Class 12 Engineering Maths - Determinants

Find the value of determinant $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$ .

$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

$=2(-1)-4(-5)=-2+20=18$

Find the Adjoint matrix of the matrix $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4 \end{vmatrix}$

Given matrix:

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

Cofactor matrix:

$cof(A) = \begin{bmatrix}6 & -2 & -3\\ 1 & -5 & 3\\ -5 & 4 & -1 \end{bmatrix}$

Adjoint matrix is the transpose of the cofactor matrix:

$adj(A) = \begin{bmatrix}6 & 1 & -5\\ -2 & -5 & 4\\ -3 & 3 & -1 \end{bmatrix}$

If $\omega$ is one of the imaginary cube roots of unity, find the value of
$\begin{vmatrix}1 & \omega^{3} & \omega^{2}\\ \omega^{3} & 1 &\omega \\ \omega^{2} & \omega & 1\end{vmatrix}$ .

$\Delta =\left| \begin{matrix} 1 & { \omega }^{ 3 } & { \omega }^{ 2 } \\ { \omega }^{ 3 } & 1 & \omega \\ { \omega }^{ 2 } & \omega & 1 \end{matrix} \right|$

we know

${ \omega }^{ 3 }=1$

$\Rightarrow \Delta =\left| \begin{matrix} 1 & 1 & { \omega }^{ 2 } \\ 1 & 1 & \omega \\ { \omega }^{ 2 } & \omega & 1 \end{matrix} \right| \\ { R }_{ 1 }={ R }_{ 1 }-{ R }_{ 2 }\\ \Rightarrow \Delta =\left| \begin{matrix} 0 & 0 & { \omega }^{ 2 }-\omega \\ 1 & 1 & \omega \\ { \omega }^{ 2 } & \omega & 1 \end{matrix} \right|$

Expanding along

${R}_{1}$

$\Rightarrow \Delta =0+0+({ \omega }^{ 2 }-\omega )\{ \omega -{ \omega }^{ 2 }\} \\ \Rightarrow \Delta =-{ \left( { \omega }^{ 2 }-\omega \right) }^{ 2 }\\ \Rightarrow \Delta =-\left( { \omega }^{ 4 }+{ \omega }^{ 2 }-2{ \omega }^{ 3 } \right) \\ \Rightarrow \Delta =-\left( \omega +{ \omega }^{ 2 }-2 \right)$

$1+\omega +{ \omega }^{ 2 }=0$

$\Rightarrow \Delta =-(-1-2)\\ \Rightarrow \Delta =3$

Find determinant $Dc=\begin{vmatrix} 1 & 1 & -2 \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix}$

$D=\left| \begin{matrix} 1 & 1 & -2 \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{matrix} \right| \\ D=1(2+3)-1(-1-6)-2(-1+4)\\ D=5+7-6\\ D=6$

Find the relation between $a$ and $b$ . If the points $P(1,2),Q(0,0)$ and $P(a,b)$ are collinear

If the points are collinear

$\left| \begin{matrix} 0 & 0 & 1 \\ 1 & 2 & 1 \\ a & b & 1 \end{matrix} \right| =0\\ 0(2-b)-0(1-)+1(b-2a)=0\\ b-2a=0\\ b=2a$

If the matrix $A$ = $\left[ {\begin{array}{*{20}{c}}6 & x & 2\\2 & { - 1} & 2\\{ - 10} & 5 & 2\end{array}} \right]$ is a singular matrix. Find the value of x.

$\begin{array}{l}\left| A \right| = 0\\6\left( { - 2 - 10} \right) - x\left( {4 + 20} \right) + \left( {10 - 10} \right)\\ = 0\\so,\\ - 72 = 2 4x\\x = - 3\end{array}$

Find $'x'$ if
$\left| {\begin{array}{*{20}{c}}4&x&6\\2&3&4\\1&1&1\end{array}} \right| = 10$

$\left| {\begin{array}{*{20}{c}}4&x&6\\2&3&4\\1&1&1\end{array}} \right| = 10$

we solve this matrix

$=>4(3-4)-x(2-4)+6(2-3)=10$

$=>4\times (-1)-x\times (-2)+6\times (-1)=10$

$=>-4+2x-6=10$

$=>2x=10+10$

$=>x=10$

If $\triangle = \begin{vmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{vmatrix}$ , find it's value.

$\triangle =\begin{vmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{vmatrix}$

$=3(-6-1)-5(4-1)+7(2+3)$

$=-21-15+35=-1$

How do I find
$A=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right] =\left| A \right| =?$

$A=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right]$

$\displaystyle \therefore\left| A \right| =1(3\times 1-(-2\times 0)-2(-1-0)+(-2)(2-0)$

$= 1(3)+2-4$

$= 5-4$

$= 1$

1119348_1166866_ans_4989e9fb3ce2439f8f7248441ea83950.jpeg

Find ad-joint of
$A=\begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$

1226908_1297897_ans_88544e019a7d4da9a61b2a9e7f9be2ea.jpg

Find the values of x, if

$\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}$

$\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}$

$\Rightarrow x+1(x+2)-(x-1)(x-3)=12+1$

$\Rightarrow x^2+3x+2-x^2+4x-3=13$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Find the values of x, if
$\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & 5 \\ 8 & 3 \end{vmatrix}$

$\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & 5 \\ 8 & 3 \end{vmatrix}$

$\Rightarrow 2x^2-40=18-40$

$\Rightarrow x^2=9$

$\Rightarrow x=\pm 3$

Expand: $\begin{vmatrix} 1 & -7 & 3 \\ 5 & -6 & 0 \\ 1 & 2 & -3 \end{vmatrix}$

$\begin{vmatrix} 1 & -7 & 3 \\ 5 & -6 & 0 \\ 1 & 2 & -3 \end{vmatrix}$

$=1\left(18-0\right)+7\left(-15-0\right)+3\left(10+6\right)$

$=18-105+48$

$=-39$

Find the values of x, if
$\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}$

$\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}$

$\Rightarrow x+1(x+2)-(x-1)(x-3)=12+1$

$\Rightarrow x^2+3x+2-x^2+4x-3=13$

$\Rightarrow 7x=14$

$\Rightarrow x=2$

Show $\begin{vmatrix} ax & by & cz\\ x^2 & y^2 & z^2\\ 1 & 1 & 1 \end{vmatrix}=\begin{vmatrix} a & b & c\\ x & y & z\\ yz & zx & xy \end{vmatrix}$

$\begin{vmatrix} ax & by & cz \\ x^{ 2 } & y^{ 2 } & z^{ 2 } \\ 1 & 1 & 1 \end{vmatrix}$
Multiplying

${ R }_{ 3 }$ by

$xyz$

$=\cfrac { 1 }{ xyz } \begin{vmatrix} ax & by & cz \\ x^{ 2 } & y^{ 2 } & z^{ 2 } \\ xyz & xyz & xyz \end{vmatrix}$
Taking

$x,y$ and

$z$ common from

${ C }_{ 1 },{ C }_{ 2 }$ and

${ C }_{ 3 }$ respectively, we get

$=\cfrac { xyz }{ xyz } \begin{vmatrix} a & b & c \\ x & y & z \\ yz & xz & xy \end{vmatrix}=\begin{vmatrix} a & b & c \\ x & y & z \\ yz & zx & xy \end{vmatrix}$

For a fixed positive integer n, if $D=\begin{vmatrix} n! & (n+1)! & (n+2)!\\ (n+1)! & (n+2)! & (n+3)!\\ (n+2)! & (n+3)! & (n+4)! \end{vmatrix}$ then show
$\left [\frac {D}{(n!)^3}-4\right ]$ is divisible by n.

$D=\begin{vmatrix} n! & (n+1)! & (n+2)! \\ (n+1)! & (n+2)! & (n+3)! \\ (n+2)! & (n+3)! & (n+4)! \end{vmatrix}$
Taking

$\left( n! \right) ,\left( n+1 \right) !$ and

$\left( n+2 \right) !$ common from

${ C }_{ 1 },{ C }_{ 2 }$ and

${ C }_{ 3 }$ , we get

$D=\left( n! \right) \left( n+1 \right) !\left( n+2 \right) !\begin{vmatrix} 1 & 1 & 1 \\ n+1 & \left( n+2 \right) & \left( n+3 \right) \\ \left( n+2 \right) \left( n+1 \right) & \left( n+3 \right) \left( n+2 \right) & \left( n+4 \right) \left( n+3 \right) \end{vmatrix}$
Applying

${ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 },{ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }$

$D=\left( n! \right) \left( n+1 \right) !\left( n+2 \right) !\begin{vmatrix} 1 & 0 & 0 \\ n+1 & 1 & 2 \\ \left( n+2 \right) \left( n+1 \right) & 2\left( n+2 \right) & 2\left( n+5 \right) \end{vmatrix}\\ =\left( n! \right) \left( n+1 \right) !\left( n+2 \right) !\left( 2n+2 \right)$
Hence

$\left[ \frac { D }{ (n!)^{ 3 } } -4 \right] =\left[ \frac { \left( n! \right) \left( n+1 \right) !\left( n+2 \right) !\left( 2n+2 \right) }{ { \left( n! \right) }^{ 3 } } -4 \right] =\left[ \frac { \left( n+1 \right) \left( n+1 \right) \left( n+2 \right) \left( 2n+2 \right) }{ 1 } -4 \right] \\ =\left( 2\left( { n }^{ 4 }+5{ n }^{ 3 }+9{ n }^{ 2 }+7n+2 \right) -4 \right) =2\left( { n }^{ 4 }+5{ n }^{ 3 }+9{ n }^{ 2 }+7n \right)$

If $u=ax^2+2bxy+cy^2, u'=a'x^2+2b'xy+c'y^2$ , then prove that $\begin{vmatrix} y^2 & -xy & x^2\\ a & b & c\\ a' & b' & c' \end{vmatrix}=\begin{vmatrix} ax+by & bx+cy\\ a'x+b'y & b'x+c'y \end{vmatrix}=-\frac {1}{y}\begin{vmatrix} u & u'\\ ax+by & a'x+b'y \end{vmatrix}$

Consider first part

$\begin{vmatrix} y^{ 2 } & -xy & x^{ 2 } \\ a & b & c \\ a' & b' & c' \end{vmatrix}$

Expanding along the first row,

${ y }^{ 2 }\begin{vmatrix} b & c \\ b' & c' \end{vmatrix}+xy\begin{vmatrix} a & c \\ a' & c' \end{vmatrix}+x^{ 2 }\begin{vmatrix} a & b \\ a' & b' \end{vmatrix}$

$={ y }^{ 2 }(bc'-b'c)+xy(ac'-a'c)+{ x }^{ 2 }(ab'-a'b)$

Consider second part,

$\begin{vmatrix} ax+by & bx+cy \\ a'x+b'y & b'x+c'y \end{vmatrix}$

$=(ax+by)(b'x+c'y)-(bx+cy)(a'x+b'y)$

$=ab'x^2+ac'xy+bb'xy+bc'y^2 -(a'bx^2+bb'xy+a'cxy+b'cy^2)$

$={ y }^{ 2 }(bc'-b'c)+xy(ac'-a'c)+{ x }^{ 2 }(ab'-a'b)$

Consider third part,

$-\cfrac { 1 }{ y } \begin{vmatrix} u & u' \\ ax+by & a'x+b'y \end{vmatrix}$

$=-\cfrac { 1 }{ y } \begin{vmatrix} ax^{ 2 }+2bxy+cy^{ 2 } & a'x^{ 2 }+2b'xy+c'y^{ 2 } \\ ax+by & a'x+b'y \end{vmatrix}$

$=-\cfrac { 1 }{ y } [(ax^{ 2 }+2bxy+cy^{ 2 })(a'x+b'y)-(a'x^{ 2 }+2b'xy+c'y^{ 2 })(ax+by)]$

$=-\cfrac { 1 }{ y } [(aa'{ x }^{ 3 }+ab'{ x }^{ 2 }y+2a'b{ x }^{ 2 }y+2bb'{ xy }^{ 2 }+ac'{ xy }^{ 2 }+b'c{ y }^{ 3 })$

$-(aa'{ x }^{ 3 }+a'by{ x }^{ 2 }+2b'a{ x }^{ 2 }y+2bb'{ xy }^{ 2 }+ac'{ xy }^{ 2 }+bc'{ y }^{ 3 })]$

$=-\cfrac { 1 }{ y } [ab'{ x }^{ 2 }y+2a'b{ x }^{ 2 }y+2bb'{ xy }^{ 2 }+ac'{ xy }^{ 2 }+b'c{ y }^{ 3 }-a'by{ x }^{ 2 }-2b'a{ x }^{ 2 }y-2bb'{ xy }^{ 2 }-ac'{ xy }^{ 2 }-bc'{ y }^{ 3 }]$

$=-[ab'{ x }^{ 2 }+2a'b{ x }^{ 2 }+2bb'{ xy }+ac'{ xy }+b'c{ y }^{ 2 }-a'b{ x }^{ 2 }-2b'a{ x }^{ 2 }-2bb'{ xy }-ac'{ xy }-bc'{ y }^{ 2 }]$

$=-[-ab'{ x }^{ 2 }+a'b{ x }^{ 2 }+ac'{ xy }+b'c{ y }^{ 2 }-ac'{ xy }-bc'{ y }^{ 2 }]$

$={ y }^{ 2 }(bc'-b'c)+xy(ac'-a'c)+{ x }^{ 2 }(ab'-a'b)$

Hence, all the three parts are equal

If $\omega$ is one of the imaginary cube roots of unity, find the value of
$\begin{vmatrix}1 & \omega & \omega^{2}\\ \omega & \omega^{2} & 1\\ \omega^{2} & 1 & \omega\end{vmatrix}$ .

$\Delta =\left| \begin{matrix} 1 & \omega & { \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right| \\ { R }_{ 1 }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }\\ \Rightarrow \Delta =\left| \begin{matrix} 1+\omega +{ \omega }^{ 2 } & 1+\omega +{ \omega }^{ 2 } & 1+\omega +{ \omega }^{ 2 } \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right| \\ \Rightarrow \Delta =\left| \begin{matrix} 0 & 0 & 0 \\ \omega & { \omega }^{ 2 } & 1 \\ { \omega }^{ 2 } & 1 & \omega \end{matrix} \right|$

All the elements of one row are

$0$

$\Rightarrow \Delta=0$

Prove the following :
$\left| \begin{matrix} 2ab & a^{ 2 } & { b }^{ 2 } \\ a^{ 2 } & { b }^{ 2 } & 2ab \\ { b }^{ 2 } & 2ab & a^{ 2 } \end{matrix} \right| =-{ \left( { a }^{ 3 }+{ b }^{ 3 } \right) }^{ 2 }.$

Apply

${C}_{1}+{C}_{2}+{C}_{3}$ and take out

${ \left( a+b \right) }^{ 2 }$ common from

${C}_{1}$

$\triangle ={ \left( a+b \right) }^{ 2 }\begin{vmatrix} 1 & { a }^{ 2 } & { b }^{ 2 } \\ 1 & { b }^{ 2 } & 2ab \\ 1 & 2ab & { a }^{ 2 } \end{vmatrix}$
Apply

${R}_{2}-{R}_{1}$ ;

${R}_{3}-{R}_{1}$ to make two zeros.

$\triangle ={ \left( a+b \right) }^{ 2 }\begin{vmatrix} 1 & { a }^{ 2 } & { b }^{ 2 } \\ 0 & { b }^{ 2 }-{ a }^{ 2 } & b\left( 2a-b \right) \\ 0 & a\left( 2b-a \right) & { a }^{ 2 }-{ b }^{ 2 } \end{vmatrix}$

$={ \left( a+b \right) }^{ 2 }\begin{vmatrix} { b }^{ 2 }-{ a }^{ 2 } & b\left( 2a-b \right) \\ a\left( 2b-a \right) & { a }^{ 2 }-{ b }^{ 2 } \end{vmatrix}$

$={ \left( a+b \right) }^{ 2 }\left[ -{ \left( { a }^{ 2 }-b^{ 2 } \right) }^{ 2 }-ab\left( 2b-a \right) \left( 2a-b \right) \right]$

$={ -\left( a+b \right) }^{ 2 }\left[ { \left( { a }^{ 2 }-b^{ 2 } \right) }^{ 2 }+4{ a }^{ 2 }{ b }^{ 2 }-2ab\left( { a }^{ 2 }+b^{ 2 } \right) +{ a }^{ 2 }{ b }^{ 2 } \right]$

$={ -\left( a+b \right) }^{ 2 }\left[ { \left( { a }^{ 2 }-b^{ 2 } \right) }^{ 2 }-2ab\left( { a }^{ 2 }+b^{ 2 } \right) +{ a }^{ 2 }{ b }^{ 2 } \right]$

$={ -\left( a+b \right) }^{ 2 }{ \left[ { a }^{ 2 }+b^{ 2 }-ab \right] }^{ 2 }=-{ \left( { a }3+{ b }^{ 3 } \right) }^{ 2 }$

Let A be the matrix of order $3 \times 3$ such that $|A| = 1,$ $B = 2A^{-1}$ and $C = \frac{(adj A)}{\sqrt[3]{2}}$ , then the value of $|AB^2 . C^3|$ is
[Note : |A| represent determinant value of matrix A.]

$|A|=1$

$adj(A)=|A|^{n-1}=1^{3-1}=1$

$A^{-1}=\dfrac{adj(A)}{|A|}$

$=\dfrac{1}{1}=1$

$B=2A^{-1}=2\times 1=2$

$C=\dfrac{adj(A)}{\sqrt[3]{2}}=\dfrac{1}{\sqrt[3]{2}}$

$\therefore |AB^2C^3|=1\times 2^2\times \dfrac{1}{2}=2$

Prove the following :
$\left| \begin{matrix} \dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ c } & c & c \\ a & \dfrac { { b }^{ 2 }+{ c }^{ 2 } }{ a } & a \\ b & b & \dfrac { { c }^{ 2 }+{ a }^{ 2 } }{ b } \end{matrix} \right| =4abc$

Multiply

${R}_{1}$ ,

${R}_{2}$ and

${R}_{3}$ by

$c$ ,

$a$
and

$b$ respectively and hence divide

$\triangle$ by

$abc$

$\therefore \Delta =\cfrac { 1 }{ abc } \begin{vmatrix} { a }^{ 2 }+{ b }^{ 2 } & { c }^{ 2 } & { c }^{ 2 } \\ { a }^{ 2 } & { b }^{ 2 }+{ { c }^{ 2 } } & { a }^{ 2 } \\ { b }^{ 2 } & { b }^{ 2 } & c^{ 2 }+{ a^{ 2 } } \end{vmatrix}$
Now apply

${ R }_{ 1 }-\left( { R }_{ 2 }+{ R }_{ 3 } \right)$

$\therefore \Delta =\dfrac { 1 }{ abc } \begin{vmatrix} 0 & { -2b }^{ 2 } & { -2a }^{ 2 } \\ { a }^{ 2 } & { b }^{ 2 }+{ { c }^{ 2 } } & { a }^{ 2 } \\ { b }^{ 2 } & { b } ^{ 2 } & c^{ 2 }+{ a^{ 2 } } \end{vmatrix}$

$=-\cfrac { 2 }{ abc } \begin{vmatrix} 0 & { b }^{ 2 } & { a }^{ 2 } \\ { a }^{ 2 } & { b }^{ 2 }+{ { c }^{ 2 } } & { a }^{ 2 } \\ { b }^{ 2 } & { b }^{ 2 } & c^{ 2 }+{ a^{ 2 } } \end{vmatrix}$
Apply

${R}_{2}-{R}_{1}$ and

${R}_{3}-{R}_{1}$

$=-\cfrac { 2 }{ abc } \begin{vmatrix} 0 & { b }^{ 2 } & { a }^{ 2 } \\ { a }^{ 2 } & { { c }^{ 2 } } & 0 \\ { b }^{ 2 } & { b }^{ 2 } & { c^{ 2 } } \end{vmatrix}$ Expand.

$=-\cfrac { 2 }{ abc } \left[ 0-b^{ 2 }\left( { a }^{ 2 }{ c }^{ 2 }-0 \right) +{ a }^{ 2 }\left( 0-{ b }^{ 2 }{ c }^{ 2 } \right) \right]$

$=-\cfrac { 2 }{ abc } \left( -2{ a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 } \right) =4abc$

If A = $\begin{bmatrix} a \\[0.3em] b \\[0.3em] -a \end{bmatrix}_{3\times1} \begin{bmatrix} a & b & -a \\[0.3em] \end{bmatrix}_{1\times3}$ then find wheather $A^{-1}$ exists or not.

A will be 3 X 3 matrix.

$\therefore \ \ \ \ A = \begin{bmatrix} a^2 & ab & -a^2 \\[0.3em] ba & b^2 & ba \\[0.3em] -a^2 & -ab & a^2 \end{bmatrix}$
Clearly |A| = 0 because

$R_1$ and

$R_3$ are identical. Therefore A is singular and hence its inverse does not exists.

Solve: $\begin{vmatrix}x & x^2 & yz\\ y & y^2 & zx\\ z & z^2 & xy\end{vmatrix} = (x - y)(y - z)(z - x)(xy + yz + zx)$ .

Now,

$\begin{vmatrix}x & x^2 & yz\\ y & y^2 & zx\\ z & z^2 & xy\end{vmatrix}$ .

[

$R_2'=R_2-R_1$ and

$R_3'=R_3-R_1$ ]

$=\begin{vmatrix}x & x^2 & yz\\ y-x & y^2-x^2 & z(x-y)\\ z-x & z^2-x^2 & (x-z)y\end{vmatrix}$

$=(y-x)(z-x)\begin{vmatrix}x & x^2 & yz\\ 1 & y+x & -z\\ 1 & z+x &-y\end{vmatrix}$

[

$R_3'=R_3-R_2$ ]

$=(y-x)(z-x)\begin{vmatrix}x & x^2 & yz\\ 1 & y+x & -z\\ 0 & z-y &z-y\end{vmatrix}$

$=(y-x)(z-x)(z-y)\begin{vmatrix}x & x^2 & yz\\ 1 & y+x & -z\\ 0 & 1 &1\end{vmatrix}$

$=(y-x)(z-x)(z-y)\{x(x+y+z)-1(x^2-yz)\}$

$=(x-y)(y-z)(z-x)(xy+yz+zx)$

Find the value of $x$ for which the determinant $\left| {\begin{array}{*{20}{c}}{2x - 3}&{x - 2}&{x - 1}\\{x - 2}&{2x - 2}&x\\{x - 1}&x&{2x - 1}\end{array}} \right|$ vanishes if

$R_1\rightarrow R_1-(R_2+R_3)$

$\left| {\begin{array}{*{20}{c}}{0}&{-2x}&{-2x}\\{x-2}&{2x-2}&{x}\\{x-1}&{x}&{2x-1} \end{array}} \right|=-(-2x)[(2x-2)(2x-1)-x^{2} ] +(-2x)[(x-2)x-(2x-2)(x-1)]=0$

$4x(x-1)^{2}=0$

$x=0,1,1$

Show that:
$\left| {\begin{array}{*{20}{c}}a&b&c\\{a - b}&{b - c}&{c - a}\\{b + c}&{c + a}&{a + b}\end{array}} \right| = {a^3} + {b^3} + {c^3} - 3abc$

1092223_1179723_ans_5b04e268ca384e8691f2bba802681199.png

Find the maximum value of
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin { \theta } & 1 \\ 1 & 1 & 1+\cos { \pi } \end{vmatrix}$

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin { \theta } & 1 \\ 1 & 1 & 1 \end{vmatrix}$

$\Rightarrow \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin { \theta } & 1 \\ 1 & 1 & 0 \end{vmatrix}\Rightarrow \begin{vmatrix} 0 & 1 & 1 \\ 0 & 1+\sin { \theta } & 1 \\ 1 & 1 & 0 \end{vmatrix}$

$\Rightarrow \ 1(1-1-\sin \theta )\Rightarrow \ -\sin \theta$

minimum value

$=1$

Find the equation of line passing through the points $(3,2)$ and $(-1,3)$ by using determinants.

Equation of line through

$(3,2)$ &

$(-1,3)$ is

$\begin{vmatrix} x-3 & y-2 \\ -1-3 & 3-2 \end{vmatrix}=0$

$\Rightarrow \begin{vmatrix} x-3 & y-2 \\ -4 & 1 \end{vmatrix}=0$

$x-3+4(y-2)=0$

$\Rightarrow x+4y=11$

Find the determinant of the matrix
$\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$

$\left| \begin{matrix} 3 & 1 \\ 5 & 2 \end{matrix} \right|=6-5=1$

Find the value of the determinant $\begin{vmatrix} 2i & -3i \\ i^3 & -2i^ s \end{vmatrix}$ where $i=\sqrt{-i}$

$\begin{vmatrix} 2i & -3i \\ { i }^{ 3 } & -2i^{ 8 } \end{vmatrix}\\ =2i\left( -2i^{ 8 } \right) -{ i }^{ 3 }\left( -3i \right) \\ =-4i+3{ i }^{ 4 }\quad \left( \because { i }^{ 8 }={ \left( { i }^{ 4 } \right) }^{ 2 }=1 \right) \\ =3-4i\quad \left( { i }^{ 4 }=1 \right)$

prove that $\left| \begin{matrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{matrix} \right| ={ \left( x-1 \right) }^{ 2 }\left( x+9 \right)$

$\Delta =\left| \begin{matrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{matrix} \right| \\ { C }_{ 1 }+{ C }_{ 2 }+{ C }_{ 3 }\\ \Delta =\left| \begin{matrix} x+9 & 3 & 5 \\ x+9 & x+2 & 5 \\ x+9 & 3 & x+4 \end{matrix} \right| \\ =(x+9)\left| \begin{matrix} 1 & 3 & 5 \\ 1 & x+2 & 5 \\ 1 & 3 & x+4 \end{matrix} \right| \\ { R }_{ 2 }={ R }_{ 2 }-{ R }_{ 1 }\quad \quad \quad { R }_{ 3 }={ R }_{ 3 }-{ R }_{ 1 }\\ \Delta =(x+9)\left| \begin{matrix} 1 & 3 & 5 \\ 0 & x-1 & 0 \\ 0 & 0 & x-1 \end{matrix} \right|$

expanding along

${ C }_{ 1 }$ we have

$\Delta =(x+9)\times 1\left| { (x-1) }^{ 2 }-0 \right| ={ (x-1) }^{ 2 }(x+9)$

If $\begin{vmatrix} 2 & -4 \\ 9 & d-3 \end{vmatrix} =4$ , then find the value of d.

$\begin{vmatrix} 2 & -4 \\ 9 & d-3 \end{vmatrix} =4$

$2(d-3)-(-4\times 9)=4$

$2d-6+36=4$

$2d=-26$

$d=-13$

$\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$

$\Rightarrow \cos 15^{\circ}\cos 75^{\circ}-\sin {75}^{\circ}\sin 15^{\circ}$

$\Rightarrow \cos(75+15)$

$\Rightarrow \cos 90=0$

Find the value of $x$ if
$\begin{vmatrix} 3 & 4 &1 \\0 & x & 8\\ 3 & -1 & 4\end{vmatrix}=0$

$\begin{vmatrix} 3 & 4 &1 \\0 & x & 8\\ 3 & -1 & 4\end{vmatrix}=0$

$\Rightarrow\,3\left(4x+8\right)-4\left(0-24\right)+1\left(0-3x\right)=0$

$\Rightarrow\,12x+24+96-3x=0$

$\Rightarrow\,9x+120=0$

$\Rightarrow\,9x=-120$

$\Rightarrow\,x=-\dfrac{120}{9}=-\dfrac{40}{3}$

$\therefore\,x=-\dfrac{40}{3}$

For what value of x the matrix A is singular?
$A= \begin{bmatrix} 1+x & 7 \\ 3-x & 8 \end{bmatrix}$

$A=\begin{bmatrix}1+x&7\\3-x&8\end{bmatrix}$

Its is given that

$A$ is singular.

$\therefore$

$|A|=0$

$\begin{vmatrix}1+x&7\\3-x&8\end{vmatrix}=0$

$\Rightarrow$

$8(1+x)-7(3-x)=0$

$\Rightarrow$

$8+8x-21+7x=0$

$\Rightarrow$

$-13+15x=0$

$\Rightarrow$

$15x=13$

$\Rightarrow$

$x=\dfrac{13}{15}$

Evaluate the following determinant :
$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$

$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$

Expanding det. on both sides

$a(bc-f^2)-h(hc-gf)+g(hf-gb)$

$abc-af^2-ch^2+gfh+ghf-bg^2$

$abc-af^2-bg^2-ch^2$

$\text { Prove that: }\left|\begin{array}{ccc}a+b+2 c & a & b \\c & b+c+2 a & b \\c & a & c+a+2 b\end{array}\right|=2(a+b+c)^{3}\\$

$\text { LHS }\left|\begin{array}{ccc}a+b+2 c & a & b \\c & b+c+2 a & b \\c & a & c+a+2 b\end{array}\right|\\\\$

Applying

$C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ , we get

$\\=\left|\begin{array}{ccc}2(a+b+c) & a & b \\2(a+b+c) & b+c+2 a & b \\2(a+b+c) & a & c+a+2 b\end{array}\right|\\\\$

Taking

$2(a+b+c)$ common from

$C_{1}$ , we get

$=2(a+b+c)\left|\begin{array}{ccc}1 & a & b \\1 & b+c+2 a & b \\1 & a & c+a+2 b\end{array}\right|\\\\$

Applying

$R_{2} \rightarrow R_{2}-R_{1}$ and

$R_{3} \rightarrow R_{3}-R_{1},$ we get

$=2(a+b+c)\left|\begin{array}{ccc}1 & a & b \\0 & a+b+c & 0 \\0 & 0 & a+b+c\end{array}\right|\\$

Taking

$(a+b+c)$ common from

$R_{2}$ and

$R_{3},$ we get

$=2(a+b+c)^{3}\left|\begin{array}{lll}1 & a & b \\0 & 1 & 0 \\0 & 0 & 1\end{array}\right|\\$

Expanding along

$C_{1}$ , we get

$=2(a+b+c)^{3}[1-0]=2(a+b+c)^{3}=\mathrm{RHS}\\$

If the point $(x,y),(a,0),0,b)$ are collinear, prove that
$\cfrac { x }{ a } +\cfrac { y }{ b } =1$

As points are collinear

$\begin{vmatrix} x & y & 1 \\ a & 0 & 1 \\ 0 & b & 1 \end{vmatrix}=0$

$\Rightarrow$

$bx+ay=ab$

$\cfrac { x }{ a } +\cfrac { y }{ b } =1$

If $D_r=\begin{vmatrix} 2^{r-1} & 2(3^{r-1}) & 4(5^{r-1})\\ x & y & z\\ 2^n-1 & 3^n-1 & 5^n-1 \end{vmatrix}$ then prove that $\displaystyle \sum_{r=1}^nD_r=0$

$\displaystyle \sum_{r=1}^nD_r=\begin{vmatrix} \sum_{r=1}^n2^{r-1} & \sum_{r=1}^n2(3^{r-1}) & \sum_{r=1}^n4(5^{r-1})\\ x & y & z\\ 2^n-1 & 3^n-1 & 5^n-1 \end{vmatrix}$

$=\begin{vmatrix} 1+2+...2^{n-1} & 2\left \{1+3....3^{n-1}\right \} & 4(1+5+....5^{n-1})\\ x & y & z\\ 2^n-1 & 3^n-1 & 5^n-1 \end{vmatrix}$

$=\begin{vmatrix} \frac {2^n-1}{2-1} & \frac {2(3^n-1)}{3-1} & \frac {4(5^n-1)}{5-1}\\ x & y & z\\ 2^n-1 & 3^n-1 & 5^n-1 \end{vmatrix}=\begin{vmatrix} { 2 }^{ n-1 } & { 3 }^{ n-1 } & { 5 }^{ n-1 } \\ x & y & z \\ { 2 }^{ n-1 } & { 3 }^{ n-1 } & { 5 }^{ n-1 } \end{vmatrix}\quad=0$
Since

$R_1 = R_3$

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is $3$ and the trace of $A^3$ is $-18$ , then the value of the determinant of $A$ is ______.

Let

$A = \begin{bmatrix} a & b \\ c & 3 - a \end{bmatrix}$

$|A| = 3a - a^2 - bc$

$A^2 = \begin{bmatrix} a & b \\ c & 3 - a \end{bmatrix} \begin{bmatrix} a & b \\ c & 3 - a \end{bmatrix} = \begin{bmatrix} a^2 + bc & 3b \\ 3c & bc + (3 - a)^2 \end{bmatrix}$

$A^3 = \begin{bmatrix} a^2 + bc & 3b \\ 3c & bc + (3 - a)^2 \end{bmatrix} \begin{bmatrix} a & b \\ c & 3-a \end{bmatrix}$

Trace of

$(A^3) = a^3 + abc + 3bc + 3bc - abc + 3 (3 - a)^2 - a(3 - a)^2 = -18$

$\Rightarrow 3a - a^2 - bc = 5$

Determinants - Class 12 Engineering Maths - Extra Questions

Find the value of determinant |24−5−1|\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}.

Find the Adjoint matrix of the matrix |123232334|\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4 \end{vmatrix}

If ω\omega is one of the imaginary cube roots of unity, find the value of|1ω3ω2ω31ωω2ω1|\begin{vmatrix}1 & \omega^{3} & \omega^{2}\\ \omega^{3} & 1 &\omega \\ \omega^{2} & \omega & 1\end{vmatrix}.

Find determinant Dc=|11−21−232−1−1|Dc=\begin{vmatrix} 1 & 1 & -2 \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix}

Find the relation between aa and bb. If the points P(1,2),Q(0,0)P(1,2),Q(0,0) and P(a,b)P(a,b) are collinear

If the matrix AA = [6x22−12−1052]\left[ {\begin{array}{*{20}{c}}6 & x & 2\\2 & { - 1} & 2\\{ - 10} & 5 & 2\end{array}} \right]is a singular matrix. Find the value of x.

Find ′x′'x' if|4x6234111|=10\left| {\begin{array}{*{20}{c}}4&x&6\\2&3&4\\1&1&1\end{array}} \right| = 10

If △=|3572−31112|\triangle = \begin{vmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{vmatrix} , find it's value.

How do I find A=[12−2−1300−21]=|A|=?A=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right] =\left| A \right| =?

Find ad-joint of A=[cosxsinx−sinxcosx]A=\begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}

Find the values of x, if |x+1x−1x−3x+2|=|4−113|\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}

Find the values of x, if |2x58x|=|6583|\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & 5 \\ 8 & 3 \end{vmatrix}

Expand:|1−735−6012−3| \begin{vmatrix} 1 & -7 & 3 \\ 5 & -6 & 0 \\ 1 & 2 & -3 \end{vmatrix}

Find the values of x, if |x+1x−1x−3x+2|=|4−113|\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}

Show |axbyczx2y2z2111|=|abcxyzyzzxxy|\begin{vmatrix} ax & by & cz\\ x^2 & y^2 & z^2\\ 1 & 1 & 1 \end{vmatrix}=\begin{vmatrix} a & b & c\\ x & y & z\\ yz & zx & xy \end{vmatrix}

For a fixed positive integer n, if D=|n!(n+1)!(n+2)!(n+1)!(n+2)!(n+3)!(n+2)!(n+3)!(n+4)!|D=\begin{vmatrix} n! & (n+1)! & (n+2)!\\ (n+1)! & (n+2)! & (n+3)!\\ (n+2)! & (n+3)! & (n+4)! \end{vmatrix} then show[D(n!)3−4]\left [\frac {D}{(n!)^3}-4\right ] is divisible by n.

If ω\omega is one of the imaginary cube roots of unity, find the value of|1ωω2ωω21ω21ω|\begin{vmatrix}1 & \omega & \omega^{2}\\ \omega & \omega^{2} & 1\\ \omega^{2} & 1 & \omega\end{vmatrix}.

Prove the following :|2aba2b2a2b22abb22aba2|=−(a3+b3)2.\left| \begin{matrix} 2ab & a^{ 2 } & { b }^{ 2 } \\ a^{ 2 } & { b }^{ 2 } & 2ab \\ { b }^{ 2 } & 2ab & a^{ 2 } \end{matrix} \right| =-{ \left( { a }^{ 3 }+{ b }^{ 3 } \right) }^{ 2 }.

Let A be the matrix of order 3×3 3 \times 3 such that |A|=1,|A| = 1, B=2A−1B = 2A^{-1} and C=(adjA)3√2C = \frac{(adj A)}{\sqrt[3]{2}}, then the value of |AB2.C3||AB^2 . C^3| is [Note : |A| represent determinant value of matrix A.]

Prove the following :|a2+b2cccab2+c2aabbc2+a2b|=4abc\left| \begin{matrix} \dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ c } & c & c \\ a & \dfrac { { b }^{ 2 }+{ c }^{ 2 } }{ a } & a \\ b & b & \dfrac { { c }^{ 2 }+{ a }^{ 2 } }{ b } \end{matrix} \right| =4abc

If A = [ab−a]3×1[ab−a]1×3\begin{bmatrix} a \\[0.3em] b \\[0.3em] -a \end{bmatrix}_{3\times1} \begin{bmatrix} a & b & -a \\[0.3em] \end{bmatrix}_{1\times3} then find wheather A−1A^{-1} exists or not.

Solve: |xx2yzyy2zxzz2xy|=(x−y)(y−z)(z−x)(xy+yz+zx)\begin{vmatrix}x & x^2 & yz\\ y & y^2 & zx\\ z & z^2 & xy\end{vmatrix} = (x - y)(y - z)(z - x)(xy + yz + zx).

Find the value of xx for which the determinant |2x−3x−2x−1x−22x−2xx−1x2x−1|\left| {\begin{array}{*{20}{c}}{2x - 3}&{x - 2}&{x - 1}\\{x - 2}&{2x - 2}&x\\{x - 1}&x&{2x - 1}\end{array}} \right| vanishes if

Show that:|abca−bb−cc−ab+cc+aa+b|=a3+b3+c3−3abc\left| {\begin{array}{*{20}{c}}a&b&c\\{a - b}&{b - c}&{c - a}\\{b + c}&{c + a}&{a + b}\end{array}} \right| = {a^3} + {b^3} + {c^3} - 3abc

Find the maximum value of |11111+sinθ1111+cosπ|\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin { \theta } & 1 \\ 1 & 1 & 1+\cos { \pi } \end{vmatrix}

Find the equation of line passing through the points (3,2)(3,2) and (−1,3)(-1,3) by using determinants.

Find the determinant of the matrix[3152]\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}

Find the value of the determinant |2i−3ii3−2is|\begin{vmatrix} 2i & -3i \\ i^3 & -2i^ s \end{vmatrix} where i=√−ii=\sqrt{-i}

prove that |x+1352x+2523x+4|=(x−1)2(x+9)\left| \begin{matrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{matrix} \right| ={ \left( x-1 \right) }^{ 2 }\left( x+9 \right)

If |2−49d−3|=4\begin{vmatrix} 2 & -4 \\ 9 & d-3 \end{vmatrix} =4, then find the value of d.

|cos15∘sin15∘sin75∘cos75∘|\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}

Find the value of xx if|3410x83−14|=0 \begin{vmatrix} 3 & 4 &1 \\0 & x & 8\\ 3 & -1 & 4\end{vmatrix}=0

For what value of x the matrix A is singular? A=[1+x73−x8]A= \begin{bmatrix} 1+x & 7 \\ 3-x & 8 \end{bmatrix}

Evaluate the following determinant : |ahghbfgfc| \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}

Prove that: |a+b+2cabcb+c+2abcac+a+2b|=2(a+b+c)3\text { Prove that: }\left|\begin{array}{ccc}a+b+2 c & a & b \\c & b+c+2 a & b \\c & a & c+a+2 b\end{array}\right|=2(a+b+c)^{3}\\

If the point (x,y),(a,0),0,b)(x,y),(a,0),0,b) are collinear, prove thatxa+yb=1\cfrac { x }{ a } +\cfrac { y }{ b } =1

If Dr=|2r−12(3r−1)4(5r−1)xyz2n−13n−15n−1|D_r=\begin{vmatrix} 2^{r-1} & 2(3^{r-1}) & 4(5^{r-1})\\ x & y & z\\ 2^n-1 & 3^n-1 & 5^n-1 \end{vmatrix} then prove that n∑r=1Dr=0\displaystyle \sum_{r=1}^nD_r=0

The trace of a square matrix is defined to be the sum of its diagonal entries. If AA is a 2×22 \times 2 matrix such that the trace of AA is 33 and the trace of A3A^3 is −18-18, then the value of the determinant of AA is ______.

Class 12 Engineering Maths Extra Questions

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Find the value of determinant $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$ .

Find the Adjoint matrix of the matrix $\begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 2 \\ 3 & 3 & 4 \end{vmatrix}$

If $\omega$ is one of the imaginary cube roots of unity, find the value of
$\begin{vmatrix}1 & \omega^{3} & \omega^{2}\\ \omega^{3} & 1 &\omega \\ \omega^{2} & \omega & 1\end{vmatrix}$ .

Find determinant $Dc=\begin{vmatrix} 1 & 1 & -2 \\ 1 & -2 & 3 \\ 2 & -1 & -1 \end{vmatrix}$

Find the relation between $a$ and $b$ . If the points $P(1,2),Q(0,0)$ and $P(a,b)$ are collinear

If the matrix $A$ = $\left[ {\begin{array}{*{20}{c}}6 & x & 2\\2 & { - 1} & 2\\{ - 10} & 5 & 2\end{array}} \right]$ is a singular matrix. Find the value of x.

Find $'x'$ if
$\left| {\begin{array}{*{20}{c}}4&x&6\\2&3&4\\1&1&1\end{array}} \right| = 10$

If $\triangle = \begin{vmatrix} 3 & 5 & 7 \\ 2 & -3 & 1 \\ 1 & 1 & 2 \end{vmatrix}$ , find it's value.

How do I find
$A=\left[ \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right] =\left| A \right| =?$

Find ad-joint of
$A=\begin{bmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{bmatrix}$

Find the values of x, if

$\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}$

Find the values of x, if
$\begin{vmatrix} 2x & 5 \\ 8 & x \end{vmatrix}=\begin{vmatrix} 6 & 5 \\ 8 & 3 \end{vmatrix}$

Expand: $\begin{vmatrix} 1 & -7 & 3 \\ 5 & -6 & 0 \\ 1 & 2 & -3 \end{vmatrix}$

Find the values of x, if
$\begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix}=\begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix}$

Show $\begin{vmatrix} ax & by & cz\\ x^2 & y^2 & z^2\\ 1 & 1 & 1 \end{vmatrix}=\begin{vmatrix} a & b & c\\ x & y & z\\ yz & zx & xy \end{vmatrix}$

For a fixed positive integer n, if $D=\begin{vmatrix} n! & (n+1)! & (n+2)!\\ (n+1)! & (n+2)! & (n+3)!\\ (n+2)! & (n+3)! & (n+4)! \end{vmatrix}$ then show
$\left [\frac {D}{(n!)^3}-4\right ]$ is divisible by n.

If $\omega$ is one of the imaginary cube roots of unity, find the value of
$\begin{vmatrix}1 & \omega & \omega^{2}\\ \omega & \omega^{2} & 1\\ \omega^{2} & 1 & \omega\end{vmatrix}$ .

Prove the following :
$\left| \begin{matrix} 2ab & a^{ 2 } & { b }^{ 2 } \\ a^{ 2 } & { b }^{ 2 } & 2ab \\ { b }^{ 2 } & 2ab & a^{ 2 } \end{matrix} \right| =-{ \left( { a }^{ 3 }+{ b }^{ 3 } \right) }^{ 2 }.$

Let A be the matrix of order $3 \times 3$ such that $|A| = 1,$ $B = 2A^{-1}$ and $C = \frac{(adj A)}{\sqrt[3]{2}}$ , then the value of $|AB^2 . C^3|$ is
[Note : |A| represent determinant value of matrix A.]

Prove the following :
$\left| \begin{matrix} \dfrac { { a }^{ 2 }+{ b }^{ 2 } }{ c } & c & c \\ a & \dfrac { { b }^{ 2 }+{ c }^{ 2 } }{ a } & a \\ b & b & \dfrac { { c }^{ 2 }+{ a }^{ 2 } }{ b } \end{matrix} \right| =4abc$

If A = $\begin{bmatrix} a \\[0.3em] b \\[0.3em] -a \end{bmatrix}_{3\times1} \begin{bmatrix} a & b & -a \\[0.3em] \end{bmatrix}_{1\times3}$ then find wheather $A^{-1}$ exists or not.

Solve: $\begin{vmatrix}x & x^2 & yz\\ y & y^2 & zx\\ z & z^2 & xy\end{vmatrix} = (x - y)(y - z)(z - x)(xy + yz + zx)$ .

Find the value of $x$ for which the determinant $\left| {\begin{array}{*{20}{c}}{2x - 3}&{x - 2}&{x - 1}\\{x - 2}&{2x - 2}&x\\{x - 1}&x&{2x - 1}\end{array}} \right|$ vanishes if

Show that:
$\left| {\begin{array}{*{20}{c}}a&b&c\\{a - b}&{b - c}&{c - a}\\{b + c}&{c + a}&{a + b}\end{array}} \right| = {a^3} + {b^3} + {c^3} - 3abc$

Find the maximum value of
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin { \theta } & 1 \\ 1 & 1 & 1+\cos { \pi } \end{vmatrix}$

Find the equation of line passing through the points $(3,2)$ and $(-1,3)$ by using determinants.

Find the determinant of the matrix
$\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$

Find the value of the determinant $\begin{vmatrix} 2i & -3i \\ i^3 & -2i^ s \end{vmatrix}$ where $i=\sqrt{-i}$

prove that $\left| \begin{matrix} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{matrix} \right| ={ \left( x-1 \right) }^{ 2 }\left( x+9 \right)$

If $\begin{vmatrix} 2 & -4 \\ 9 & d-3 \end{vmatrix} =4$ , then find the value of d.

$\begin{vmatrix} \cos15^{\circ} & \sin15^{\circ} \\ \sin75^{\circ} & \cos75^{\circ} \end{vmatrix}$

Find the value of $x$ if
$\begin{vmatrix} 3 & 4 &1 \\0 & x & 8\\ 3 & -1 & 4\end{vmatrix}=0$

For what value of x the matrix A is singular?
$A= \begin{bmatrix} 1+x & 7 \\ 3-x & 8 \end{bmatrix}$

Evaluate the following determinant :
$\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix}$

$\text { Prove that: }\left|\begin{array}{ccc}a+b+2 c & a & b \\c & b+c+2 a & b \\c & a & c+a+2 b\end{array}\right|=2(a+b+c)^{3}\\$

If the point $(x,y),(a,0),0,b)$ are collinear, prove that
$\cfrac { x }{ a } +\cfrac { y }{ b } =1$

If $D_r=\begin{vmatrix} 2^{r-1} & 2(3^{r-1}) & 4(5^{r-1})\\ x & y & z\\ 2^n-1 & 3^n-1 & 5^n-1 \end{vmatrix}$ then prove that $\displaystyle \sum_{r=1}^nD_r=0$

The trace of a square matrix is defined to be the sum of its diagonal entries. If $A$ is a $2 \times 2$ matrix such that the trace of $A$ is $3$ and the trace of $A^3$ is $-18$ , then the value of the determinant of $A$ is ______.