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CBSE Questions for Class 12 Commerce Maths Matrices Quiz 4 - MCQExams.com

If A=[0110], then A5=
  • I
  • O
  • A
  • A2
If A=[1203] and B=[3 1], then BA=
  • [3003]
  • [3   0]
  • [3   3]
  • [0  3]
If A=[ahghbfgfc], then A is 

  • a nilpotent matrix
  • an involutory matrix
  • a symmetric matrix
  • an idempotent matrix
 If A = \begin{bmatrix} x & 1\\ 1 & 0 \end{bmatrix} and A^{2} is identity matrix, then x=
  • 1
  • -1
  • \pm 1
  • 0
L=\left[\begin{array}{lll} 2 & 3 & 5\\ 4 & 1 & 2\\ 1 & 2 & 1 \end{array}\right] =P+Q, P  is a symmetric matrix, {Q} is a skew-symmetric matrix then {P} is equal to
  • \left[\begin{array}{lll} 3 & 5 & 6\\ 5 & 6 & 4\\ 9 & 4 & 3 \end{array}\right]
  • \left[\begin{array}{lll} 2 & 3.5 & 3\\ 3.5 & 1 & 2\\ 3 & 2 & 1 \end{array}\right]
  • \left[\begin{array}{lll} 6 & 5 & 4\\ 3 & 6 & 3\\ 5 & 2 & 5 \end{array}\right]
  • \left[\begin{array}{lll} 6 & 5 & 4\\ 4 & 5 & 3\\ 3 & 4 & 3 \end{array}\right]
lf \mathrm{A}=\left[\begin{array}{ll} 2 & -1\\ 3 & -2 \end{array}\right], then \mathrm{A}^{5}=
  • I
  • A
  • -A
  • A^{2}
 If  I=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} and E =\begin{bmatrix} 0 & 1\\ 0 & 0 \end{bmatrix}, then \left ( 2I+3E \right )^{3}= 
  • 8I+ 18E
  • 4I+ 36E
  • 8I +36E
  • 2I+ 3E
Let A and B be 3\times3 matrices such that A^{T}=-A, \, B^{T}=B, then matrix (\lambda AB+3BA) is a skew symmertric matrix for
  • \lambda =3
  • \lambda =-3
  • \lambda =3 or \lambda =-3
  • \lambda =3 and \lambda =-3
If in a square matrix A=\left[ { a }_{ ij } \right] , we find that { a }_{ ij }={ a }_{ ji }\quad \forall \quad i,j , then A is
  • Symmetric 
  • Skew Symmetric
  • Idempotent
  • none of these
\mathrm{A}: If \mathrm{A}=\left\{\begin{array}{ll} 1 & -1\\ -1 & 1 \end{array}\right\} and \mathrm{B}=\left\{\begin{array}{ll} 2 & 2\\ 2 & 2 \end{array}\right\}, then \mathrm{A}\mathrm{B}=0 
\mathrm{R}: If \mathrm{A}\mathrm{B}=0\Rightarrow  \mathrm{A} or \mathrm{B} need not be null matrices.
 The correct answer is 
  • Both A and R are true, R is correct explanation to A
  • Both A and R are true but R is not correct explanation to A
  • A is true, R is false
  • A is false, R is true
If P =  \begin{bmatrix} 1\\ 3\\ 4\end{bmatrix} , Q = \begin{bmatrix} 2 & -1&5 \end{bmatrix} then PQ = 
  • \begin{bmatrix}2 & -1 & 5\\ 6& -3& 15\\ 8& -4 & 20\end{bmatrix}
  • \begin{bmatrix}2 & -3 & 20\end{bmatrix}
  • \begin{bmatrix}2\\ -3\\ 20\end{bmatrix}
  • [19]
If \mathrm{A}=\left(\begin{array}{lll} x & 1 & 4\\ -1 & 0 & 7\\ -4 & -7 & 0 \end{array}\right) such that \mathrm{A}^{\mathrm{T}}=-\mathrm{A}, then \mathrm{x}=
  • -1
  • 0
  • 1
  • 4
\mathrm{If}\mathrm{A}=\left[\begin{array}{lll} 2 & x-3 & x-2\\ 3 & -2 & -1\\ 4 & -1 & -5 \end{array}\right] is a symmetric matrix then \mathrm{x}
  • 0
  • 3
  • 6
  • 8
\begin{bmatrix} 10 & 20 & 30 \\ 20 & 45 & 80 \\ 30 & 80 & 171 \end{bmatrix}=\begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 3 & 4 & 1 \end{bmatrix}\begin{bmatrix} x & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix} then {x}=
  • 10
  • 20
  • 30
  • 40
A=\begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}  then A^{2}-A=
  • \begin{bmatrix} 3 & 0 & 0\\ 0 & 1 & -1\\ 0 & 5 & 4 \end{bmatrix}
  • \begin{bmatrix} 1 & 2 & -1\\ -3 & 1 & 1\\ 3 & 5 & -4 \end{bmatrix}
  • \begin{bmatrix} 3 & 1 & -1\\ -3 & 1 & -1\\ 3 & 5 & 4 \end{bmatrix}
  • \begin{bmatrix} 3 & -1 & 1\\ 3 & -1 & 1\\ 3 & 5 & 4 \end{bmatrix}
1\mathrm{f}\mathrm{A}=\left[\begin{array}{lll} 4 & 1 & 0\\ 1 & -2 & 2 \end{array}\right],\ \mathrm{B}=\left[\begin{array}{lll} 2 & 0 & -1\\ 3 & 1 & 4 \end{array}\right]\mathrm{C}=\left[\begin{array}{l} 1\\ 2\\ -1 \end{array}\right] and (3\mathrm{B}-2\mathrm{A})\mathrm{C}+2\mathrm{X}=0 then \mathrm{X} is equal to
  • {\displaystyle \frac{1}{2}} \begin{bmatrix} 3\\ 13 \end{bmatrix}
  • {\displaystyle \frac{1}{2}}\begin{bmatrix} 3\\ -13 \end{bmatrix}
  • {\displaystyle \frac{1}{2}} \begin{bmatrix} -3\\ 13 \end{bmatrix}
  • \begin{bmatrix} 3\\ -13 \end{bmatrix}
A=\begin{bmatrix} 1 & -3 & -4 \\ -1 & 3 & 4 \\ 1 & -3 & -4 \end{bmatrix} and \mathrm{A}^{2}=\lambda I then \lambda=
  • 0
  • 1
  • \dfrac{1}{2}

  • -2
A : \begin{vmatrix} 0 &p-e  & e-r\\  e-p& 0  &r-p \\  r-e& p-r & 0 \end{vmatrix} =0
R : The determinant of a skew symmetric matrix is zero 
The correct answer is
  • Both A and R are true R is correct explanation to A
  • Both A and R are true but R is not correct explanation to A
  • A is true R is false
  • A is false R is true
A=\left[\begin{array}{lll} 2 & 2 & 1\\ 1 & 2 & 1\\ 3 & 4 & 2 \end{array}\right] then (\mathrm{A}-\mathrm{I}) (\mathrm{A}-2I)=
  • \left[\begin{array}{lll} 5 & 6 & -2\\ 1 & 0 & 7\\ 0 & 1 & 1 \end{array}\right]
  • \left[\begin{array}{lll} 5 & 6 & 3\\ 4 & 6 & 2\\ 7 & 10 & 7 \end{array}\right]
  • \left[\begin{array}{lll} 1 & 0 & 7\\ 7 & 10 & 7\\ 4 & 6 & 2 \end{array}\right]
  • \left[\begin{array}{lll} -1 & 1 & 2\\ 3 & 4 & 1\\ 1 & 1 & 2 \end{array}\right]
lf  A=\begin{bmatrix} 1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2 \end{bmatrix} and  B=\begin{bmatrix} 10 & -4 & -1 \\ -11 & 5 & 0 \\ 9 & -5 & 1 \end{bmatrix} then 
  • \mathrm{A}\mathrm{B}=\mathrm{B}\mathrm{A}
  • \mathrm{A}\mathrm{B}=-\mathrm{A}\mathrm{B}
  • \mathrm{A}\mathrm{B}=2\mathrm{B}\mathrm{A}
  • \mathrm{A}\mathrm{B}=3\mathrm{B}\mathrm{A}
A=\begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix} then A^{3}-35A=

  • \mathrm{A}
  • 2\mathrm{A}
  • 3\mathrm{A}
  • 4\mathrm{A}
A=\left[\begin{array}{ll} 2 & 1\\ 3 & 0 \end{array}\right] then \mathrm{A}^{2}+2\mathrm{A}+I=
  • \left[\begin{array}{ll} 12 & 4\\ 12 & 4 \end{array}\right]
  • \left[\begin{array}{ll} 12 & -4\\ 4 & 12 \end{array}\right]
  • \left[\begin{array}{ll} 4 & 12\\ 12 & 4 \end{array}\right]
  • \left[\begin{array}{ll} 4 & 12\\ -12 & -4 \end{array}\right]
Let \left[\begin{array}{ll} 2 & -2\\-2 & \ 5 \end{array}\right]=\left[\begin{array}{ll} 1 & 0\\ -1 & 1 \end{array}\right]\left[\begin{array}{ll} 2 & 0\\ 0 & x \end{array}\right]\left[\begin{array}{ll} 1 & -1\\ 0 & 1 \end{array}\right], then the value of x is
  • -3
  • -2
  • 0
  • 3
If A=\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}, then additive inverse of A is
  • A^{T}
  • A^{-1}
  • -A
  • -A^{-1}
If A and B are two matrices such that AB = B and BA = A, then A^2 + B^2 is equal to
  • 2AB
  • 2BA
  • A + B
  • AB
If A \times \begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix} = [1 \ \ 2], then A =
  • \left[\dfrac {1}{2} \ \ \ 1\right]
  • \begin{bmatrix} 1 & 2\\ 1 & 0 \end{bmatrix}
  • \begin{bmatrix} 2 & 1\\ 1 & 5 \end{bmatrix}
  • \left[1  \ \ \ \dfrac {1}{2}  \right]
If A is a non-singular matrix, then
  • A^{-1} is symmetric if A is symmeteric
  • A^{-1} is skew-symmetric if A is symmeteric
  • \left| { A }^{ -1 } \right| =\left| A \right|
  • \left| { A }^{ -1 } \right| ={ \left| A \right|  }^{ -1 }
\begin{bmatrix}a\ \ b\end{bmatrix} x \begin{bmatrix}x\\y \end{bmatrix} =           
  • \begin{bmatrix}ax+ay +bx+by \end{bmatrix}
  • \begin{bmatrix}ax \\ by \end{bmatrix}
  • \begin{bmatrix}ax +by \end{bmatrix}
  • \begin{bmatrix}ax\ \ by \end{bmatrix}
lf \left[3x^{2}+10xy+5y^{2} \right]=\begin{bmatrix}x & y \end{bmatrix}A\begin{bmatrix} x\\ y\end{bmatrix}, and {A} is a symmetric matrix then \mathrm{A}=
  • \left[\begin{array}{ll} 3 & 10\\ 10 & 5 \end{array}\right]
  • \left[\begin{array}{ll} 10 & 3\\ 5 & 10 \end{array}\right]
  • \left[\begin{array}{ll} +3 & -5\\ -5 & +5 \end{array}\right]
  • \left[\begin{array}{ll} 3 & 5\\ 5 & 5 \end{array}\right]
If A = \begin{bmatrix} 1 & -1\\ 2 & -1 \end{bmatrix}; B = \begin{bmatrix} 1 & 1\\ 4 & -1 \end{bmatrix}, then A^2 + B^2 =
  • 2 I
  • 4 I
  • \begin{bmatrix} 7 & 0\\ 0 & 7 \end{bmatrix}
  • \begin{bmatrix} 1 & -1\\ 0 & 5 \end{bmatrix}
0:0:1


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Practice Class 12 Commerce Maths Quiz Questions and Answers