Let $$\alpha $$ and $$\beta $$ be the roots of the equation x2 + x + 1 = 0. Then for y $$ \ne $$ 0 in R, $$$\left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr

$$\alpha $$ and $$\beta $$ are the roots of the equation x2 + x + 1 = 0. $$ \therefore $$ $$\alpha $$ = $$\omega $$ and $$\beta $$ = $${\omega ^2}$$ $$\left| {\matrix{ {y + 1} & \alpha & \beta \cr \alpha & {y + \beta } & 1 \cr \beta & 1 & {y + \alpha } \cr } } \right|$$ = $$\left| {\matrix{ {y + 1} & \omega & {{\omega ^2}} \cr \omega & {y + {\omega ^2}} & 1 \cr {{\omega ^2}} & 1 & {y + \omega } \cr } } \right|$$ C1 $$ \to $$ C1 + C2 + C3 = $$\left|

Let a, b, c $$ \in $$ R be all non-zero and satisfy a3 + b3 + c3 = 2. If the matrix A = $$\left( {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right)$$ satisfies ATA

Given, $${a^3} + {b^3} + {c^3} = 2$$ $${A^T}A = I$$ $$ \Rightarrow \left[ {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right]\left[ {\matrix{ a & b & c \cr b & c & a \cr c & a & b \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr } } \right]$$ $$ \Rightarrow {a^2} + {b^2} + {c^2} = 1$$ and ab + bc + ca = 0 Now (a + b + c)2 = 1 $$ \Rightarrow (a + b + c) = \pm 1$$ So, $${a^3} + {b^3} + {

If the matrix $$A = \left( {\matrix{ 0 & 2 \cr K & { - 1} \cr } } \right)$$ satisfies $$A({A^3} + 3I) = 2I$$, then the value of K is :

Given matrix $$A = \left[ {\matrix{ 0 & 2 \cr k & { - 1} \cr } } \right]$$ $${A^4} + 3IA = 2I$$ $$ \Rightarrow {A^4} = 2I - 3A$$ Also characteristic equation of A is $$|A - \lambda I|\, = 0$$ $$ \Rightarrow \left| {\matrix{ {0 - \lambda } & 2 \cr k & { - 1 - \lambda } \cr } } \right| = 0$$ $$ \Rightarrow \lambda + {\lambda ^2} - 2k = 0$$ $$ \Rightarrow A + {A^2} = 2K.I$$ $$ \Rightarrow {A^2} = 2KI - A$$ $$ \Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK$$ Put $${A^2} = 2KI - A$$ and $$

Let $$\alpha $$ be a root of the equation x2 + x + 1 = 0 and the matrix A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & {{\alpha ^2}} \cr 1 & {{\alpha ^2}} &a

x2 + x + 1 = 0 $$ \Rightarrow $$ x = $${{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$ or $${{ - 1 - i\sqrt 3 } \over 2}$$ = $${\omega ^2}$$ Let $$\alpha $$ = $$\omega $$ $$ \therefore $$ A = $${1 \over {\sqrt 3 }}\left[ {\matrix{ 1 & 1 & 1 \cr 1 & \omega & {{\omega ^2}} \cr 1 & {{\omega ^2}} & {{\omega ^4}} \cr } } \right]$$ A2 = $${1 \over 3}\left[ {\matrix{ 3 & 0 & 0 \cr 0 & 0 & 3 \cr 0 & 3 & 0 \cr } } \right]$$ = $$\left[ {\matrix{ 1 & 0 &

Let A be a 3 $$ \times $$ 3 matrix such that adj A = $$\left[ {\matrix{ 2 & { - 1} & 1 \cr { - 1} & 0 & 2 \cr 1 & { - 2} & { - 1} \cr } } \right]$$ and B = adj(adj A). If |A| =

$$adj\,A = \left[ {\matrix{ 2 & { - 1} & 1 \cr { - 1} & 0 & 2 \cr 1 & { - 2} & { - 1} \cr } } \right]$$ $$B = adj\,(adj\,A)$$ $$ = |A{|^{n - 2}}A$$ $$ = |A{|^{3 - 2}}.A$$ [As here n = 3] $$ = |A|.A$$ .....(1) Now, $$|adj\,A| = \left[ {\matrix{ 2 & { - 1} & 1 \cr { - 1} & 0 & 2 \cr 1 & { - 2} & { - 1} \cr } } \right]$$ $$ = + 2(0 + 4) + 1(1 - 2) + 1(2 - 0)$$ $$ = 8 - 1 + 2$$ $$ = 9$$ Also we know, |adj A| = |A|n$$-$$1 $$ \therefore $$ Here |adj A| =

Matrices and Determinants — Page 27 | JEE Mathematics

Q261

Let

\alpha

and

\beta

be the roots of the equation x2 + x + 1 = 0. Then for y

\ne

0 in R,

\left| \begin{array}{lll}{y + 1} & \alpha & \beta \\ \alpha & {y + \beta } & 1 \\ \beta & 1 & {y + \alpha } \end{array} \right|

$ is equal to

A y(y2 – 1)

B y(y2 – 3)

C y3

D y3 – 1

Correct Answer

Option C

Solution

$\alpha$ and $\beta$ are the roots of the equation x2 + x + 1 = 0. $\therefore$ $\alpha$ = $\omega$ and $\beta$ =

{\omega ^2}

\left| \begin{array}{lll}{y + 1} & \alpha & \beta \\ \alpha & {y + \beta } & 1 \\ \beta & 1 & {y + \alpha } \end{array} \right|

=

\left| \begin{array}{lll}{y + 1} & \omega & {{\omega ^2}} \\ \omega & {y + {\omega ^2}} & 1 \\ {{\omega ^2}} & 1 & {y + \omega } \end{array} \right|

C1 $\to$ C1 + C2 + C3 =

\left| \begin{array}{lll}{y + 1 + \omega + {\omega ^2}} & \omega & {{\omega ^2}} \\ {y + 1 + \omega + {\omega ^2}} & {y + {\omega ^2}} & 1 \\ {y + 1 + \omega + {\omega ^2}} & 1 & {y + \omega } \end{array} \right|

=

\left| \begin{array}{lll}y & \omega & {{\omega ^2}} \\ y & {y + {\omega ^2}} & 1 \\ y & 1 & {y + \omega } \end{array} \right|

As

1 + \omega + {\omega ^2}

= 0 =

y\left| \begin{array}{lll}1 & \omega & {{\omega ^2}} \\ 1 & {y + {\omega ^2}} & 1 \\ 1 & 1 & {y + \omega } \end{array} \right|

R2 $\to$ R2 - R1 R3 $\to$ R3 - R1 =

y\left| \begin{array}{lll}1 & \omega & {{\omega ^2}} \\ 0 & {y + {\omega ^2} - \omega } & {1 - {\omega ^2}} \\ 0 & {1 - \omega } & {y + \omega - {\omega ^2}} \end{array} \right|

= y

\left[ {\left( {y + {\omega ^2} - \omega } \right)\left( {y + \omega - {\omega ^2}} \right) - \left( {1 - {\omega ^2}} \right)\left( {1 - \omega } \right)} \right]

= y(y2) = y3

Q262

Let a, b, c

\in

R be all non-zero and satisfy a3 + b3 + c3 = 2. If the matrix A =

\left( \begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b \end{array} \right)

satisfies ATA = I, then a value of abc can be :

A 3

B

{1 \over 3}

C -

{1 \over 3}

D

{2 \over 3}

Correct Answer

Option B

Solution

Given,

{a^3} + {b^3} + {c^3} = 2

{A^T}A = I

\Rightarrow \left[ \begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b \end{array} \right]\left[ \begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b \end{array} \right] = \left[ \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]

\Rightarrow {a^2} + {b^2} + {c^2} = 1

and ab + bc + ca = 0 Now (a + b + c)2 = 1

\Rightarrow (a + b + c) = \pm 1

So,

{a^3} + {b^3} + {c^3} - 3abc

\Rightarrow (a + b + c)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)

\Rightarrow \pm 1(1 - 0) = \pm 1

\therefore 2 - 3abc = \pm 1

\Rightarrow 3abc = 2 \pm 1 = 3,1

\Rightarrow abc = 1,{1 \over 3}

Q263

If

{a_r} = \cos {{2r\pi } \over 9} + i\sin {{2r\pi } \over 9}

, r = 1, 2, 3, ....., i =

\sqrt { - 1}

, then the determinant

\left| \begin{array}{lll}{{a_1}} & {{a_2}} & {{a_3}} \\ {{a_4}} & {{a_5}} & {{a_6}} \\ {{a_7}} & {{a_8}} & {{a_9}} \end{array} \right|

is equal to :

A a2a6

-

a4a8

B a9

C a1a9

-

a3a7

D a5

Correct Answer

Option C

Solution

{a_r} = {e^{{{i2\pi r} \over 9}}}

, r = 1, 2, 3, ......, a1, a2, a3, ..... are in G.P.

\left| \begin{array}{lll}{{a_1}} & {{a_2}} & {{a_3}} \\ {{a_n}} & {{a_5}} & {{a_6}} \\ {{a_7}} & {{a_8}} & {{a_9}} \end{array} \right| = \left| \begin{array}{lll}{{a_1}} & {a_1^2} & {a_1^3} \\ {a_1^4} & {a_1^5} & {a_1^6} \\ {a_1^7} & {a_1^8} & {a_1^9} \end{array} \right|

= {a_1}\,.\,a_1^4\,.\,a_1^7\left| \begin{array}{lll}1 & {{a_1}} & {a_1^2} \\ 1 & {{a_1}} & {a_1^2} \\ 1 & {{a_1}} & {a_1^2} \end{array} \right| = 0

Now,

{a_1}{a_9} - {a_3}{a_7} = {a_1}^{10} - {a_1}^{10} = 0

Q264

Consider the system of linear equations

-

x + y + 2z = 0 3x

-

ay + 5z = 1 2x

-

2y

-

az = 7 Let S1 be the set of all a

\in

R for which the system is inconsistent and S2 be the set of all a

\in

R for which the system has infinitely many solutions. If n(S1) and n(S2) denote the number of elements in S1 and S2 respectively, then

A n(S1) = 2, n(S2) = 2

B n(S1) = 1, n(S2) = 0

C n(S1) = 2, n(S2) = 0

D n(S1) = 0, n(S2) = 2

Correct Answer

Option C

Solution

\Delta = \left| \begin{array}{lll}{ - 1} & 1 & 2 \\ 3 & { - a} & 5 \\ 2 & { - 2} & { - a} \end{array} \right|

= - 1({a^2} + 10) - 1( - 3a - 10) + 2( - 6 + 2a)

= - {a^2} - 10 + 3a + 10 - 12 + 4a

\Delta = - {a^2} + 7a - 12

\Delta = - [{a^2} - 7a + 12]

\Delta = - [(a - 3)(a - 4)]

{\Delta _1} = \left| \begin{array}{lll}0 & 1 & 2 \\ 1 & { - a} & 5 \\ 7 & { - 2} & { - a} \end{array} \right|

= a + 35 - 4 + 14a

=

15a + 31

Now,

{\Delta _1} = 15a + 31

For inconsistent

\Delta

= 0 $\therefore$ a = 3, a = 4 and for a = 3 and 4,

\Delta

1 $\ne$ 0 n(S1) = 2 For infinite solution :

\Delta

= 0 and

\Delta

1 =

\Delta

2 =

\Delta

3 = 0 Not possible $\therefore$ n(S2) = 0

Q265

If the matrix

A = \left( \begin{array}{ll}0 & 2 \\ K & { - 1} \end{array} \right)

satisfies

A({A^3} + 3I) = 2I

, then the value of K is :

A

{1 \over 2}

B

-

{1 \over 2}

C

-

1

D 1

Correct Answer

Option A

Solution

Given matrix

A = \left[ \begin{array}{ll}0 & 2 \\ k & { - 1} \end{array} \right]

{A^4} + 3IA = 2I

\Rightarrow {A^4} = 2I - 3A

Also characteristic equation of A is

|A - \lambda I|\, = 0

\Rightarrow \left| \begin{array}{ll}{0 - \lambda } & 2 \\ k & { - 1 - \lambda } \end{array} \right| = 0

\Rightarrow \lambda + {\lambda ^2} - 2k = 0

\Rightarrow A + {A^2} = 2K.I

\Rightarrow {A^2} = 2KI - A

\Rightarrow {A^4} = 4{K^2}I + {A^2} - 4AK

Put

{A^2} = 2KI - A

and

{A^4} = 2I - 3A

2I - 3A = 4{K^2}I + 2KI - A - 4AK

\Rightarrow I(2 - 2K - 4{K^2}) = A(2 - 4K)

\Rightarrow - 2I(2{K^2} + K - 1) = 2A(1 - 2K)

\Rightarrow - 2I(2K - 1)(K + 1) = 2A(1 - 2K)

\Rightarrow (2K - 1)(2A) - 2I(2K - 1)(K + 1) = 0

\Rightarrow (2K - 1)[2A - 2I(K + 1)] = 0

\Rightarrow K = {1 \over 2}

Q266

Let

A = \left[ \begin{array}{ll}i & { - i} \\ { - i} & i \end{array} \right],i = \sqrt { - 1}

. Then, the system of linear equations

{A^8}\left[ \begin{array}{ll}x \\ y \end{array} \right] = \left[ \begin{array}{ll}8 \\ {64} \end{array} \right]

has :

A Exactly two solutions

B Infinitely many solutions

C A unique solution

D No solution

Correct Answer

Option D

Solution

A = \left[ \begin{array}{ll}i & { - i} \\ { - i} & i \end{array} \right]

{A^2} = \left[ \begin{array}{ll}i & { - i} \\ { - i} & i \end{array} \right]\left[ \begin{array}{ll}i & { - i} \\ { - i} & i \end{array} \right] = \left[ \begin{array}{ll}{ - 2} & 2 \\ 2 & { - 2} \end{array} \right] = 2\left[ \begin{array}{ll}{ - 1} & 1 \\ 1 & { - 1} \end{array} \right]

{A^4} = 4\left[ \begin{array}{ll}{ - 1} & 1 \\ 1 & { - 1} \end{array} \right]\left[ \begin{array}{ll}{ - 1} & 1 \\ 1 & { - 1} \end{array} \right] = 4\left[ \begin{array}{ll}2 & { - 2} \\ { - 2} & 2 \end{array} \right] = 8\left[ \begin{array}{ll}1 & { - 1} \\ { - 1} & 1 \end{array} \right]

{A^8} = 64\left[ \begin{array}{ll}1 & { - 1} \\ { - 1} & 1 \end{array} \right]\left[ \begin{array}{ll}1 & { - 1} \\ { - 1} & 1 \end{array} \right] = 64\left[ \begin{array}{ll}2 & { - 2} \\ { - 2} & 2 \end{array} \right] = 128\left[ \begin{array}{ll}1 & { - 1} \\ { - 1} & 1 \end{array} \right]

128\left[ \begin{array}{ll}1 & { - 1} \\ { - 1} & 1 \end{array} \right]\left[ \begin{array}{ll}x \\ y \end{array} \right] = \left[ \begin{array}{ll}8 \\ {64} \end{array} \right]

128\left[ \begin{array}{ll}{x - y} \\ { - x + y} \end{array} \right] = \left[ \begin{array}{ll}8 \\ {64} \end{array} \right]

\Rightarrow 128(x - y) = 8

\Rightarrow x - y = {1 \over {16}}

.... (1) and

128( - x + y) = 64 \Rightarrow x - y = {{ - 1} \over 2}

.... (2) $\Rightarrow$ No solution (from eq. (1) & (2))

Q267

Let A and B be 3

\times

3 real matrices such that A is symmetric matrix and B is skew-symmetric matrix. Then the system of linear equations (A2B2

-

B2A2) X = O, where X is a 3

\times

1 column matrix of unknown variables and O is a 3

\times

1 null matrix, has :

A no solution

B exactly two solutions

C infinitely many solutions

D a unique solution

Correct Answer

Option C

Solution

AT = A, BT = $-$ B Let A2B2 $-$ B2A2 = P PT = (A2B2 $-$ B2A2)T = (A2B2)T $-$ (B2A2)T = (B2)T (A2)T $-$ (A2)T (B2)T = B2A2 $-$ A2B2 $\Rightarrow$ P is skew-symmetric matrix

\left[ \begin{array}{lll}0 & a & b \\ { - a} & 0 & c \\ { - b} & { - c} & 0 \end{array} \right]\left[ \begin{array}{ll}x \\ y \\ z \end{array} \right] = \left[ \begin{array}{ll}0 \\ 0 \\ 0 \end{array} \right]

$\therefore$ ay + bz = 0 ..... (1) $-$ ax + cz = 0 .... (2) $-$ bx $-$ cy = 0 ..... (3) From equation 1, 2, 3

\Delta

= 0 &

\Delta

1 =

\Delta

2 =

\Delta

3 = 0 $\therefore$ equation have infinite number of solution

Q268

Let

\alpha

be a root of the equation x2 + x + 1 = 0 and the matrix A =

{1 \over {\sqrt 3 }}\left[ \begin{array}{lll}1 & 1 & 1 \\ 1 & \alpha & {{\alpha ^2}} \\ 1 & {{\alpha ^2}} & {{\alpha ^4}} \end{array} \right]

then the matrix A31 is equal to

A A2

B A

C I3

D A3

Correct Answer

Option D

Solution

x2 + x + 1 = 0 $\Rightarrow$ x =

{{ - 1 + i\sqrt 3 } \over 2}

= $\omega$ or

{{ - 1 - i\sqrt 3 } \over 2}

=

{\omega ^2}

Let $\alpha$ = $\omega$ $\therefore$ A =

{1 \over {\sqrt 3 }}\left[ \begin{array}{lll}1 & 1 & 1 \\ 1 & \omega & {{\omega ^2}} \\ 1 & {{\omega ^2}} & {{\omega ^4}} \end{array} \right]

A2 =

{1 \over 3}\left[ \begin{array}{lll}3 & 0 & 0 \\ 0 & 0 & 3 \\ 0 & 3 & 0 \end{array} \right]

=

\left[ \begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array} \right]

Now A4 =

\left[ \begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right]

= I $\therefore$ A31 = A28A3 = A3

Q269

Let A be a 3

\times

3 matrix such that adj A =

\left[ \begin{array}{lll}2 & { - 1} & 1 \\ { - 1} & 0 & 2 \\ 1 & { - 2} & { - 1} \end{array} \right]

and B = adj(adj A). If |A| =

\lambda

and |(B-1)T| =

\mu

, then the ordered pair, (|

\lambda

|,

\mu

) is equal to :

A (3, 81)

B

\left( {9,{1 \over 9}} \right)

C

\left( {3,{1 \over {81}}} \right)

D

\left( {9,{1 \over {81}}} \right)

Correct Answer

Option C

Solution

adj\,A = \left[ \begin{array}{lll}2 & { - 1} & 1 \\ { - 1} & 0 & 2 \\ 1 & { - 2} & { - 1} \end{array} \right]

B = adj\,(adj\,A)

= |A{|^{n - 2}}A

= |A{|^{3 - 2}}.A

[As here n = 3]

= |A|.A

.....(1) Now,

|adj\,A| = \left[ \begin{array}{lll}2 & { - 1} & 1 \\ { - 1} & 0 & 2 \\ 1 & { - 2} & { - 1} \end{array} \right]

= + 2(0 + 4) + 1(1 - 2) + 1(2 - 0)

= 8 - 1 + 2

= 9

Also we know, |adj A| = |A|n $-$ 1 $\therefore$ Here |adj A| = |A|3 $-$ 1 = |A|2 $\therefore$ |A|2 = 9 $\Rightarrow$ |A| = $\pm$ 3 Given, |A| = $\lambda$ $\therefore$ $\lambda$ = $\pm$ 3 $\Rightarrow$ | $\lambda$ | = 3 From (1) B = ( $\pm$ 3)A Given, |(B $-$ 1)T| = $\mu$ $\Rightarrow$ |(BT) $-$ 1| = $\mu$ $\Rightarrow$

{1 \over {|{B^T}|}} = \mu

$\Rightarrow$

{1 \over {|B|}} = \mu

[As |BT| = |B|] $\Rightarrow$

{1 \over {| \pm 3A|}} = \mu

$\Rightarrow$

{1 \over {{{\left( { \pm 3} \right)}^3}\left| A \right|}} = \mu

[As |KA| = Kn |A|]

\Rightarrow {1 \over {( \pm 27)|A|}} = \mu

\Rightarrow \mu = {1 \over {( \pm 27)( \pm 3)}} = \pm {1 \over {81}}

$\therefore$

(|\lambda |,\,\mu ) = \left( {3,\, \pm {1 \over {81}}} \right)

Here correct option will be

\left( {3,\,{1 \over {81}}} \right)

Q270

Let A = [aij] and B = [bij] be two 3 × 3 real matrices such that bij = (3)(i+j-2)aji, where i, j = 1, 2, 3. If the determinant of B is 81, then the determinant of A is:

A 3

B

{1 \over 3}

C

{1 \over 9}

D

{1 \over {81}}

Correct Answer

Option C

Solution

|B| =

\left| \begin{array}{lll}{{b_{11}}} & {{b_{12}}} & {{b_{13}}} \\ {{b_{21}}} & {{b_{22}}} & {{b_{23}}} \\ {{b_{31}}} & {{b_{32}}} & {{b_{33}}} \end{array} \right|

=

\left| \begin{array}{lll}{{3^0}{a_{11}}} & {{3^1}{a_{12}}} & {{3^2}{a_{13}}} \\ {{3^1}{a_{21}}} & {{3^2}{a_{22}}} & {{3^3}{a_{23}}} \\ {{3^2}{a_{31}}} & {{3^3}{a_{32}}} & {{3^4}{a_{33}}} \end{array} \right|

=

{3.3^2}\left| \begin{array}{lll}{{a_{11}}} & {{3^1}{a_{12}}} & {{3^2}{a_{13}}} \\ {{a_{21}}} & {{3^1}{a_{22}}} & {{3^2}{a_{23}}} \\ {{a_{31}}} & {{3^1}{a_{32}}} & {{3^2}{a_{33}}} \end{array} \right|

=

{3.3^2}{.3.3^2}\left| \begin{array}{lll}{{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \end{array} \right|

= 36.|A| $\therefore$ 36.|A| = 81 $\Rightarrow$ |A| =

{1 \over 9}