If a + x = b + y = c + z + 1, where a, b, c, x, y, z are non-zero distinct real numbers, then $$\left| {\matrix{ x & {a + y} & {x + a} \cr y & {b + y} & {y + b} \cr z & {c + y} &am

$$\left| {\matrix{ x & {a + y} & {x + a} \cr y & {b + y} & {y + b} \cr z & {c + y} & {z + c} \cr } } \right|$$ C3 $$ \to $$ C3 – C1 = $$\left| {\matrix{ x & {a + y} & a \cr y & {b + y} & b \cr z & {c + y} & c \cr } } \right|$$ C2 $$ \to $$ C2 – C3 = $$\left| {\matrix{ x & y & a \cr y & y & b \cr z & y & c \cr } } \right|$$ R3 $$ \to $$ R3 – R1, R2 $$ \to $$ R2 – R1 = $$\left| {\matrix{ x & y & a \cr {y - x} & 0 &

Let m and M be respectively the minimum and maximum values of $$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr

$$\left| {\matrix{ {{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \cr {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$ R1 $$ \to $$ R1 – R2, R2 $$ \to $$ R2 – R3 $$\left| {\matrix{ { - 1} & 1 & 0 \cr 1 & 0 & { - 1} \cr {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \cr } } \right|$$ = –1(sin2 x) – 1(1 + sin2x + cos2 x) = - sin2x - 2 $$ \therefore $$ minimum value when sin2x = 1

Let $$\theta = {\pi \over 5}$$ and $$A = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$. If B = A + A4 , then det (B) :

$$A = \left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$ A2 = $$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$$$\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$$ $$ \Rightarrow $$ A2 = $$\left[ {\matrix{ {\cos 2\theta } & {\sin 2\theta } \cr { - \sin 2\theta } & {\cos 2\theta } \cr } } \right]$$ S

Let A = {X = (x, y, z)T: PX = 0 and x2 + y2 + z2 = 1} where $$P = \left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]$$, then the set A :

Let $$X = \left[ {\matrix{ x \cr y \cr z \cr } } \right]$$ PX = O $$\left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$ x + 2y + z = 0........(1) -2x + 3y – 4z = 0....(2) x + 9y - z = 0..........(3) from (1) & (3) $$ \Rightarrow $$ 2x+11y =0 from (1) & (2) $$ \Rightarrow $$ 2x + 11y = 0 from (2) & (3) –6x –33y = 0 $$ \Rightarrow $$

If $$\Delta $$ = $$\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {2x - 3} & {3x - 4} & {4x - 5} \cr {3x - 5} & {5x - 8} & {10x - 17} \cr } } \right|$$ = Ax3 + Bx2 + Cx + D

$$\Delta $$ = $$\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {2x - 3} & {3x - 4} & {4x - 5} \cr {3x - 5} & {5x - 8} & {10x - 17} \cr } } \right|$$ R2 $$ \to $$ R2 – R1 R3 $$ \to $$ R3 – R2 = $$\left| {\matrix{ {x - 2} & {2x - 3} & {3x - 4} \cr {x - 1} & {x - 1} & {x - 1} \cr {x - 2} & {2\left( {x - 2} \right)} & {6\left( {x - 2} \right)} \cr } } \right|$$ = $$\left( {x - 1} \right)\left( {x - 2} \right)\left| {\matrix{ {x - 2} & {2x - 3}

Matrices and Determinants — Page 6 | JEE Mathematics

Q51

If the minimum and the maximum values of the function

f:\left[ {{\pi \over 4},{\pi \over 2}} \right] \to R

, defined by

f\left( \theta \right) = \left| \begin{array}{lll}{ - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \\ { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \\ {12} & {10} & { - 2} \end{array} \right|

are m and M respectively, then the ordered pair (m,M) is equal to :

A

\left( {0,2\sqrt 2 } \right)

B (-4, 0)

C (-4, 4)

D (0, 4)

Correct Answer

Option B

Solution

Given

f\left( \theta \right) = \left| \begin{array}{lll}{ - {{\sin }^2}\theta } & { - 1 - {{\sin }^2}\theta } & 1 \\ { - {{\cos }^2}\theta } & { - 1 - {{\cos }^2}\theta } & 1 \\ {12} & {10} & { - 2} \end{array} \right|

C1 $\to$ C1 – C2 , C3 $\to$ C3 + C2 =

\left| \begin{array}{lll}1 & { - 1 - {{\sin }^2}\theta } & { - {{\sin }^2}\theta } \\ 1 & { - 1 - {{\cos }^2}\theta } & { - {{\cos }^2}\theta } \\ 2 & {10} & 8 \end{array} \right|

C2 $\to$ C2 – C3 =

\left| \begin{array}{lll}1 & { - 1} & { - {{\sin }^2}\theta } \\ 1 & { - 1} & { - {{\cos }^2}\theta } \\ 2 & 2 & 8 \end{array} \right|

= 1(2cos2 $\theta$ – 8) + (8 + 2cos2 $\theta$ ) – 4sin2 $\theta$ = 4cos2 $\theta$ - 4cos2 $\theta$ = 4 cos 2 $\theta$ $\theta$

\in

\left[ {{\pi \over 4},{\pi \over 2}} \right]

$\Rightarrow$ 2 $\theta$

\in

\left[ {{\pi \over 2},{\pi }} \right]

$\Rightarrow$ cos 2 $\theta$

\in

[-1, 0] $\Rightarrow$ 4cos 2 $\theta$

\in

[-4, 0] $\Rightarrow$

f\left( \theta \right)

\in

[-4, 0] $\therefore$ (m, M) = (–4, 0)

Q52

Let

\lambda \in

R . The system of linear equations 2x1 - 4x2 +

\lambda

x3 = 1 x1 - 6x2 + x3 = 2

\lambda

x1 - 10x2 + 4x3 = 3 is inconsistent for:

A exactly one positive value of

\lambda

B exactly one negative value of

\lambda

C exactly two values of

\lambda

D every value of

\lambda

Correct Answer

Option B

Solution

D =

\left| \begin{array}{lll}2 & { - 4} & \lambda \\ 1 & { - 6} & 1 \\ \lambda & { - 10} & 4 \end{array} \right|

= 0 $\Rightarrow$ $\lambda$ = 3,

- {2 \over 3}

D1 =

\left| \begin{array}{lll}1 & { - 4} & \lambda \\ 2 & { - 6} & 1 \\ 3 & { - 10} & 4 \end{array} \right|

= 14 + 4(5) + $\lambda$ (–2) = –2 $\lambda$ + 6 D2 =

\left| \begin{array}{lll}2 & 1 & \lambda \\ 1 & 2 & 1 \\ \lambda & 3 & 4 \end{array} \right|

= –2( $\lambda$ – 3)( $\lambda$ + 1) D3 =

\left| \begin{array}{lll}2 & { - 4} & 1 \\ 1 & { - 6} & 2 \\ \lambda & { - 10} & 3 \end{array} \right|

= – 2 $\lambda$ + 6 When , $\lambda$ = 3 then D = D1 = D2 = D3 = 0 $\Rightarrow$ Infinite many solution When $\lambda$ =

- {2 \over 3}

then D1, D2, D3 none of them is zero so equations are inconsistant. $\therefore$ $\lambda$ =

- {2 \over 3}

Q53

If a + x = b + y = c + z + 1, where a, b, c, x, y, z are non-zero distinct real numbers, then

\left| \begin{array}{lll}x & {a + y} & {x + a} \\ y & {b + y} & {y + b} \\ z & {c + y} & {z + c} \end{array} \right|

is equal to :

A y(b – a)

B y(a – b)

C y(a – c)

D 0

Correct Answer

Option B

Solution

\left| \begin{array}{lll}x & {a + y} & {x + a} \\ y & {b + y} & {y + b} \\ z & {c + y} & {z + c} \end{array} \right|

C3 $\to$ C3 – C1 =

\left| \begin{array}{lll}x & {a + y} & a \\ y & {b + y} & b \\ z & {c + y} & c \end{array} \right|

C2 $\to$ C2 – C3 =

\left| \begin{array}{lll}x & y & a \\ y & y & b \\ z & y & c \end{array} \right|

R3 $\to$ R3 – R1, R2 $\to$ R2 – R1 =

\left| \begin{array}{lll}x & y & a \\ {y - x} & 0 & {b - a} \\ {z - x} & 0 & {c - a} \end{array} \right|

= (–y)[(y – x) (c – a) – (b – a) (z – x)] Given, a + x = b + y = c + z + 1 = (–y)[(a – b) (c – a) + (a – b) (a – c – 1)] = (–y)[(a – b) (c – a) + (a – b) (a – c) + b – a) = –y(b – a) = y(a – b)

Q54

Let m and M be respectively the minimum and maximum values of

\left| \begin{array}{lll}{{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \\ {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \\ {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \end{array} \right|

Then the ordered pair (m, M) is equal to :

A (–3, –1)

B (–4, –1)

C (1, 3)

D (–3, 3)

Correct Answer

Option A

Solution

\left| \begin{array}{lll}{{{\cos }^2}x} & {1 + {{\sin }^2}x} & {\sin 2x} \\ {1 + {{\cos }^2}x} & {{{\sin }^2}x} & {\sin 2x} \\ {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \end{array} \right|

R1 $\to$ R1 – R2, R2 $\to$ R2 – R3

\left| \begin{array}{lll}{ - 1} & 1 & 0 \\ 1 & 0 & { - 1} \\ {{{\cos }^2}x} & {{{\sin }^2}x} & {1 + \sin 2x} \end{array} \right|

= –1(sin2 x) – 1(1 + sin2x + cos2 x) = - sin2x - 2 $\therefore$ minimum value when sin2x = 1 m = - 2 - 1 = -3 $\therefore$ Maximum value when sin2x = –1 M = -2 + 1 = -1 $\therefore$ (m, M) = (–3, –1)

Q55

The values of

\lambda

and

\mu

for which the system of linear equations x + y + z = 2 x + 2y + 3z = 5 x + 3y +

\lambda

z =

\mu

has infinitely many solutions are, respectively:

A 6 and 8

B 5 and 8

C 5 and 7

D 4 and 9

Correct Answer

Option B

Solution

For infinite many solutions D = D1 = D2 = D3 = 0 Now D =

\left| \begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \lambda \end{array} \right|

= 0 $\Rightarrow$ 1. (2 $\lambda$ – 9) –1.( $\lambda$ – 3) + 1.(3 – 2) = 0 $\Rightarrow$ $\lambda$ = 5 Now D1 =

\left| \begin{array}{lll}2 & 1 & 1 \\ 5 & 2 & 3 \\ \mu & 3 & 5 \end{array} \right|

= 0 $\Rightarrow$ 2(10 – 9) –1(25 – 3 $\mu$ ) + 1(15 – 2 $\mu$ ) = 0 $\Rightarrow$ $\mu$ = 8

Q56

Let

\theta = {\pi \over 5}

and

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

. If B = A + A4 , then det (B) :

A lies in (1, 2)

B lies in (2, 3).

C is zero.

D is one.

Correct Answer

Option A

Solution

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

A2 =

\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]

$\Rightarrow$ A2 =

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

Similarly, An =

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

$\therefore$ B = A + A4 =

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

+

\left[ \begin{array}{ll}{\cos 4\theta } & {\sin 4\theta } \\ { - \sin 4\theta } & {\cos 4\theta } \end{array} \right]

=

\left[ \begin{array}{ll}{\cos 4\theta + \cos \theta } & {\sin 4\theta + \sin \theta } \\ { - \sin 4\theta - \sin \theta } & {\cos 4\theta + \cos \theta } \end{array} \right]

detB = (cos4 $\theta$ + cos $\theta$ )2 + (sin4 $\theta$ + sin $\theta$ )2 = cos24 $\theta$ + cos2 $\theta$ + 2cos4 $\theta$ cos $\theta$ + sin24 $\theta$ + sin2 $\theta$ + 2sin4 $\theta$ –sin $\theta$ = 2 + 2 ( cos4 $\theta$ cos $\theta$ + sin4 $\theta$ sin $\theta$ ) $\Rightarrow$ detB = 2 + 2 cos3 $\theta$ at $\theta$ =

{\pi \over 5}

detB = 2 + 2cos

{{3\pi } \over 5}

= 2(1 - sin18) = 2(1 -

{{\sqrt 5 - 1} \over 4}

) = 2

\left( {{{5 - \sqrt 5 } \over 4}} \right)

=

{{{5 - \sqrt 5 } \over 2}}

\simeq

1.385 $\therefore$ detB

\in

(1, 2)

Q57

Let A = {X = (x, y, z)T: PX = 0 and x2 + y2 + z2 = 1} where

P = \left[ \begin{array}{lll}1 & 2 & 1 \\ { - 2} & 3 & { - 4} \\ 1 & 9 & { - 1} \end{array} \right]

, then the set A :

A is an empty set.

B contains more than two elements.

C contains exactly two elements.

D is a singleton.

Correct Answer

Option C

Solution

Let

X = \left[ \begin{array}{ll}x \\ y \\ z \end{array} \right]

PX = O

\left[ \begin{array}{lll}1 & 2 & 1 \\ { - 2} & 3 & { - 4} \\ 1 & 9 & { - 1} \end{array} \right]\left[ \begin{array}{ll}x \\ y \\ z \end{array} \right] = \left[ \begin{array}{ll}0 \\ 0 \\ 0 \end{array} \right]

x + 2y + z = 0........(1) -2x + 3y – 4z = 0....(2) x + 9y - z = 0..........(3) from (1) & (3) $\Rightarrow$ 2x+11y =0 from (1) & (2) $\Rightarrow$ 2x + 11y = 0 from (2) & (3) –6x –33y = 0 $\Rightarrow$ 2x +11y = 0 putting value of x in (1), we get –7y + 2z = 0 Now

{\left( {{{11y} \over 2}} \right)^2} + {y^2} + {\left( {{{7y} \over 2}} \right)^2} = 1

y2(121 + 1 + 49) = 4 y2(171) = 4

y = \pm {2 \over {\sqrt {171} }}

\Rightarrow x = \pm {7 \over {\sqrt {171} }}

\Rightarrow z = \pm {{11} \over {\sqrt {171} }}

$\therefore$ So, there are 2 solution set of (x, y, z)

Q58

If

\Delta

=

\left| \begin{array}{lll}{x - 2} & {2x - 3} & {3x - 4} \\ {2x - 3} & {3x - 4} & {4x - 5} \\ {3x - 5} & {5x - 8} & {10x - 17} \end{array} \right|

= Ax3 + Bx2 + Cx + D, then B + C is equal to :

A -1

B -3

C 9

D 1

Correct Answer

Option B

Solution

\Delta

=

\left| \begin{array}{lll}{x - 2} & {2x - 3} & {3x - 4} \\ {2x - 3} & {3x - 4} & {4x - 5} \\ {3x - 5} & {5x - 8} & {10x - 17} \end{array} \right|

R2 $\to$ R2 – R1 R3 $\to$ R3 – R2 =

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

=

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

C1 $\to$ C1 - C2 C2 $\to$ C2 - C3 =

\left( {x - 1} \right)\left( {x - 2} \right)\left| \begin{array}{lll}{ - x + 1} & { - x + 1} & {3x - 4} \\ 0 & 0 & 1 \\ { - 1} & { - 4} & 6 \end{array} \right|

= -(x - 1)(x - 2)[-4(1 - x) + 1(1 - x)] = -(x2 - 3x + 2)[3x - 3] = -3x3 + 9x2 - 6x + 3x2 - 9x + 6 = -3x3 + 12x2 - 15x + 6 = Ax3 + Bx2 + Cx + D $\therefore$ A = -3, B = 12, C = -15 $\therefore$ B + C = 12 – 15 = – 3

Q59

If the system of linear equations 2x + 2ay + az = 0 2x + 3by + bz = 0 2x + 4cy + cz = 0, where a, b, c

\in

R are non-zero distinct; has a non-zero solution, then:

A

{1 \over a},{1 \over b},{1 \over c}

are in A.P.

B a + b + c = 0

C a, b, c are in G.P.

D a,b,c are in A.P.

Correct Answer

Option A

Solution

For non-zero solution

\left| \begin{array}{lll}2 & {2a} & a \\ 2 & {3b} & b \\ 2 & {4c} & c \end{array} \right| = 0

$\Rightarrow$

\left| \begin{array}{lll}1 & {2a} & a \\ 0 & {3b - 2a} & {b - a} \\ 0 & {4c - 2a} & {c - a} \end{array} \right| = 0

$\Rightarrow$ (3b – 2a) (c –a) – (b – a) (4c – 2a) = 0 $\Rightarrow$ 2ac = bc + ab $\Rightarrow$

{2 \over b} = {1 \over a} + {1 \over c}

$\therefore$

{1 \over a},{1 \over b},{1 \over c}

are in A.P.

Q60

For which of the following ordered pairs (

\mu

,

\delta

), the system of linear equations x + 2y + 3z = 1 3x + 4y + 5z =

\mu

4x + 4y + 4z =

\delta

is inconsistent ?

A (1, 0)

B (4, 3)

C (4, 6)

D (3, 4)

Correct Answer

Option B

Solution

For inconsistent system we need

\Delta

= 0 and atleast one of

\Delta

x,

\Delta

y,

\Delta

z $\ne$ 0 $\therefore$

\Delta

=

\left| \begin{array}{lll}1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 4 & 4 \end{array} \right|

= 0

\Delta

x =

\left| \begin{array}{lll}1 & 2 & 3 \\ \mu & 4 & 5 \\ \delta & 4 & 4 \end{array} \right|

= (-4) - 2( $\mu$ - 5 $\delta$ ) + 3(4 $\mu$ - 4 $\delta$ ) $\Rightarrow$ 2 $\mu$ $\ne$ $\delta$ + 2 ....(

1) Only ( $\mu$ , $\delta$ ) = (4, 3) does satisfy the equation (1).