Math Chapter Wise

EnglishTGT 2011

Q421.A skew-symmetric matrix cannot be of rank–

A1 B2 C3 D4

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EnglishTGT 2011

Q422.Two matrices A and B are said to be anti commute if –

AAB=BA BAB2 =A2 B CAB=–BA DA B B A =

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EnglishTGT 2011

Q423.Which of the following statement is true–

A. Adj BImage/formula option B (see source) CIf A and B are square matrices of degree n, then Adj (AB) = Adj DAll of the above are false

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EnglishTGT 2011

Q424.If each element of 3 × 3 matrix is multiplying by '3' then determinant of new matrix will be–

A3(detA) B9(detA) C27(detA) D(detA)3

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EnglishTGT 2011

Q425.If system of equation x+2y–3z=2 (K+3)z=3, (2K+1) y+z = 2 has a solution, then the value of 'K' is–

A–3 B−1/2 C1 DNone of these

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EnglishTGT 2013

Q426.Find the value of the given determinant– 13 3 2 5 5 15 26 5 10 3 65 15 5 +

A5 B5 3(6 5) C5 3 D5 3( 6 5)/+

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EnglishTGT 2013

Q427.If 5 a 2 1 a 2 3 2 = then the value of 'a' will–

A±1 B±2 C±3 D±4

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EnglishTGT 2013

Q428.The value of the determinant a b c 2a 2a 2b b c a 2b 2c 2c c a b − − is–

A(a + b + c) B(a + b + c)2 C(a + b + c)3 D0

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EnglishUP PCS (Pre) 1997

Q429.Let A be a square matrix and At be its transpose matrix. Then A–At, is:

ASymmetric matrix BSkew-symmetric matrix CZero matrix DIdentity matrix

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EnglishUP PCS (Pre) 1997

Q430.A square matrix A is invertible if and only if:

Ait has a non- zero element Bdeterminant of A is zero Cdeterminant of A is non-zero Dhas all elements not equal to zero

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EnglishUP PCS (Pre) 1997

Q431.If w3 =1, then the value of the determinant 2 1 w w w w 1 w 1 w is:

A0 B1 C–1 DNone of these

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EnglishUP PCS (Pre) 1997

Q432.If A is a matrix a b, c d then:

AA (adjA) = |A|I B|A–1 | = (|A|)–1 C|adjA–1 | = |A| D|adj A| = |A–1 |

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EnglishTGT 2010

Q433.If A = [x y z], a h g B h b f g f c = and x C y z =, then ABC

A[ax2 – 2hxy + by2 – 2gzx –2fyz + cz2 ] B[ax2 + 2hxy + by2 + 2gzx +2fyz + z2 ] C[ax2 + 2hxy + by2 + 2gzx +2fyz + cz2 ] D[abc + 2fgh – af2 – bg2 – ch2 ]

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EnglishTGT 2010

Q434.The value of determinant 13 16 19 A 14 17 20 is 15 18 21 =

A10 B15 C–15 D0

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EnglishTGT 2010

Q435.The values of x,y,z by Cramer's rule from the equation x+y+z= 4, 2x–y+2z = 5 and x–2y–z =–3 are

A1,1,2 B1,1,3 C1,2,2 DNone of these

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EnglishTGT 2009

Q436.Number of solutions of the given equations. x2–x3=1, –x1+2x3 = – 2, x1–2x2 = 3 will be–

A0 B1 C2 DInfinite

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EnglishTGT 2009

Q437.D = 1 1 1 1 1 x 1 1 1 1 y +, where x ≠ 0, y ≠ 0 then D.

AIs not divisible by 'x' and 'y' both. BIs divisible by 'x' and 'y' both. CIs divisible by 'x' but not 'y' DIs divisible by 'y' but not 'x'

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EnglishTGT 2009

Q438.A= 5 5 0 5, 0 0 5 α if |A2 | = 25 then the value of |α| will be.

A52 B1 C1/5 D5

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EnglishTGT 2009

Q439.The system of equations. –2x + y + z = a, x–2y + z = b, x + y – 2z = c have no solution–

Aa + b + c = 0 Ba + b = c Ca – b – c = 0 DNone of these

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EnglishTGT 2009

Q440.If A2 – A + I = 0, then the inverse of matrix A is–

AI–A BA–I CA DA+I

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