Let the matrix A and B be defined as A - begin bmatrix 382111 - lalpha lend bmatrix and - B - begin bmatrix 3 - 11 -7 - 3lendbmatrix - If wet 2 A - 5 B -1-2- then the number of distinct possible real values of - isA- 2B- 1C- 3D- 0

The correct option is B 1det (2A^5 B^ 1) = 2 2^2 |A|^5/|B| = 2 ⇒4(3 α 2)^5/2 = 2 ⇒ (3 α 2)^5 = 1 ⇒ 3 α 2 = 1 3 α = 1 ⇒α = 1/3 ...

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Byju's Answer

Standard XII

Mathematics

Adjoint of a Matrix

Let the matri...

Question

Let the matrix A and B be defined as $A = [\begin{matrix} 3 & 2 1 & α \end{matrix}] a n d B = [\begin{matrix} 3 & 1 7 & 3 \end{matrix}] . I f d e t (2 A^{5} B^{- 1}) = - 2,$ then the number of distinct possible real values of $α$ is

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Solution

The correct option is B 1
$d e t (2 A^{5} B^{- 1}) = - 2 \frac{2^{2} | A |^{5}}{| B |} = - 2 \Rightarrow \frac{4 (3 α - 2)^{5}}{2} = - 2 \Rightarrow (3 α - 2)^{5} = - 1 \Rightarrow 3 α - 2 = - 1 3 α = 1 \Rightarrow α = \frac{1}{3}$

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Let the matrix A and B be defined as A=[321α]and B=[3173].If det(2A5B−1)=−2, then the number of distinct possible real values of α is

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Let the matrix A and B be defined as $A = [\begin{matrix} 3 & 2 1 & α \end{matrix}] a n d B = [\begin{matrix} 3 & 1 7 & 3 \end{matrix}] . I f d e t (2 A^{5} B^{- 1}) = - 2,$ then the number of distinct possible real values of $α$ is

The correct option is B 1
$d e t (2 A^{5} B^{- 1}) = - 2 \frac{2^{2} | A |^{5}}{| B |} = - 2 \Rightarrow \frac{4 (3 α - 2)^{5}}{2} = - 2 \Rightarrow (3 α - 2)^{5} = - 1 \Rightarrow 3 α - 2 = - 1 3 α = 1 \Rightarrow α = \frac{1}{3}$