Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $A=\left[ \begin{matrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{matrix} \right] $ and $C _{ij}$ is cofactor of $a _{ij}$ in $A$, then value of $|A|$ is given by

  1. $a _{11}C _{31}+a _{12}C _{32}+a _{13}C _{33}$
  2. $a _{11}C _{11}+a _{12}C _{21}+a _{13}C _{31}$
  3. $a _{21}C _{11}+a _{22}C _{21}+a _{23}C _{31}$
  4. $a _{11}C _{11}+a _{21}C _{21}+a _{31}C _{31}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$, then the value of $\lambda$ is

  1. 8

  2. 27

  3. 1

  4. -1
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $D=\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}$


determinant of cofactors is

$D^{c}=\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}=D^2$

$\begin{vmatrix} { a }^{ 2 }+{ \lambda  }^{ 2 } & ab+c\lambda  & ca-b\lambda  \ ab-c\lambda  & { b }^{ 2 }+{ \lambda  }^{ 2 } & bc+a\lambda  \ ca+b\lambda  & bc-a\lambda  & { c }^{ 2 }+{ \lambda  }^{ 2 } \end{vmatrix}\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$

$\Rightarrow D^3= { \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$ -------(1)
Now,
$D=\begin{vmatrix} \lambda  & c & -b \ -c & \lambda  & a \ b & -a & \lambda  \end{vmatrix}$

 $=\lambda(\lambda^2+a^2)-c(-\lambda c-ab)-b(ac-b\lambda)$
 $=\lambda(\lambda^2+a^2+b^2+c^2)$
from (1)
$\left(\lambda(\lambda^2+a^2+b^2+c^2)\right)^3={ \left( 1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right)  }^{ 3 }$
comparing on both sides gives
$\lambda^3=1$ and $\lambda^2=1$
$\therefore \lambda=1$
Hence, option C.

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

$\begin{vmatrix}a^2 + x^2 & ab - cx & ac + bx\ ab+ cx & b^2 + x^2 & bc - ax\ ac - bx & bc + ax & c^2 + x^2\end{vmatrix} =$

  1. $ \begin{vmatrix}x & b & -c\\ -a & x & c\\ a & -b & x\end{vmatrix}^2$
  2. $ \begin{vmatrix}x & -b & c\\ a & x & -c\\ -a & b & x\end{vmatrix}^2$
  3. $ \begin{vmatrix}x & c & -b\\ -c & x & a\\ b & -a & x\end{vmatrix}^2$
  4. $ \begin{vmatrix}x & -c & b\\ c & x & -a\\ -b & a & x\end{vmatrix}^2$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

Let $D = \begin{vmatrix} x & c & -b\ -c & x & a\ b & -a & x\end{vmatrix}$
Cofactors of 1st Row of D are
$x^2 + a^2 ,  ab + cx, ac - bx$
Cofactor of 2nd Row of D are
$ab - cx,x^2 + b^2, ax + bc$
and cofactors of 3rd row of D are
$ax + bx, bc - ax, x^2 + c^2$
$\therefore $ Determinant of cofactors of D is
$D^c

= \begin{vmatrix}x^2 +a^2 & ab + cx & ac - bx\ ab - cx &

x^2 + b^2 & ax + bc\ ac + bx & bc - ax & x^2 +

c^2\end{vmatrix}$
$= \begin{vmatrix}a^2 + x^2 & ab- cx & ac

+ bx\ ab + cx & b^2 + x^2 & bc - ax\ ac - bx & ax +

bc & c^2 + x^2\end{vmatrix}$              (Rows interchanging into columns)
$= D^2$
$= \begin{vmatrix} x & c &-b^2 \ -c

& x & a\ b & -a & x\end{vmatrix}^2$            

($\because D^c = D^2$, D is third order determinant)
Hence,
$\begin{vmatrix}a^2

+ x^2 & ab - cx & ac + bx\ ab+cx & b^2 + x^2 & bc -

ax\ ac - bx & ax + bc &c^2 + x^2 \end{vmatrix}

=  \begin{vmatrix}x & c & -b\ -c & x & a\ b & -a

& x\end{vmatrix}^2$

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\Delta = \begin{vmatrix}a _1 & b _1 & c _1\a _2 & b _2 & c _2\a _3 & b _3 & c _3\end{vmatrix}$ and $A _1, B _1, C _1$ denote the co-factors of $a _1, b _1, c _1$ respectively, then the value of the determinant $\begin{vmatrix}A _1 & B _1 & C _1\A _2 & B _2 & C _2\ A _3 & B _3 & C _3\end{vmatrix}$ is

  1. $\Delta$
  2. $\Delta^2$
  3. $\Delta^3$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $A=\begin{bmatrix} a & c & b\ b & a & c\ c & b & a\end{bmatrix}$ then the cofactor of $a _{32}$ in $A+A^T$ is?

  1. $-(2a(b+c)-(b+c)^2)$
  2. $ac-b^2$
  3. $a^2-bc$
  4. $2a(a+c)-(a+c)^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A + A^T is a symmetric matrix. The element a32 of A+A^T is the sum of the (3,2) element of A and A^T, which is b + c. The cofactor of this element in the resulting matrix involves calculating the minor of the (3,2) position.

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

$\displaystyle A _{1},B _{1},C _{1}$ are respectively the co-factors of $\displaystyle a _{1},b _{1},c _{1}$ of the determinant $\displaystyle \Delta = \begin{vmatrix}a _{1} &b _{1}  &c _{1} \a _{2}  &b _{2}  &c _{2} \a _{3} &b _{3}  &c _{3}\end{vmatrix}$ then $\displaystyle \begin{vmatrix}B _{2} &C _{2} \B _{3} &C _{3}\end{vmatrix}$ equals

  1. $\displaystyle a _{1}a _{3}\Delta $
  2. $\displaystyle (a _{1}-b _{1})\Delta $
  3. $\displaystyle a _{1} \Delta $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For $\Delta =\begin{vmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{vmatrix}$ 
Let $R=\begin{vmatrix} { A } _{ 1 } & { B } _{ 1 } & { C } _{ 1 } \ { A } _{ 2 } & { B } _{ 2 } & { C } _{ 2 } \ { A } _{ 3 } & { B } _{ 2 } & { C } _{ 3 } \end{vmatrix}$ is the matrix of cofactor 
Then ${ a } _{ 1 }\Delta =\begin{vmatrix} { B } _{ 2 } & { C } _{ 2 } \ { B } _{ 2 } & { C } _{ 3 } \end{vmatrix}$

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $\Delta =\begin{vmatrix} a _1 & b _1 & c _1 \ a _2 & b _2 & c _2 \ a _3 & b _3 & c _3\end{vmatrix}$ and $A _2, B _2, C _2$ are respectively cofactors of $a _2, b _2, c _2$ then $a _1A _2 + b _1B _2 + c _1C _2$ is equal to

  1. $-\Delta$

  2. 0
  3. $\Delta$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta =\begin{vmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{vmatrix}\ { A } _{ 2 }=-\begin{vmatrix} { b } _{ 1 } & { c } _{ 1 } \ { b } _{ 3 } & { c } _{ 3 } \end{vmatrix}={ b } _{ 3 }{ c } _{ 1 }-{ b } _{ 1 }{ c } _{ 3 }\ B _{ 2 }=\begin{vmatrix} { a } _{ 1 } & { c } _{ 1 } \ { a } _{ 3 } & { c } _{ 3 } \end{vmatrix}={ c } _{ 3 }{ a } _{ 1 }-{ c } _{ 1 }{ a } _{ 3 }\ C _{ 3 }=-\begin{vmatrix} { a } _{ 1 } & { b } _{ 1 } \ { a } _{ 3 } & { b } _{ 3 } \end{vmatrix}={ a } _{ 3 }{ b } _{ 1 }-{ a } _{ 1 }{ b } _{ 3 }\ \therefore { a } _{ 1 }{ A } _{ 2 }+{ b } _{ 1 }B _{ 2 }+{ c } _{ 1 }C _{ 3 }={ a } _{ 1 }{ b } _{ 3 }{ c } _{ 1 }-{ a } _{ 1 }{ b } _{ 1 }{ c } _{ 3 }+{ a } _{ 1 }{ b } _{ 1 }{ c } _{ 3 }-{ a } _{ 3 }{ b } _{ 1 }{ c } _{ 1 }+{ a } _{ 3 }{ b } _{ 1 }{ c } _{ 1 }-{ a } _{ 1 }{ b } _{ 3 }{ c } _{ 1 }=0$

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

If $A = (a _{ij})$ is a $4\times 4$ matrix and $C _{ij}$ is the co-factor of the element $a _{ij}$ in Det (A), then the expression $a _{11}C _{11} + a _{12}C _{12} + a _{13}C _{13} + a _{14}C _{14}$ equals

  1. $0$
  2. $-1$
  3. $1$
  4. $Det. (A)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The sum of the products of elements of any row and their corresponding cofactors is equal to the determinant of the matrix.

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

Let $A = [a _{ij}] _{n\times n}$ be a square matirx and let $c _{ij}$ be cofactor of $a _{ij}$ in A. If $C = [c _{ij}]$, then

  1. $|C|=|A|$
  2. $|C|=|A|^{n-1}$
  3. $|C|=|A|^{n-2}$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$C\rightarrow$ Cofactor matrix

$AdjA= \left (C \right )^{^{T}}$
But det of $AdjA= Det \quad of  \quad C$
Because they are transpore of each other .
$\Rightarrow\left  | AdjA \right | = \left | C \right |= \left | A \right |^{n-1} $
Option-B

Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

$\begin{vmatrix}1+i & 1-i & i \ 1-i & i & 1+i\ i & 1+i & 1-i\end{vmatrix}$ (where $i=\sqrt {-1}$ ) equals

  1. $7 + 4i$
  2. $7 - 4i$
  3. $4 + 7i$
  4. $4 - 7i$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{vmatrix} 1+i & 1-i & i \ 1-i & i & 1+i \ i & 1+i & 1-i \end{vmatrix}\$


$ =\left( 1+i \right) \begin{vmatrix} i & 1+i \ 1+i & 1-i \end{vmatrix}-\left( 1-i \right) \begin{vmatrix} 1-i & 1+i \ i & 1-i \end{vmatrix}+i\begin{vmatrix} 1-i & i \ i & 1+i \end{vmatrix}\$

$ =\left( 1+i \right) \left( i+1-\left( 1-1+2i \right)  \right) -\left( 1-i \right) \left( 1-1-2i-i+1 \right) +i\left( 1+1+1 \right) \ $

$=\left( 1+i \right) \left( 1-i \right) -\left( 1-i \right) \left( 1-3i \right) +3i\$

$ =1+1-1+3+3i+i+3i$

 $=4+7i$