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Gujarati JEE Mathematics : શ્રેણિક

Multiple Choice Questions

21. જો w એ 1નાં અવાસ્તવિક ઘનમૂળ હોય અને $bold A bold space bold equals bold space open square brackets table row bold w bold 0 row bold 0 bold w end table close square brackets bold space bold ત ો bold space bold A to the power of bold 37 bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space$

A
0
-A
A²

22.

bold જ ો bold space bold A bold space bold equals bold space open vertical bar table row bold 5 cell bold 5 bold alpha end cell bold alpha row bold 0 bold alpha cell bold 3 bold alpha end cell row bold 0 bold 0 bold 5 end table close vertical bar bold space bold તથ ા bold space bold vertical line bold A to the power of bold 2 bold vertical line bold space bold equals bold space bold 25 bold space bold હ ો ય bold space bold ત ો bold space bold space bold vertical line bold alpha bold vertical line bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold. bold.

1
25
$bold 1 over bold 5$
5

bold 1 over bold 5

Tips: -

|A²| = 25,

|A|² = 25, આથી |A| = ± 5

$bold therefore open vertical bar table row bold 5 cell bold 5 bold alpha end cell bold alpha row bold 0 bold alpha cell bold 3 bold alpha end cell row bold 0 bold 0 bold 5 end table close vertical bar bold space bold equals bold space bold plus-or-minus bold space bold 5$

25α = ± 5

α = ± $bold 1 over bold 5$

આથી |α| = $bold 1 over bold 5$

23. $open square brackets table row bold 5 cell bold minus bold 7 end cell row cell bold minus bold 2 end cell bold 3 end table close square brackets bold space bold X bold space bold equals bold space open square brackets table row cell bold minus bold 16 end cell cell bold minus bold 6 end cell row bold 7 bold 2 end table close square brackets$ થાય તેવો 2×2 શ્રેણિક X = ........

$open square brackets table row bold 1 cell bold minus bold 4 end cell row cell bold minus bold 3 end cell bold 2 end table close square brackets$
$open square brackets table row bold 1 cell bold minus bold 4 end cell row bold 3 cell bold minus bold 2 end cell end table close square brackets$
$open square brackets table row cell bold minus bold 1 end cell bold 4 row cell bold minus bold 3 end cell bold 2 end table close square brackets$
$open square brackets table row cell bold minus bold 1 end cell bold 4 row bold 3 cell bold minus bold 2 end cell end table close square brackets$

24.

bold P bold space bold equals bold space open square brackets table row bold sinθ bold cosθ row cell bold minus bold cosθ end cell bold sinθ end table close square brackets bold space bold ત ો bold space bold P to the power of bold minus bold 1 end exponent bold space bold equals bold space bold. bold. bold. bold. bold. bold space

-PT
-P
PT
P

25.

જો $open square brackets table row bold 1 cell bold minus bold cotθ end cell row bold cotθ bold 1 end table close square brackets bold space open square brackets table row bold 1 bold cotθ row cell bold minus bold cotθ end cell bold 1 end table close square brackets to the power of bold minus bold 1 end exponent bold space bold equals bold space open square brackets table row bold a cell bold minus bold b end cell row bold b bold a end table close square brackets bold space bold ત ો bold space bold a bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold space bold અન ે bold space bold b bold equals bold space bold semicolon bold space bold theta bold element of bold left curly bracket bold kπ bold space bold vertical line bold k bold element of bold space bold Z bold right curly bracket$

cos2θ, -sin2θ
-cos2θ, -sin2θ
-cos2θ, sin2θ
1,1

26. 3×3 શ્રેણિકો M અને N માટે નીચેનામાંથી કયાં વિધાન સત્ય નથી ?

M અને N એ સંમિત અથવા વિસંમિત શ્રેણિકો હોય, તો અનુક્રમે N^TMN એ સંમિત અથવા વિસંમિત શ્રેણિક છે.
MN - NM એ વિસંમિત શ્રેણિકો હોય, જ્યાં M અને N સંમિત શ્રેણિક છે.
જો M અને N એ સંમિત શ્રેણિકો હોય, તો MN એ સંમિત શ્રેણિક છે.
(adjM)(adjN) = adj(MN) જ્યાં M અને N સામાન્ય શ્રેણિકો છે.

27. જો 3×3 શ્રેણિક A માટે |A| = 3 તો |adj A| = .......

3
-3
9
27

28.

ધારો કે $bold A bold space bold equals bold space open square brackets table row bold 1 bold 0 bold 0 row bold 2 bold 1 bold 0 row bold 3 bold 2 bold 1 end table close square brackets$ અને U₁, U₂, U₃ એ એવા સ્તંભ શ્રેણિકો છે કે જેથી $bold AU subscript bold 1 bold space bold equals bold space open square brackets table row bold 1 row bold 0 row bold 0 end table close square brackets bold semicolon bold space bold AU subscript bold 2 bold space bold equals bold space open square brackets table row bold 2 row bold 3 row bold 0 end table close square brackets bold semicolon bold space bold equals bold space bold AU subscript bold 3 bold space bold equals bold space open square brackets table row bold 2 row bold 3 row bold 1 end table close square brackets$ જો U એ 3×3 શ્રેણિક હોય કે જેના સ્તંંભ અનુક્રમે U₁, U₂, U₃ છે તો |U| = ........

3
-3
2
$bold 3 over bold 2$

29.

bold જ ો bold space bold space bold left square bracket bold space bold 1 bold space bold x bold space bold 1 bold right square bracket bold space open square brackets table row bold 1 bold 3 bold 2 row bold 2 bold 5 bold 1 row bold 15 bold 3 bold 2 end table close square brackets bold space open square brackets table row bold 1 row bold 2 row bold x end table close square brackets bold space bold equals bold space bold 0 bold space bold ત ો bold space bold x bold space bold equals bold space bold. bold. bold. bold. bold. bold. bold. bold. bold space

2, 14
2, 14
-2, 14
-2, -14

30.

જો શ્રેણિક $bold A bold space bold equals bold space open square brackets table row bold 1 bold 2 bold 2 row bold 1 bold 2 cell bold minus bold 2 end cell row bold a bold 2 bold b end table close square brackets$ એ AA^T = 91 નું સમાધાન કરે છે, તો ક્રમયુક્ત (a,b) = .......

(2, 1)
(2, -1)
(-2, -1)
(-2, 1)

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Gujarati JEE Mathematics : શ્રેણિક

Multiple Choice Questions

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