Question

$\begin{bmatrix} 1 & 0 & 0 \\ 0 & cosα & sinα \\ 0 & sinα & -cosα \end{bmatrix}$

Answer 1

Answer:

Here is your answer. .

-1

Step-by-step explanation:

1(-sin square alpha-cos square alpha)

=-1(1)

=1

Answer 2

Given :     $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}$

To Find :  A⁻¹

Solution:

A⁻¹  =  AdjA / | A |

$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}$

| A |  = 1 (-cos²α - sin²α)

=> | A | = -(cos²α + sin²α)

=> | A | = -1

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

A₁₁  = (-1)¹⁺¹ ((cosα)*(-cosα) - (sinα)*(sinα)) = -1

A₁₂ =  (-1)¹⁺² (0*(-cosα) - (0)*(sinα)) = 0

A₁₃ = (-1)¹⁺³(0*(sinα) - (0)*(cosα)) = 0

A₂₁  = (-1)²⁺¹ (0*(-cosα) - (sinα)*(0)) = 0

A₂₂ =  (-1)²⁺² (1*(-cosα) - (0)*(0)) = -cosα

A₂₃ = (-1)²⁺³(1*(sinα) - (0)*(0)) = -sinα

A₃₁  = (-1)³⁺¹ ((0)*(sinα) - (0)*(cosα)) = 0

A₃₂ =  (-1)³⁺² (1*(sinα) - (0)*(0)) = -sinα

A₃₃ = (-1)³⁺³(1*(cosα) - 0*(0)) = cosα

$Adj A = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -cos\alpha & -sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{bmatrix}$      

A⁻¹  =  $- \begin{bmatrix} -1 & 0 & 0 \\ 0 & -cos\alpha & -sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{bmatrix}$

=> A⁻¹    = $\begin{bmatrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{bmatrix}$

    

और सीखें

निम्नलिखित सारणिकों के अवयवों के उपसारणिक एवं सहखण्ड ज्ञात कीजिए

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दिए गए शीर्ष बिन्दुओं वाले त्रिभुजों का क्षेत्रफल ज्ञात कीजिए

brainly.in/question/16386350

A⁻¹

https://brainly.in/question/16386940

A⁻¹ ज्ञात कीजिए

https://brainly.in/question/16386935