IIT JEE & NEET Question Bank with Solution: Matrices And Determinant: Mathematics

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Q.1. If $open square brackets table row 4 2 row 3 4 end table close square brackets apostrophe space then space colon space open vertical bar adj. straight A close vertical bar$

A) 6

B) 16

C) 10

D) -6

Explanation-

$straight A equals open square brackets table row 4 2 row 3 4 end table close square brackets space space space space space space space space space space space space space space space space therefore adj space straight A equals space open square brackets table row 4 cell negative 3 end cell row cell negative 2 end cell 4 end table close square brackets therefore open vertical bar adj space straight A close vertical bar equals left parenthesis 4 right parenthesis left parenthesis 4 right parenthesis minus open parentheses negative 3 close parentheses open parentheses negative 2 close parentheses equals 16 minus 6 equals 10 space space space space space space space space space space space space space... left parenthesis straight c right parenthesis space$

Q.2. If A= $open square brackets table row cosα sinα row cell negative sinα end cell cosα end table close square brackets comma space then space straight A squared equals$

A) $open square brackets table row cell cos 2 alpha end cell cell sin 2 alpha end cell row cell sin 2 alpha end cell cell negative cos 2 alpha end cell end table close square brackets$

B) $open square brackets table row cell cos 2 alpha end cell cell negative sin 2 alpha end cell row cell negative sin 2 alpha end cell cell cos 2 alpha end cell end table close square brackets$

C) $open square brackets table row cell cos 2 straight alpha end cell cell sin 2 straight alpha end cell row cell negative sin 2 straight alpha end cell cell cos 2 straight alpha end cell end table close square brackets$

D) $open square brackets table row cell negative cos 2 straight alpha end cell cell sin 2 straight alpha end cell row cell sin 2 straight alpha end cell cell negative cos 2 straight alpha end cell end table close square brackets$

Explanation-

$Since space straight A squared equals straight A. straight A equals open square brackets table row cosα sinα row cell negative sinα end cell cosα end table close square brackets open square brackets table row cosα sinα row cell negative sinα end cell cosα end table close square brackets equals open square brackets table row cell cos 2 straight alpha end cell cell sin 2 straight alpha end cell row cell negative sin 2 straight alpha end cell cell cos 2 straight alpha end cell end table close square brackets$

Q.3. If the matrix $open square brackets table row 0 1 cell negative 2 end cell row cell negative 1 end cell 0 3 row straight lambda cell negative 3 end cell 0 end table close square brackets$ is singular, then $lambda$ =

A) 3

B) -5

C) 4

D) 2

Explanation-

The matrix A= $open square brackets table row 0 1 cell negative 2 end cell row cell negative 1 end cell 0 3 row straight lambda cell negative 3 end cell 0 end table close square brackets$ is singular $rightwards double arrow$ |A|=0

$rightwards double arrow 0 minus 1 left parenthesis negative 3 straight lambda right parenthesis plus left parenthesis negative 2 right parenthesis left parenthesis 3 right parenthesis equals 0 rightwards double arrow 3 straight lambda minus 6 equals 0 rightwards double arrow straight lambda equals 2$

Q.4. If $omega$ is a cube root of unity and A= $open square brackets table row 1 1 1 row 1 straight omega cell straight omega squared end cell row 1 cell straight omega squared end cell straight omega end table close square brackets comma space then space straight A to the power of negative 1 end exponent equals$

A) $1 fourth open square brackets table row 1 omega cell omega squared end cell row cell omega squared end cell 1 omega row omega cell omega squared end cell 1 end table close square brackets$

B) $1 third open square brackets table row 1 1 1 row 1 cell omega squared end cell omega row 1 omega cell omega squared end cell end table close square brackets$

C) $1 over 6 open square brackets table row 1 omega cell omega squared end cell row 1 cell omega squared end cell omega row 1 1 1 end table close square brackets$

D) $1 half open square brackets table row 1 omega cell omega squared end cell row 1 cell omega squared end cell omega row 1 1 1 end table close square brackets$

Q.5. If A= $open square brackets table row cell negative 1 end cell 0 0 row 0 cell negative 1 end cell 0 row 0 0 cell negative 1 end cell end table close square brackets comma$ then $straight A squared$ is

A) A null matrix

B) A unit matrix

C) a symmetric matrix

D) 2A

Explanation-

$straight A squared equals open square brackets table row cell negative 1 end cell 0 0 row 0 cell negative 1 end cell 0 row 0 0 cell negative 1 end cell end table close square brackets open square brackets table row cell negative 1 end cell 0 0 row 0 cell negative 1 end cell 0 row 0 0 cell negative 1 end cell end table close square brackets equals open square brackets table row 1 0 0 row 0 1 0 row 0 0 1 end table close square brackets equals straight I$

Q.6. If F(a)= $open square brackets table row cosα cell negative sinα end cell 0 row sinα cocsα 0 row 0 0 1 end table close square brackets$ and G( $beta$ )= $open square brackets table row cell cos beta end cell 0 cell sin beta end cell row 0 1 0 row cell negative sin beta end cell 0 cell c o s beta end cell end table close square brackets$ then $open square brackets straight F left parenthesis straight alpha right parenthesis straight G left parenthesis straight beta right parenthesis close square brackets to the power of negative 1 end exponent equals$

A) $straight F left parenthesis straight alpha right parenthesis minus straight G left parenthesis straight beta right parenthesis$

B) $negative straight F left parenthesis straight alpha right parenthesis minus straight G left parenthesis straight beta right parenthesis$

C) $left square bracket straight F left parenthesis straight alpha right parenthesis right square bracket to the power of negative 1 end exponent minus left square bracket straight G left parenthesis straight beta right parenthesis right square bracket to the power of negative 1 end exponent$

D) $left square bracket straight G left parenthesis straight beta right parenthesis right square bracket to the power of negative 1 end exponent left square bracket straight F left parenthesis straight alpha right parenthesis right square bracket to the power of negative 1 end exponent$

Explanation-

Since $left parenthesis AB right parenthesis to the power of negative 1 end exponent equals straight B to the power of negative 1 end exponent. straight A to the power of negative 1 end exponent$

Q.7. let A= $open square brackets straight a subscript ij close square brackets subscript straight n cross times straight n end subscript$ be a square matrix and let $straight c subscript ij$ be the cofactor of $straight a subscript ij$ in A. If C= $open square brackets straight c subscript ij close square brackets comma$ then

A) $open vertical bar straight C close vertical bar equals open vertical bar straight A close vertical bar to the power of straight n$

B) $open vertical bar straight C close vertical bar equals open vertical bar straight A close vertical bar to the power of straight n minus 1 end exponent$

C) $open vertical bar straight C close vertical bar equals open vertical bar straight A close vertical bar to the power of straight n minus 3 end exponent$

D) None of these

Q.8. If A',B' are transpose matrices of the square matrices A,B respectively, then (AB)'=

A) A'B'

B) B'A'

C) B'A

D) A'B

Q.9. The factors of $open vertical bar table row straight x straight a straight b row straight a straight x straight b row straight a straight b straight x end table close vertical bar$ , are

A) x - a, x - b and x + a + b

B) x + a, x + b and x + a + b

C) x + a, x + b and x - a - b

D) x - a, x - b and x - a - b

Q.10. If A is non-singular and (A - 2I) (A - 4I) = O, then $1 over 6 straight A space plus space 4 over 3 straight A to the power of negative 1 end exponent equals$

A) I

B) O

C) 2I

D) 6I

Explanation-

We have,

$left parenthesis straight A space minus space 2 straight I right parenthesis space left parenthesis straight A space minus space 4 straight I right parenthesis space equals space straight O rightwards double arrow straight A squared space minus space 2 straight A space minus space 4 straight A space plus space 8 straight I space equals space straight O rightwards double arrow space straight A squared space minus space 6 straight A space plus space 8 straight I space equals space straight O rightwards double arrow space straight A to the power of negative 1 end exponent left parenthesis straight A squared minus space 6 straight A space plus space 8 straight I right parenthesis space equals space straight A to the power of negative 1 end exponent space straight O rightwards double arrow space straight A space minus 6 straight I space plus space 8 straight A to the power of negative 1 end exponent space equals space straight O rightwards double arrow space straight A space plus space 8 straight A to the power of negative 1 end exponent space equals space 6 straight I space rightwards double arrow 1 over 6 straight A space plus space 4 over 3 space straight A to the power of negative 1 end exponent space equals straight I$

IIT JEE & NEET Question Bank with Solution

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Wednesday, August 18, 2021

Matrices And Determinant: Mathematics

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