Tag: minors and cofactors

Questions Related to minors and cofactors

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

  1. $0$

  2. $\Delta$

  3. $2\Delta$

  4. $\Delta^2$


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

  1. $0$

  2. $\Delta$

  3. $2\Delta$

  4. $\Delta^2$


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

  1. $0$

  2. 2

  3. $-1$

  4. 4


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

  1. $0$

  2. $-1$

  3. $1$

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


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 

  1. $\displaystyle 0$

  2. $\displaystyle -D$

  3. $\displaystyle D$

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


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

  1. $8$

  2. $6$

  3. $4$

  4. $0$


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 $\begin{vmatrix} { B } _{ 2 } & C _{ 2 } \ B _{ 3 } & C _{ 3 } \end{vmatrix}$

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

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

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

  4. $None\ of\ these$


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

  1. $\Delta$

  2. $\Delta^{2}$

  3. $\Delta^{3}$

  4. $0$


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

  1. $\Delta =7$

  2. $\Delta =343$

  3. $\Delta =-49$

  4. $\Delta =49$


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

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

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

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

  4. $\Delta _0^{2}$


Correct Option: B
Explanation:

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