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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

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

α= a² + b², β = 2ab

Tips: -

bold A to the power of bold 2 bold space bold equals bold space bold A bold space bold times bold space bold A bold space bold equals bold space open square brackets table row bold a bold b row bold b bold a end table close square brackets bold space open square brackets table row bold a bold b row bold b bold a end table close square brackets bold space bold equals bold space open square brackets table row cell bold a to the power of bold 2 bold plus bold b to the power of bold 2 end cell cell bold 2 bold ab end cell row cell bold 2 bold ab end cell cell bold a to the power of bold 2 bold plus bold b to the power of bold 2 end cell end table close square brackets open square brackets table row bold alpha bold beta row bold beta bold alpha end table close square brackets bold space bold equals bold space bold space open square brackets table row cell bold a to the power of bold 2 bold plus bold b to the power of bold 2 end cell cell bold 2 bold ab end cell row cell bold 2 bold ab end cell cell bold a to the power of bold 2 bold plus bold b to the power of bold 2 end cell end table close square brackets bold therefore bold space bold alpha bold space bold equals bold space bold a to the power of bold 2 bold space bold plus bold space bold b to the power of bold 2 bold comma bold space bold beta bold space bold equals bold space bold 2 bold ab

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

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

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