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શ્રેણિક

જો તો નીચે $bold A bold space bold equals bold space open vertical bar table row bold 0 bold 0 bold 1 row bold 0 cell bold minus bold 1 end cell bold 0 row cell bold minus bold 1 end cell bold 0 bold 0 end table close vertical bar bold space$ પૈકીનું કયું સત્ય છે ?

A² = I
A શુન્ય શ્રેણિક છે.
A^-1અસ્તિત્વ નથી
A (-1) I

ધારો કે, $bold A bold space bold equals bold space open square brackets table row bold 1 cell bold minus bold 1 end cell bold 1 row bold 2 bold 1 cell bold minus bold 3 end cell row bold 1 bold 1 bold 1 end table close square brackets bold space bold અન ે bold space bold 10 bold space bold B bold space bold equals bold space open square brackets table row bold 4 bold 2 bold 2 row cell bold minus bold 5 end cell bold 0 bold alpha row bold 1 cell bold minus bold 2 end cell bold 3 end table close square brackets bold comma$ અને જો B એ A નો વ્યસ્ત શ્રેણિક હોય તો α = ........

-1
5
-1
-2

Tips: -

અહીં B = A-1 આપેલ છે.

∴ AB = I

10(AB) = 10 I

∴ A(10 B) = 10 I

$bold therefore bold space open square brackets table row bold 1 cell bold minus bold 1 end cell bold 1 row bold 2 bold 1 cell bold minus bold 3 end cell row bold 1 bold 1 bold 1 end table close square brackets bold space open square brackets table row bold 4 bold 2 bold 2 row cell bold minus bold 5 end cell bold 0 bold alpha row bold 1 cell bold minus bold 2 end cell bold 3 end table close square brackets bold space bold equals bold space bold 10 bold space open square brackets table row bold 1 bold 0 bold 0 row bold 0 bold 1 bold 0 row bold 0 bold 0 bold 1 end table close square brackets bold therefore open square brackets table row cell bold 4 bold plus bold 5 bold plus bold 1 end cell cell bold 2 bold plus bold 0 bold minus bold 2 end cell cell bold 2 bold minus bold alpha bold plus bold 3 end cell row cell bold 8 bold minus bold 5 bold minus bold 3 end cell cell bold 4 bold plus bold 0 bold plus bold 6 end cell cell bold 4 bold plus bold alpha bold minus bold 9 end cell row cell bold 4 bold minus bold 5 bold plus bold 1 end cell cell bold 2 bold plus bold 0 bold minus bold 2 end cell cell bold 2 bold plus bold alpha bold plus bold 3 end cell end table close square brackets bold space bold equals bold space open square brackets table row bold 10 bold 0 bold 0 row bold 0 bold 10 bold 0 row bold 0 bold 0 bold 10 end table close square brackets bold space bold આથ ી bold space open square brackets table row bold 10 bold 0 cell bold 5 bold minus bold alpha end cell row bold 0 bold 10 cell bold alpha bold minus bold 5 end cell row bold 0 bold 0 cell bold alpha bold space bold plus bold 5 end cell end table close square brackets bold equals bold space open square brackets table row bold 10 bold 0 bold 0 row bold 0 bold 10 bold 0 row bold 0 bold 0 bold 10 end table close square brackets$

∴ α + 5 = 10

∴ α = 5 ( α = 5 તો 5 - α = α- 5 = 0)

જો $bold A bold space bold equals bold space open square brackets table row bold 1 bold 0 row bold 1 bold 1 end table close square brackets bold space bold અન ે bold space bold I bold space bold equals bold space open square brackets table row bold 1 bold 0 row bold 0 bold 1 end table close square brackets$ તો ગણિતિય અનુમાનના સિદ્ધાંતથી નીચેના પૈકી કયું સત્ય છે ? (n ≥ 1)

Aⁿ = nA + (n-1)I
Aⁿ - nA - (n-1)I
Aⁿ = 2^n-1 A-(n-1)I
Aⁿ = 2^n-1 = nA + (n -1)I

જો $bold A bold space bold equals bold space open square brackets table row bold 1 bold 2 row bold 3 bold 4 end table close square brackets bold space bold અન ે bold space bold B bold space bold equals bold space open square brackets table row bold a bold 0 row bold 0 bold b end table close square brackets bold comma bold space bold a bold comma bold space bold b bold element of bold N bold space bold ત ો$

અસિમિત સંખ્યામાં B મળે કે જેથી AB = BA
એક B મળે કે જેથી AB = BA
એક કરતાં વધુ પરંતુ સિમિત સંખ્યામાં B મળે જે જેથી AB = BA
AB = BA થાય, તેવો B મળે નહિ

જો $bold A bold space bold equals open square brackets table row bold a bold b row bold b bold a end table close square brackets bold space bold A to the power of bold 2 bold space bold equals bold space open square brackets table row bold alpha bold beta row bold beta bold alpha end table close square brackets bold space bold ત ો bold space bold. bold. bold. bold. bold. bold. bold. bold.$

α= 2ab, β = a² + b²
α= a² + b², β= ab
α= a² + b², β = a² - b²
α= a² + b², β = 2ab

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