If A- 1 2- 2 4 - - B- a 0- 0 b - a-b- N- Then

Explanation for correct option:Finding solution of matrix Given,A=[[ 1 2; 2 4 ]] & B=[[ a 0; 0 b ]] a,b∈ NA=[[ 1 2; 2 4 ]] & B=[[ a 0; 0 b ]] a,b∈ N0ex0ex∴ AB = ...

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Byju's Answer

Standard XII

Mathematics

Sum of n Terms

If A=[[ 1 2; ...

Question

$I f A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] & B = [\begin{matrix} a & 0 \\ 0 & b \end{matrix}] a, b \in N . T h e n$

$T h e r e e x i s t m o r e t h a n o n e b u t n u m b e r o f B ’ s s u c h t h a t A B = B A$

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$T h e r e e x i s t e x a c t l y m o r e B s u c h t h a t A B = B A$

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$T h e r e e x i s t s i n f i n i t e l y m a n y B ’ s s u c h t h a t A B = B A$

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$T h e r e c a n n o t e x i s t a n y B s u c h t h a t A B = B A$

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Solution

The correct option is C
$T h e r e e x i s t s i n f i n i t e l y m a n y B ’ s s u c h t h a t A B = B A$

Explanation for correct option:
Finding solution of matrix
Given,
$A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] & B = [\begin{matrix} a & 0 \\ 0 & b \end{matrix}] a, b \in N$
$A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] & B = [\begin{matrix} a & 0 \\ 0 & b \end{matrix}] a, b \in N ∴ A B = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] [\begin{matrix} a & 0 \\ 0 & b \end{matrix}] = [\begin{matrix} a & 2 b \\ 3 a & 4 b \end{matrix}] B A = [\begin{matrix} a & 0 \\ 0 & b \end{matrix}] [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] = [\begin{matrix} a & 2 a \\ 3 b & 4 b \end{matrix}] S in c e A B = B A, t h e n 2 a = 2 b a = b A B = B A o s p o s s i b l e f o r a l l v a l u e s o f B$
Hence, correct option is $(C)$

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IfA=1224&B=a00ba,b∈N.Then

The correct option is C ThereexistsinfinitelymanyB’ssuchthatAB=BAExplanation for correct option:Finding solution of matrix Given,A=1224&B=a00ba,b∈NA=1224&B=a00ba,b∈N∴AB=1224a00b=a2b3a4bBA=a00b1224=a2a3b4bSinceAB= BA,then2a=2ba=bAB=BAospossibleforallvaluesofBHence, correct option is (C)

$I f A = [\begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix}] & B = [\begin{matrix} a & 0 \\ 0 & b \end{matrix}] a, b \in N . T h e n$