Tag: minors and cofactors

Quizzes (0) Questions (30)

Questions Related to minors and cofactors

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

If $\Delta =\begin{vmatrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{vmatrix}$ and ${ A } _{ ij }$ is cofactors of ${ a } _{ ij }$, then the value of $\Delta $ is given by

  1. ${ a } _{ 11 }{ A } _{ 31 }+{ a } _{ 12 }{ A } _{ 32 }+{ a } _{ 13 }{ A } _{ 33 }$
  2. ${ a } _{ 11 }{ A } _{ 11 }+{ a } _{ 12 }{ A } _{ 21 }+{ a } _{ 13 }{ A } _{ 31 }$
  3. ${ a } _{ 21 }{ A } _{ 11 }+{ a } _{ 22 }{ A } _{ 12 }+{ a } _{ 23 }{ A } _{ 13 }$
  4. ${ a } _{ 11 }{ A } _{ 11 }+{ a } _{ 21 }{ A } _{ 21 }+{ a } _{ 31 }{ A } _{ 31 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Delta =\begin{vmatrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{vmatrix}$


Also given ${ A } _{ ij }$ is cofactor of ${ a } _{ ij }$
$\Delta =a _{11} A _{11}+a _{21} A _{21}+a _{31} A _{31}$

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

If $\triangle =\begin{bmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{bmatrix}$ and ${A} _{2},{B} _{2},{C} _{2}$ are respectively cofactors of ${a} _{2},{b} _{2},{c} _{2}$ then ${a} _{1}{A} _{2}+{b} _{1}{B} _{2}+{c} _{1}{C} _{2}$ is equal to ?

  1. $-\triangle$
  2. $0$
  3. $\triangle$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Co-factor of $ \displaystyle a _2 = (-1)^{i+j}\left|\begin{matrix} b _1 & c _1 \ b _3 & c _3 \end{matrix}\right| = - [b _1 \, c _3 - c _1 \, b _3] = A _2 $

i=2  j=1

of $ \displaystyle b _2 = (-1)^{2+2} \left|\begin{matrix} a _1 & c _1 \\ a _3 & c _3 \end{matrix}\right| = [a _1 \, c _3 - c _1 \, a _3] = B _2 $ 

of $ \displaystyle c _2 = (-1)^{2+3} \left|\begin{matrix} a _1 & b _1 \\ a _3 & b _3 \end{matrix}\right| = -[a _1 \, b _3 \, - b _1 \, a _3 ] = C _2 $

$ a _1 \, A _2 + b _1 \, B _2 +c _1 \, C _2 $

$ \displaystyle -a _1 \, b _1 \, c _3 + a _1 \, c _1 \, b _3 + b _1 \, a _1 \, c _3 - b _1 c _1 \, a _3 -c _1 \, a _1 \, b _3 + c _1 \, b _1 \, a _3 $

$ \displaystyle = 0 $

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 teh value os 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
B Correct answer
Explanation

Given,

$\Delta = \begin{vmatrix}a _1 & b _1 & c _1 \ a _2 & b _2 & c _2\ a _3 & b _3 & c _3\end{vmatrix}$ and $A _1, B _1, C _1$ denote the co-factors of $a _1, b _1, c _1$ respectively.
Now,
 $\begin{vmatrix}A _1 & B _1 & C _1\ A _2 & B _2 & C _2\ A _3 & B _3 & C _3\end{vmatrix}$
$=|adj \Delta|$
$=\Delta^{3-1}$
$=\Delta^2$.

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

If $\Delta  = \left| {\begin{array}{*{20}{c}}  {{a _1}}&{{b _1}}&{{c _1}} \   {{a _2}}&{{b _2}}&{{c _2}} \   {{a _3}}&{{b _3}}&{{c _3}} \end{array}} \right|$ 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 

  1. $ - \Delta $
  2. $0$
  3. $\Delta $

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\Delta =\begin{vmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \\ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \\ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{vmatrix}$

$A _2=c _3\ b _1+b _1\ c _1$

$B _2=a _1\ c _3+a _3\ b _1$

$C _2=-a _1\ b _3+a _3\ b _1$

$\to \ a _1 A _2+b _1 B _2+c _1 C _2$

$a, b, c _3+a, c, b _3+b, a, c _3-b, a, c _3-b, c, a _3-c, a, b _3+b, c, a _3$

$=0$

$B$ is correct.
Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

The value of a third order determinant is $11$, then the value of the square of the determinant formed by the cofactors will be?

  1. $11$
  2. $121$
  3. $1331$
  4. $14641$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
third order determinant = determinant of $3\times 3$ matrix $A$
given $|A|=11$
det (cofactor matrix of $A$) =set (transpare of cofactor amtrix of $A$) (transpare does not change the det)
=det(adjacent of $A$)
$\left\{det\ (cofactor\ matrix\ of\ A) \right\}^2=\left\{det\ (adjacent\ of\ A)\right\}^2$
(Using for an $n\times n\ det\ (cofactor\ matrix\ of\ A)=det\ (A)^{n-1})$
we get
$det\ (cofactor\ matrix\ of\ A)^2=\left\{det (A)^{3-1}\right\}^2$
$=(11)^{2\times 2}=11^4$
$=146.41$
Multiple choice mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

Consider the determinant, $\Delta=\begin{vmatrix} p & q & r \ x & y & z \ l & m & n \end{vmatrix}$ ${M} _{0}$ denotes the minor of an element in $i$th row and $j$th column and ${C} _{ij}$ denotes the cofactor of an element in $i$th row and $j$th column.
The value of $p.{C} _{21}+q.{C} _{22}+r.{C} _{23}$ is equal to

  1. $0$
  2. $-\Delta$
  3. $\Delta$
  4. ${\Delta}^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the property of determinants, if any element is multiplied with cofactor of corresponding element of another row, and summed up for each element of original row, the sum comes out zero $\Rightarrow (A)$

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

The cofactor of the element $4$ in the determinant $\begin{vmatrix} 1 & 3 & 5 & 1\ 2 & 3 & 4 & 2\ 8 & 0 & 1 & 1\ 0 & 2 & 1 & 1\end{vmatrix}$ is?

  1. $4$
  2. $10$
  3. $-10$
  4. $-4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The element 4 is located at the second row, third column (a_23). Its minor is obtained by deleting the second row and third column: det([[1, 3, 1], [8, 0, 1], [0, 2, 1]]). Evaluating this determinant gives 1(0 - 2) - 3(8 - 0) + 1(16 - 0) = -2 - 24 + 16 = -10. The cofactor is (-1)^(2+3) * minor = -1 * (-10) = 10.

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.