * a₂₂)

| B | = 1 ( 3 - 0) - 2(-1 - 0) - 2(2 - 0)

=> | B | = 3 + 2 - 4

=> | B | = 1

AdjB = $\begin{bmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & a_{32}\\A_{13} & A_{23} & A_{33} \end{bmatrix}$

A₁₁ = 3

A₁₂= 1

A₁₃ = 2

A₂₁ = 2

A₂₂ = 1

A₂₃= 2

A₃₁ = 6

A₃₂ = 2

A₃₃ = 5

AdjB = $\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1& 2 \\ 2 & 2 & 5 \end{bmatrix}$

B⁻¹ = $\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1& 2 \\ 2 & 2 & 5 \end{bmatrix}$

(AB)⁻¹ = B⁻¹ A⁻¹ = $\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1& 2 \\ 2 & 2 & 5 \end{bmatrix}$ $\begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$

=> (AB)⁻¹ = $\begin{bmatrix} 9-30++30 & -3 + 12 - 12 & 3-10+12 \\ 3-15+10& -1+6-4 & 1-5+4 \\ 6-30+25& -2+12-10 & 2-10+10 \end{bmatrix} = \begin{bmatrix} 9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}$