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Minors and cofactors - class-XII

Description: minors and cofactors
Number of Questions: 30
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Tags: matrices and determinants mathematics and statistics determinants and matrices determinants maths
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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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  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. ${ a } _{ 11 }{ A } _{ 11 }+{ a } _{ 21 }{ A } _{ 21 }+{ a } _{ 31 }{ A } _{ 31 }$

Show answer
Correct Option: D
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 ${ a } _{ ij }$
$\Delta =a _{11} A _{11}+a _{21} A _{21}+a _{31} A _{31}$

$A=\left{\begin{array}{ll}
8 & 9\
10 & 11
\end{array}\right}$, then cofactor of $\mathrm{a} _{12}$ is:

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. 11

  2. 10

  3. -11

  4. -10

Show answer
Correct Option: D
Explanation:

By property of cofactor (2),
$A=\begin{bmatrix}
8 & 9\
10 & 11
\end{bmatrix}$
minor of $a _{12}=M _{12}=10$
So, cofactor of $a _{12}=M _{12}(-1)^{1+2}=-M _{12}=-10$

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 ?

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $-\triangle$

  2. $0$

  3. $\triangle$

  4. $none\ of\ these$

Show answer
Correct Option: B
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 $

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-

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $\Delta$

  2. $\Delta^2$

  3. $\Delta^3$

  4. $0$

Show answer
Correct Option: B
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,
 $\begin{vmatrix}A _1 & B _1 & C _1\ A _2 & B _2 & C _2\ A _3 & B _3 & C _3\end{vmatrix}$
$=|adj \Delta|$
$=\Delta^{3-1}$
$=\Delta^2$.

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 

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $ - \Delta $

  2. $0$

  3. $\Delta $

  4. none of these

Show answer
Correct Option: B
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.

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $11$

  2. $121$

  3. $1331$

  4. $14641$

Show answer
Correct Option: D
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
$det\ (cofactor\ matrix\ of\ A)^2=\left\{det (A)^{3-1}\right\}^2$
$=(11)^{2\times 2}=11^4$
$=146.41$

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $0$

  2. $-\Delta$

  3. $\Delta$

  4. ${\Delta}^{2}$

Show answer
Correct Option: A
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 $\Rightarrow (A)$

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?

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $4$

  2. $10$

  3. $-10$

  4. $-4$

Show answer
Correct Option: A

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  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. $a _{11}C _{11}+a _{21}C _{21}+a _{31}C _{31}$

Show answer
Correct Option: A

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. 8

  2. 27

  3. 1

  4. -1

Show answer
Correct Option: C
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.

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  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$

Show answer
Correct Option: C,D
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$

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $\Delta$

  2. $\Delta^2$

  3. $\Delta^3$

  4. $0$

Show answer
Correct Option: A

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?

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $-(2a(b+c)-(b+c)^2)$

  2. $ac-b^2$

  3. $a^2-bc$

  4. $2a(a+c)-(a+c)^2$

Show answer
Correct Option: A

$\displaystyle A _{1},B _{1},C _{1}$ 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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $\displaystyle a _{1}a _{3}\Delta $

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

  3. $\displaystyle a _{1} \Delta $

  4. None of these

Show answer
Correct Option: C
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 ${ a } _{ 1 }\Delta =\begin{vmatrix} { B } _{ 2 } & { C } _{ 2 } \ { B } _{ 2 } & { C } _{ 3 } \end{vmatrix}$

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $-\Delta$

  2. 0

  3. $\Delta$

  4. none of these

Show answer
Correct Option: B
Explanation:

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

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $0$

  2. $-1$

  3. $1$

  4. $Det. (A)$

Show answer
Correct Option: D

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $|C|=|A|$

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

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

  4. none of these

Show answer
Correct Option: B
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

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

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $7 + 4i$

  2. $7 - 4i$

  3. $4 + 7i$

  4. $4 - 7i$

Show answer
Correct Option: C
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$

 $=4+7i$

If $A=\begin{bmatrix} 1 & -2 & 3 \ 4 & 0 & -1 \ -3 & 1 & 5 \end{bmatrix}$, then ${(adj. A)} _{23}$ is equal to

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $13$

  2. $-13$

  3. $5$

  4. $-5$

Show answer
Correct Option: A
Explanation:

$A=\begin{bmatrix} 1 & -2 & 3 \ 4 & 0 & -1 \ -3 & 1 & 5 \end{bmatrix}$


${(adj. A)} _{23}={C} _{32}$

So cofactor of ${a} _{32}$


${C} _{32}={(-1)}^{3+2}(-1-12)=13$

Ans: A

Consider the determinant $\Delta=\begin{vmatrix}a _1 & a _2 & a _3 \\ b _1 & b _2 & b _3 \\ c _1 & c _2 & c _3\end{vmatrix}$
$M _{ij} =$ Minor of the element of $i^{th}$ row & $j^{th}$ column.
$C _{ij} =$ Cofactor of element of $i^{th}$ row & $j^{th}$ column.

$a _3M _{13} - b _3M _{23} + c _3M _{33}$ is equal to

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $0$

  2. $4\Delta$

  3. $2\Delta$

  4. $\Delta$

Show answer
Correct Option: D
Explanation:

${ a } _{ 3 }{ M } _{ 13 }-{ b } _{ 2 }{ M } _{ 23 }+{ c } _{ 3 }{ M } _{ 33 }$


$ ={ a } _{ 3 }\begin{vmatrix} { b } _{ 1 }\quad  & { b } _{ 2 } \ { c } _{ 1 } & { c } _{ 2 } \end{vmatrix}-{ b } _{ 3 }\begin{vmatrix} { a } _{ 1 }\quad  & { a } _{ 2 } \ { c } _{ 1 } & { c } _{ 2 } \end{vmatrix}+{ c } _{ 3 }\begin{vmatrix} { a } _{ 1 }\quad  & { a } _{ 2 } \ { b } _{ 1 } & { b } _{ 2 } \end{vmatrix}$

Is equal to the expansion of $\triangle $ along ${ C } _{ 3 }$

Consider the determinant $\Delta=\begin{vmatrix}a _1 & a _2 & a _3 \\ b _1 & b _2 & b _3 \\ c _1 & c _2 & c _3\end{vmatrix}$
$M _{ij} =$ Minor of the element of $i^{th}$ row & $j^{th}$ column.
$C _{ij} =$ Cofactor of element of $i^{th}$ row & $j^{th}$ column.

$a _2.C _{12} + b _2.C _{22} + c _2.C _{32}$ is equal to

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $0$

  2. $\Delta$

  3. $2\Delta$

  4. $\Delta^2$

Show answer
Correct Option: B
Explanation:

The value of ${ a } _{ 2 }.{ C } _{ 12 }+{ b } _{ 2 }.{ C } _{ 22 }+{ C } _{ 2 }.{ C } _{ 32 }$


$ ={ a } _{ 2 }{ \left( -1 \right)  }^{ 1+2 }\begin{vmatrix} { b } _{ 1 }\quad  & { b } _{ 3 } \ { c } _{ 1 } & { c } _{ 3 } \end{vmatrix}+{ b } _{ 2 }.{ \left( -1 \right)  }^{ 2+2 }\begin{vmatrix} { a } _{ 1 }\quad  & { a } _{ 3 } \ { c } _{ 1 } & { a } _{ 3 } \end{vmatrix}+{ c } _{ 2 }.{ \left( -1 \right)  }^{ 3+2 }\begin{vmatrix} { a } _{ 1 }\quad  & { a } _{ 3 } \ { b } _{ 1 } & { b } _{ 3 } \end{vmatrix}$

$ =-{ a } _{ 2 }\begin{vmatrix} { b } _{ 1 }\quad  & { b } _{ 3 } \ { c } _{ 1 } & { c } _{ 3 } \end{vmatrix}+{ b } _{ 2 }\begin{vmatrix} { a } _{ 1 }\quad  & { a } _{ 3 } \ { c } _{ 1 } & { a } _{ 3 } \end{vmatrix}-{ c } _{ 2 }\begin{vmatrix} { a } _{ 1 }\quad  & { a } _{ 3 } \ { b } _{ 1 } & { b } _{ 3 } \end{vmatrix}$

Is same as expansion of $\triangle $ along ${ C } _{ 2 }$

Consider the determinant $\Delta=\begin{vmatrix}a _1 & a _2 & a _3 \\ b _1 & b _2 & b _3 \\ c _1 & c _2 & c _3\end{vmatrix}$
$M _{ij} =$ Minor of the element of $i^{th}$ row & $j^{th}$ column.
$C _{ij} =$ Cofactor of element of $i^{th}$ row & $j^{th}$ column.

Value of $b _1.C _{31} + b _2.C _{32} + b _3.C _{33}$ is

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $0$

  2. $\Delta$

  3. $2\Delta$

  4. $\Delta^2$

Show answer
Correct Option: A
Explanation:

Value of ${ b } _{ 1 }.{ C } _{ 31 }+{ b } _{ 2 }.C _{ 32 }+b _{ 3 }.{ C } _{ 33 }$


$={ b } _{ 1 }.{ \left( -1 \right)  }^{ 3+1 }\begin{vmatrix} { a } _{ 2 }\quad  & { a } _{ 3 } \ { b } _{ 2 } & { b } _{ 3 } \end{vmatrix}+{ b } _{ 3 }.{ \left( -1 \right)  }^{ 3+2 }\begin{vmatrix} { a } _{ 1 }\quad  & { a } _{ 3 } \ { b } _{ 1 } & { b } _{ 3 } \end{vmatrix}+{ b } _{ 3 }.{ \left( -1 \right)  }^{ 3+3 }\begin{vmatrix} { a } _{ 1 }\quad  & { a } _{ 3 } \ { b } _{ 1 } & { b } _{ 2 } \end{vmatrix}$


$={ b } _{ 1 }\left( { b } _{ 2 }{ a } _{ 3 }-{ a } _{ 2 }{ b } _{ 3 } \right) -{ b } _{ 2 }\left( { b } _{ 1 }{ a } _{ 3 }-{ a } _{ 1 }{ b } _{ 3 } \right) +{ b } _{ 3 }\left( { b } _{ 1 }{ a } _{ 3 }-{ a } _{ 1 }{ b } _{ 2 } \right)$

$=0$

$A,B,C$ are cofactors of elements, $\mathrm{a},\ \mathrm{b},\ \mathrm{c}$ in


${\begin{bmatrix}
a & b & c\
2 & 4 & 7\
-1 & 0 & 3
\end{bmatrix}}$ then the value of $(2\mathrm{A}+4\mathrm{B}+7\mathrm{C})$
is equal to

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $0$

  2. 2

  3. $-1$

  4. 4

Show answer
Correct Option: A
Explanation:

$A = 4\times 3-0\times 7 = 12$
$B = -(2\times 3-7\times (-1)) = -13$
$C = 2\times 0-4\times (-1) = 4$
$2A+4B+7C = 0$

If $\displaystyle A=\left[ { a } _{ ij } \right] $ is a $4 \times 4$ matrix and $\displaystyle { c } _{ ij }$ is the co-factor of the element $\displaystyle { a } _{ ij }$ in $\displaystyle \left| A \right| $, then the expression $\displaystyle { a } _{ 11 }{ c } _{ 11 }+{ a } _{ 12 }{ c } _{ 12 }+{ a } _{ 13 }{ c } _{ 13 }+{ a } _{ 14 }{ c } _{ 14 }$ equals

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $0$

  2. $-1$

  3. $1$

  4. $\displaystyle \left| A \right| $

Show answer
Correct Option: D
Explanation:

$A=\left[ { a } _{ ij } \right] \quad 4\times 4$

${ c } _{ ij }\rightarrow $co factor

$A=\begin{bmatrix} { a } _{ 11 } & { a } _{ 12 } & { a } _{ 13 } \\ { a } _{ 21 } & { a } _{ 22 } & { a } _{ 23 } \\ { a } _{ 31 } & { a } _{ 32 } & { a } _{ 33 } \end{bmatrix}$ co factor $=$ Minor $\times \begin{matrix} + & - & + \\ - & + & - \\ + & - & + \end{matrix}$

Co factor matrix$=\begin{bmatrix} { a } _{ 22 }{ a } _{ 33 }-{ a } _{ 23 }{ a } _{ 32 } & -{ a } _{ 21 }{ a } _{ 33 }+{ a } _{ 23 }{ a } _{ 31 } & { a } _{ 21 }{ a } _{ 32 }-{ a } _{ 22 }{ a } _{ 31 } \\ { -a } _{ 12 }{ a } _{ 33 }+{ a } _{ 13 }{ a } _{ 32 } & { a } _{ 11 }{ a } _{ 33 }-{ a } _{ 13 }{ a } _{ 31 } & -{ a } _{ 11 }{ a } _{ 32 }+{ a } _{ 12 }{ a } _{ 31 } \\ { a } _{ 12 }{ a } _{ 23 }-{ a } _{ 13 }{ a } _{ 22 } & -{ a } _{ 11 }{ a } _{ 23 }+{ a } _{ 13 }{ a } _{ 21 } & { a } _{ 11 }{ a } _{ 22 }-{ a } _{ 12 }{ a } _{ 21 } \end{bmatrix}$

${ c } _{ 11 }={ a } _{ 22 }{ a } _{ 33 }-{ a } _{ 23 }{ a } _{ 32 }$

${ c } _{ 12 }={ a } _{ 23 }{ a } _{ 31 }-{ a } _{ 21 }{ a } _{ 33 }$

${ c } _{ 13 }={ a } _{ 21 }{ a } _{ 32 }-{ a } _{ 22 }{ a } _{ 31 }$

${ a } _{ 11 }{ c } _{ 11 }+{ a } _{ 12 }{ c } _{ 12 }+{ a } _{ 13 }{ c } _{ 13 }$

$=\left| A \right| $

Parallelly For $4\times 4$ matrix

Also

${ a } _{ 11 }{ c } _{ 11 }+{ a } _{ 12 }{ c } _{ 12 }+{ a } _{ 13 }{ c } _{ 13 }+{ a } _{ 14 }{ c } _{ 14 }=\left| A \right| $

Option D

If in $\displaystyle \left[ \begin{matrix} { a } _{ 1 } \ { a } _{ 2 } \ { a } _{ 3 } \end{matrix}\begin{matrix} { b } _{ 1 } \ { b } _{ 2 } \ { b } _{ 3 } \end{matrix}\begin{matrix} { c } _{ 1 } \ { c } _{ 2 } \ { c } _{ 3 } \end{matrix} \right] $, the cofactor of $\displaystyle { a } _{ r }$ is $\displaystyle { A } _{ r }$, then $\displaystyle { c } _{ 1 }{ A } _{ 1 }+{ c } _{ 2 }{ A } _{ 2 }+{ c } _{ 3 }{ A } _{ 3 }$ is 

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $\displaystyle 0$

  2. $\displaystyle -D$

  3. $\displaystyle D$

  4. $\displaystyle { D }^{ 2 }$

Show answer
Correct Option: A
Explanation:
$A\begin{bmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \\ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \\ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{bmatrix}$
Co-factor$\begin{bmatrix} { A } _{ 1 } & { B } _{ 1 } & { C } _{ 1 } \\ { A } _{ 2 } & { B } _{ 2 } & { C } _{ 2 } \\ { A } _{ 3 } & { B } _{ 3 } & { C } _{ 3 } \end{bmatrix}$
${ A } _{ 1 }={ b } _{ 2 }{ c } _{ 3 }-{ c } _{ 2 }{ b } _{ 3 }$
${ A } _{ 2 }=-\left( { b } _{ 1 }{ c } _{ 3 }-{ c } _{ 1 }{ b } _{ 3 } \right) $
${ A } _{ 3 }={ b } _{ 1 }{ c } _{ 2 }-{ c } _{ 1 }{ b } _{ 2 }$
${ C } _{ 1 }{ A } _{ 1 }+{ C } _{ 2 }{ A } _{ 2 }+{ C } _{ 3 }{ A } _{ 3 }$
${ c } _{ 1 }{ c } _{ 3 }{ b } _{ 2 }-{ c } _{ 1 }{ c } _{ 2 }{ b } _{ 3 }-{ c } _{ 2 }{ c } _{ 3 }{ b } _{ 1 }+{ c } _{ 1 }{ c } _{ 2 }{ b } _{ 3 }$
${ c } _{ 2 }{ c } _{ 3 }{ b } _{ 1 }-{ c } _{ 1 }{ c } _{ 3 }{ b } _{ 2 }=0$
Option A

If $A=\begin{bmatrix} 3 & 2 & 4 \ 1 & 2 & 1 \ 3 & 2 & 6 \end{bmatrix}$ and $A _{ij}$ are the cofactors of $a _{ij}$, then $a _{11}A _{11}+a _{12}A _{12}+a _{13}A _{13}$ is equal to

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $8$

  2. $6$

  3. $4$

  4. $0$

Show answer
Correct Option: A
Explanation:

$a _{11}A _{11}+a _{12}A _{12}+A _{13}A _{13}$
$=3\begin{vmatrix} 2 & 1\ 2 & 6\end{vmatrix} -2\begin{vmatrix} 1 & 1\3 & 6\end{vmatrix} +4\begin{vmatrix} 1 &2 \ 3 & 2 \end{vmatrix}$
$=3(12-2)-2(6-3)+4(2-6)$
$=30-6-16$
$=8$

If ${A} _{1}, {B} _{1}, {C} _{1}..$ are respectively the co-factor of the elements ${a} _{1}, {b} _{1}, {c} _{1}$.
$\triangle =\begin{vmatrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \ a _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{vmatrix}$, then $\begin{vmatrix} { B } _{ 2 } & C _{ 2 } \ B _{ 3 } & C _{ 3 } \end{vmatrix}$

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. ${a} _{1}\triangle$

  2. ${a} _{1}{a} _{3}\triangle$

  3. $({a} _{1}+{b} _{1})\triangle$

  4. $None\ of\ these$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} { B _{ 2 } }={ a _{ 1 } }{ c _{ 3 } }-{ a _{ 3 } }{ c _{ 1 } } \ { c _{ 2 } }=-\left( { { a _{ 1 } }{ b _{ 3 } }-{ a _{ 3 } }{ b _{ 1 } } } \right)  \ { B _{ 3 } }=-\left( { { a _{ 1 } }{ c _{ 2 } }-{ a _{ 2 } }{ c _{ 1 } } } \right)  \ { c _{ 3 } }={ a _{ 1 } }{ b _{ 2 } }-{ b _{ 1 } }{ a _{ 2 } } \ \left| \begin{array}{l} { B _{ 2 } } & { C _{ 2 } } \ { B _{ 3 } } & { C _{ 3 } } \end{array} \right| =\left| \begin{array}{l} { a _{ 1 } }{ c _{ 3 } }-{ a _{ 3 } }{ c _{ 1 } } & -{ a _{ 1 } }{ b _{ 3 } }+{ a _{ 3 } }{ b _{ 1 } } \ -{ a _{ 1 } }{ c _{ 2 } }+{ a _{ 2 } }{ c _{ 1 } } & { a _{ 1 } }{ b _{ 2 } }-{ b _{ 1 } }{ a _{ 2 } } \end{array} \right|  \ =\left| \begin{array}{l} { a _{ 1 } }{ c _{ 3 } } & -{ a _{ 1 } }{ b _{ 3 } } \ -{ a _{ 1 } }{ c _{ 2 } } & { a _{ 1 } }{ b _{ 2 } } \end{array} \right| +\left| \begin{array}{l} { a _{ 1 } }{ c _{ 3 } } & { a _{ 3 } }{ b _{ 1 } } \ -{ a _{ 1 } }{ c _{ 2 } } & -{ a _{ 2 } }{ b _{ 1 } } \end{array} \right| +\left| \begin{array}{l} -{ a _{ 3 } }{ c _{ 1 } } & -{ a _{ 1 } }{ b _{ 3 } } \ { a _{ 2 } }{ c _{ 1 } } & { a _{ 1 } }{ b _{ 2 } } \end{array} \right|  \ 1\left| \begin{array}{l} -{ a _{ 3 } }{ c _{ 1 } } & { a _{ 3 } }{ b _{ 1 } } \ { a _{ 2 } }{ c _{ 1 } } & -{ a _{ 2 } }{ b _{ 1 } } \end{array} \right|  \ =a _{ 1 }^{ 2 }\left| \begin{array}{l} { c _{ 3 } } & -{ b _{ 3 } } \ -{ c _{ 2 } } & { b _{ 2 } } \end{array} \right| +{ a _{ 1 } }{ b _{ 1 } }\left| \begin{array}{l} { c _{ 3 } } & { a _{ 3 } } \ -{ c _{ 2 } } & -{ a _{ 2 } } \end{array} \right| +{ a _{ 1 } }c\left| \begin{array}{l} -{ a _{ 3 } } & -{ b _{ 3 } } \ { a _{ 2 } } & { b _{ 2 } } \end{array} \right| +{ b _{ 1 } }{ c _{ 1 } }\left| \begin{array}{l} -{ a _{ 3 } } & { a _{ 3 } } \ { a _{ 2 } } & -{ a _{ 2 } } \end{array} \right|  \ ={ a _{ 1 } }\left{ { { a _{ 1 } }\left( { { b _{ 2 } }{ c _{ 3 } }-{ b _{ 3 } }{ c _{ 2 } } } \right) -{ b _{ 1 } }\left( { { a _{ 2 } }{ c _{ 3 } }-{ a _{ 3 } }{ c _{ 2 } } } \right) +{ c _{ 1 } }\left( { { a _{ 2 } }{ b _{ 3 } }-{ a _{ 3 } }{ b _{ 2 } } } \right)  } \right}  \ ={ a _{ 1 } }\left| \begin{array}{l} { a _{ 1 } } & { b _{ 1 } } & { c _{ 1 } } \ { a _{ 2 } } & { b _{ 2 } } & { c _{ 2 } } \ { a _{ 3 } } & { b _{ 3 } } & { c _{ 3 } } \end{array} \right|  \ ={ a _{ 1 } }\Delta  \end{array}$

If $\Delta =\left| \begin{matrix} { a } _{ 1 } & { b } _{ 1 } & { c } _{ 1 } \ { a } _{ 2 } & { b } _{ 2 } & { c } _{ 2 } \ { a } _{ 3 } & { b } _{ 3 } & { c } _{ 3 } \end{matrix} \right|$ and $A _{1},B _{1},C _{1}$ denote the co-factors of $a _{1},b _{2},c _{1}$ respectively, then the value of the determinant $\left| \begin{matrix} { A } _{ 1 } & { B } _{ 1 } & { C } _{ 1 } \ { A } _{ 2 } & { B } _{ 2 } & { C } _{ 2 } \ { A } _{ 3 } & { B } _{ 3 } & { C } _{ 3 } \end{matrix} \right|$ is

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $\Delta$

  2. $\Delta^{2}$

  3. $\Delta^{3}$

  4. $0$

Show answer
Correct Option: B

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 $c _{ij}=\left ( -1 \right )^{i+j}$ (determinant obtained by deleting ith row and jth column), then $\begin{vmatrix} c _{11} & c _{12} & c _{13}\ c _{21} & c _{22} & c _{23}\ c _{31} & c _{32} & c _{33} \end{vmatrix}=\Delta ^{2}$



If $\begin{vmatrix} 1 & x & x^{ 2 } \ x & x^{ 2 } & 1 \ x^{ 2 } & 1 & x \end{vmatrix}=7$ and $\Delta =\begin{vmatrix}
x^{3}-1 & 0 & x-x^{4}\
0 & x-x^{4} & x^{3}-1\
x-x^{4} & x^{3}-1 & 0
\end{vmatrix}$, then

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $\Delta =7$

  2. $\Delta =343$

  3. $\Delta =-49$

  4. $\Delta =49$

Show answer
Correct Option: D
Explanation:

For $\begin{vmatrix} 1 & x & x^{ 2 } \ x & x^{ 2 } & 1 \ x^{ 2 } & 1 & x \end{vmatrix}=7$
$\begin{vmatrix} c _{ 11 } & c _{ 12 } & c _{ 13 } \ c _{ 21 } & c _{ 22 } & c _{ 23 } \ c _{ 31 } & c _{ 32 } & c _{ 33 } \end{vmatrix}=\begin{vmatrix} x^{ 3 }-1 & 0 & x-x^{ 4 } \ 0 & x-x^{ 4 } & x^{ 3 }-1 \ x-x^{ 4 } & x^{ 3 }-1 & 0 \end{vmatrix}$
$\Delta ={ 7 }^{ 2 }=49$

Let $\Delta _0=\begin{bmatrix}a _{11} & a _{12}  & a _{13}\a _{21}  & a _{22} &a _{23} \ a _{31} & a _{32} & a _{33}\end{bmatrix}$ (where $\Delta _0 \neq  0$) and let $\Delta _1$ denote the determinant formed by the cofactors of elements of $\Delta _0$ and $\Delta _2$ denote the determinant formed by the cofactor at $\Delta _1$ and so on $\Delta _n$ denotes the determinant formed by the cofactors at $\Delta _{n-1}$ then the determinant value of $\Delta _{n}$ is

mathematics and statistics determinants minors and cofactors determinants and matrices matrices and determinants

  1. $\Delta _0^{2n}$

  2. $\Delta _0^{2^n}$

  3. $\Delta _0^{n^2}$

  4. $\Delta _0^{2}$

Show answer
Correct Option: B
Explanation:

$\Delta _1=\Delta ^2 _0,\Delta _2=\Delta ^2 _1=\Delta ^{2^2} _0$
$\therefore \Delta _n=\Delta ^{2n} _0$

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