If $$\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$$ = (a + b + c) (x + a + b + c)2, x $$ \ne $$

$$\left| {\matrix{ {a - b - c} & {2a} & {2a} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$$ R1 $$ \to $$ R1 + R2 + R3 $$ = \left| {\matrix{ {a + b + c} & {a + b + c} & {a + b + c} \cr {2b} & {b - c - a} & {2b} \cr {2c} & {2c} & {c - a - b} \cr } } \right|$$ $$ = \left( {a + b + c} \right)\left| {\matrix{ 1 & 0 & 0 \cr {2b} & { - \left( {a + b + c} \right)} & 0 \cr {2c} & {2c} & {c - a - b} \cr

Let d $$ \in $$ R, and $$A = \left[ {\matrix{ { - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \cr 1 & {\left( {\sin \theta } \right) + 2} & d \cr 5 & {\left( {2\sin \theta

$$\det A = \left| {\matrix{ { - 2} & {4 + d} & {\sin \theta - 2} \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \cr } } \right|$$ (R1 $$ \to $$ R1 + R3 $$-$$ 2R2) $$ = \left| {\matrix{ 1 & 0 & 0 \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \cr } } \right|$$ = (2 + sin $$\theta $$) ( 2 + 2d $$-$$ sin$$\theta $$) $$-$$ d(2sin$$\theta $$ $$-$$ d) = 4 + 4d $$-$$ 2sin$$\theta $$

If $$A = \left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$$, then the matrix A–50 when $$\theta $$ = $$\pi \over 12$$, is equal to :

(A$$-$$50) = (A$$-$$1)50 We know, A$$-$$1 = $${{adjA} \over {\left| A \right|}}$$ $$\left| A \right|$$ = cos2$$\theta $$ + sin2$$\theta $$ = 1 cofactor of A = $$\left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$$ Adjoint of A = Transpose of cofactor matrix $$ \therefore $$ Adj A = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$ $$ \therefore $$ A$$-$$1 = $$\left[ {\

Let $$A$$ be a matrix such that $$A.\left[ {\matrix{ 1 & 2 \cr 0 & 3 \cr } } \right]$$ is a scalar matrix and |3A| = 108. Then A2 equals :

According to questions, A. $$\left[ {\matrix{ 1 & 2 \cr 0 & 3 \cr } } \right]$$ = $$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$ \Rightarrow $$ A = $$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$\left[ {\matrix{ 1 & 2 \cr 0 & 3 \cr } } \right]^{-1}$$ $$ \Rightarrow $$ A = $$1 \over 3$$$$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$$ $$\left[ {\matrix{ 3 & {-2} \cr 0 & 1 \cr } } \right]$$ $$ \

Matrices and Determinants — Page 25 | JEE Mathematics

Q241

The number of values of

\theta

\in

(0,

\pi

) for which the system of linear equations x + 3y + 7z = 0

-

x + 4y + 7z = 0 (sin3

\theta

)x + (cos2

\theta

)y + 2z = 0. has a non-trival solution, is -

A two

B one

C four

D three

Correct Answer

Option A

Solution

\left| \begin{array}{lll}1 & 3 & 7 \\ { - 1} & 4 & 7 \\ {\sin 3\theta } & {\cos 2\theta } & 2 \end{array} \right| = 0

(8 $-$ 7 cos 2 $\theta$ ) $-$ 3( $-$ 2 $-$ 7 sin 3 $\theta$ ) +7 ( $-$ cos 2 $\theta$ $-$ 4 sin 3 $\theta$ ) = 0 14 $-$ 7 cos 2 $\theta$ + 21 sin 3 $\theta$ $-$ 7 cos 2 $\theta$ $-$ 28 sin 3 $\theta$ = 0 14 $-$ 7 sin 3 $\theta$ $-$ 14 cos 2 $\theta$ = 0 14 $-$ 7 (3 sin $\theta$ $-$ 4 sin3 $\theta$ ) $-$ 14 (1 $-$ 2 sin2 $\theta$ ) = 0 $-$ 21 sin $\theta$ + 28 sin3 $\theta$ + 28 sin2 $\theta$ = 0 7 sin $\theta$ [ $-$ 3 + 4 sin2 $\theta$ + 4 sin $\theta$ ] = 0 sin $\theta$ , 4 sin2 $\theta$ + 6 sin $\theta$ $-$ 2 sin $\theta$ $-$ 3 = 0 2 sin $\theta$ (2 sin $\theta$ + 3) $-$ 1 (2 sin $\theta$ + 3) = 0 sin $\theta$ =

{{ - 3} \over 2}

; sin $\theta$ =

{1 \over 2}

Hence, 2 solutions in (0, $\pi$ )

Q242

Let A be a 3

\times

3 matrix such that A2

-

5A + 7I = 0 Statement - I : A

-

1 =

{1 \over 7}

(5I

-

A). Statement - II : The polynomial A3

-

2A2

-

3A + I can be reduced to 5(A

-

4I). Then :

A Statement-I is true, but Statement-II is false.

B Statement-I is false, but Statement-II is true.

C Both the statements are true.

D Both the statements are false

Correct Answer

Option C

Solution

Given, A2 $-$ 5A + 7I = 0 $\Rightarrow$ A2 $-$ 5A = $-$ 7I $\Rightarrow$ AAA $-$ 1 $-$ 5AA $-$ 1 = $-$ 7IA $-$ 1 $\Rightarrow$ AI $-$ 5I = $-$ 7A $-$ 1 $\Rightarrow$ A $-$ 5I = $-$ 7A $-$ 1 $\Rightarrow$ A $-$ 1 =

{1 \over 7}

(5I $-$ A) Hence, statement 1 is true.

Now A3 $-$ 2A2 $-$ 3A + I = A(A2) $-$ 2A2 $-$ 3A + I = A(5A $-$ 7I) $-$ 2A2 $-$ 3A + I = 5A2 $-$ 7A $-$ 2A2 $-$ 3A + I = 3A2 $-$ 10A + I = 3(5A $-$ 7I) $-$ 10A + I = 15A $-$ 21A $-$ 10A + I = 5A $-$ 20I = 5(A $-$ 4I) So, statement 2 is also correct.

Q243

If

\left| \begin{array}{lll}{a - b - c} & {2a} & {2a} \\ {2b} & {b - c - a} & {2b} \\ {2c} & {2c} & {c - a - b} \end{array} \right|

= (a + b + c) (x + a + b + c)2, x

\ne

0, then x is equal to :

A –2(a + b + c)

B 2(a + b + c)

C abc

D –(a + b + c)

Correct Answer

Option A

Solution

\left| \begin{array}{lll}{a - b - c} & {2a} & {2a} \\ {2b} & {b - c - a} & {2b} \\ {2c} & {2c} & {c - a - b} \end{array} \right|

R1 $\to$ R1 + R2 + R3

= \left| \begin{array}{lll}{a + b + c} & {a + b + c} & {a + b + c} \\ {2b} & {b - c - a} & {2b} \\ {2c} & {2c} & {c - a - b} \end{array} \right|

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

$=$ (a + b + c) (a + b + c)2 $\Rightarrow$ x $=$ $-$ 2(a + b + c)

Q244

Let d

\in

R, and

A = \left[ \begin{array}{lll}{ - 2} & {4 + d} & {\left( {\sin \theta } \right) - 2} \\ 1 & {\left( {\sin \theta } \right) + 2} & d \\ 5 & {\left( {2\sin \theta } \right) - d} & {\left( { - \sin \theta } \right) + 2 + 2d} \end{array} \right],

\theta \in \left[ {0,2\pi } \right]

If the minimum value of det(A) is 8, then a value of d is -

A

-

7

B

2\left( {\sqrt 2 + 2} \right)

C

-

5

D

2\left( {\sqrt 2 + 1} \right)

Correct Answer

Option C

Solution

\det A = \left| \begin{array}{lll}{ - 2} & {4 + d} & {\sin \theta - 2} \\ 1 & {\sin \theta + 2} & d \\ 5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \end{array} \right|

(R1 $\to$ R1 + R3 $-$ 2R2)

= \left| \begin{array}{lll}1 & 0 & 0 \\ 1 & {\sin \theta + 2} & d \\ 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \end{array} \right|

= (2 + sin $\theta$ ) ( 2 + 2d $-$ sin $\theta$ ) $-$ d(2sin $\theta$ $-$ d) = 4 + 4d $-$ 2sin $\theta$ + 2sin $\theta$ + 2dsin $\theta$ $-$ sin2 $\theta$ $-$ 2dsin $\theta$ + d2 d2 + 4d + 4 $-$ sin2 $\theta$ = (d + 2)2 $-$ sin2 $\theta$ For a given d, minimum value of det(A) = (d + 2)2 $-$ 1 = 8 $\Rightarrow$ d = 1 or $-$ 5

Q245

If

A = \left[ \begin{array}{ll}{\cos \theta } & { - \sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array} \right]

, then the matrix A–50 when

\theta

=

\pi \over 12

, is equal to :

A

\left[ \begin{array}{ll}{ {{\sqrt 3 } \over 2}} & { - {1 \over 2}} \\ {{{ 1} \over 2}} & {{{\sqrt 3 } \over 2}} \end{array} \right]

B

\left[ \begin{array}{ll}{{1 \over 2}} & -{{{\sqrt 3 } \over 2}} \\ {{{\sqrt 3 } \over 2}} & {{{ - 1} \over 2}} \end{array} \right]

C

\left[ \begin{array}{ll}{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \\ -{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \end{array} \right]

D

\left[ \begin{array}{ll}{{1 \over 2}} & {{{\sqrt 3 } \over 2}} \\ {-{{\sqrt 3 } \over 2}} & {{{ 1} \over 2}} \end{array} \right]

Correct Answer

Option C

Solution

(A $-$ 50) = (A $-$ 1)50 We know, A $-$ 1 =

{{adjA} \over {\left| A \right|}}

\left| A \right|

= cos2 $\theta$ + sin2 $\theta$ = 1 cofactor of A =

\left[ \begin{array}{ll}{\cos \theta } & { - \sin \theta } \\ {\sin \theta } & {\cos \theta } \end{array} \right]

Adjoint of A = Transpose of cofactor matrix $\therefore$ Adj A =

\left[ \begin{array}{ll}{\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \end{array} \right]

$\therefore$ A $-$ 1 =

\left[ \begin{array}{ll}{\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \end{array} \right]

$\therefore$ (A $-$ 1)2 =

\left[ \begin{array}{ll}{\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \end{array} \right]\left[ \begin{array}{ll}{\cos \theta } & {\sin \theta } \\ { - \sin \theta } & {\cos \theta } \end{array} \right]

=

\left[ \begin{array}{ll}{\cos 2\theta } & {\sin 2\theta } \\ { - \sin 2\theta } & {\cos 2\theta } \end{array} \right]

Similarly, (A $-$ 1)3 =

\left[ \begin{array}{ll}{\cos 3\theta } & {\sin 3\theta } \\ { - \sin 3\theta } & {\cos 3\theta } \end{array} \right]

: : : (A $-$ 1)50 =

\left[ \begin{array}{ll}{\cos 50\theta } & {\sin 50\theta } \\ { - \sin 50\theta } & {\cos 50\theta } \end{array} \right]

when $\theta$ =

{\pi \over {12}}

then

{A^{ - 50}}

=

\left[ \begin{array}{ll}{\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \\ { - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \end{array} \right]

=

\left[ \begin{array}{ll}{{{\sqrt 3 } \over 2}} & {{1 \over 2}} \\ { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \end{array} \right]

Note:

\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}

Q246

Let A and B be two invertible matrices of order 3

\times

3. If det(ABAT) = 8 and det(AB–1) = 8, then det (BA–1 BT) is equal to :

A

{1 \over 4}

B 16

C

{1 \over {16}}

D 1

Correct Answer

Option C

Solution

{\left| A \right|^2}.\left| B \right| = 8

and

{{\left| A \right|} \over {\left| B \right|}} = 8 \Rightarrow \left| A \right| = 4

and

\left| B \right| = {1 \over 2}

$\therefore$ det(BA $-$ 1. BT)

= {1 \over 4} \times {1 \over 4} = {1 \over {16}}

Q247

The value of k

\in

R, for which the following system of linear equations 3x

-

y + 4z = 3, x + 2y

-

3z =

-

2 6x + 5y + kz =

-

3, has infinitely many solutions, is :

A 3

B

-

5

C 5

D

-

3

Correct Answer

Option B

Solution

\left| \begin{array}{lll}3 & { - 1} & 4 \\ 1 & 2 & { - 3} \\ 6 & 5 & k \end{array} \right| = 0

$\Rightarrow$ 3(2k + 15) + K + 18 $-$ 28 = 0 $\Rightarrow$ 7k + 35 = 0 $\Rightarrow$ k = $-$ 5

Q248

Let

f(x) = \left| \begin{array}{lll}a & { - 1} & 0 \\ {ax} & a & { - 1} \\ {a{x^2}} & {ax} & a \end{array} \right|,\,a \in R

. Then the sum of the squares of all the values of a, for which

2f'(10) - f'(5) + 100 = 0

, is

A 117

B 106

C 125

D 136

Correct Answer

Option C

Solution

f(x) = \left| \begin{array}{lll}a & { - 1} & 0 \\ {ax} & a & { - 1} \\ {a{x^2}} & {ax} & a \end{array} \right|,\,a \in R

f(x) = a({a^2} + ax) + 1({a^2}x + a{x^2})

= a{(x + a)^2}

f'(x) = 2a(x + a)

Now,

2f'(10) - f'(5) + 100 = 0

\Rightarrow 2.\,2a(10 + a) - 2a(5 + a) + 100 = 0

\Rightarrow 2a(a + 15) + 100 = 0

\Rightarrow {a^2} + 15a + 50 = 0

\Rightarrow a = - 10,\, - 5

$\therefore$ Sum of squares of values of a = 125.

Q249

The following system of linear equations 7x + 6y – 2z = 0 3x + 4y + 2z = 0 x – 2y – 6z = 0, has

A no solution

B infinitely many solutions, (x, y, z) satisfying y = 2z

C infinitely many solutions, (x, y, z) satisfying x = 2z

D only the trivial solution

Correct Answer

Option C

Solution

Given 7x + 6y – 2z = 0 .......(1) 3x + 4y + 2z = 0 ......(2) x – 2y – 6z = 0 .......(3)

\Delta

=

\left| \begin{array}{lll}7 & 6 & { - 2} \\ 3 & 4 & 2 \\ 1 & { - 2} & { - 6} \end{array} \right|

= 7(–24 + 4) – 6(–18 – 2) – 2(–6 – 4) = 0 $\therefore$

\Delta

= 0 The system of equation has infinite non-trival solution.

Also adding equation (1) and 3 $\times$ (3), we get 10x = 20z $\Rightarrow$ x = 2z

Q250

Let

A

be a matrix such that

A.\left[ \begin{array}{ll}1 & 2 \\ 0 & 3 \end{array} \right]

is a scalar matrix and |3A| = 108. Then A2 equals :

A

\left[ \begin{array}{ll}4 & { - 32} \\ 0 & {36} \end{array} \right]

B

\left[ \begin{array}{ll}{36} & 0 \\ { - 32} & 4 \end{array} \right]

C

\left[ \begin{array}{ll}4 & 0 \\ { - 32} & {36} \end{array} \right]

D

\left[ \begin{array}{ll}{36} & { - 32} \\ 0 & 4 \end{array} \right]

Correct Answer

Option D

Solution

According to questions, A.

\left[ \begin{array}{ll}1 & 2 \\ 0 & 3 \end{array} \right]

=

\left[ \begin{array}{ll}\lambda & 0 \\ 0 & \lambda \end{array} \right]

$\Rightarrow$ A =

\left[ \begin{array}{ll}\lambda & 0 \\ 0 & \lambda \end{array} \right]

\left[ \begin{array}{ll}1 & 2 \\ 0 & 3 \end{array} \right]^{-1}

$\Rightarrow$ A =

1 \over 3

\left[ \begin{array}{ll}\lambda & 0 \\ 0 & \lambda \end{array} \right]

\left[ \begin{array}{ll}3 & {-2} \\ 0 & 1 \end{array} \right]

$\Rightarrow$ A =

\left[ \begin{array}{ll}\lambda & 0 \\ 0 & \lambda \end{array} \right]

\left[ \begin{array}{ll}1 & { - {2 \over 3}} \\ 0 & {{1 \over 3}} \end{array} \right]

$\Rightarrow$ A =

\left[ \begin{array}{ll}\lambda & { - {2 \over 3}\lambda } \\ 0 & {{\lambda \over 3}} \end{array} \right]

As

\left| {3A} \right|

= 108 $\Rightarrow$ 108 =

\left| \begin{array}{ll}{3\lambda } & { - 2\lambda } \\ 0 & \lambda \end{array} \right|

$\Rightarrow$ 3 $\lambda$ 2 = 108 $\Rightarrow$ $\lambda$ 2 = 36 $\Rightarrow$ $\lambda$ = $\pm$ 6 When $\lambda$ = +6 then A =

\left[ \begin{array}{ll}6 & { - 4} \\ 0 & 2 \end{array} \right]

$\Rightarrow$ A2 =

\left[ \begin{array}{ll}{36} & { - 32} \\ 0 & 4 \end{array} \right]

For $\lambda$ = -6 A =

\left[ \begin{array}{ll}{ - 6} & 4 \\ 0 & { - 2} \end{array} \right]

$\Rightarrow$ A2 =

\left[ \begin{array}{ll}{36} & { - 32} \\ 0 & 4 \end{array} \right]