The system of linear equations $$\matrix{ {x + \lambda y - z = 0} \cr {\lambda x - y - z = 0} \cr {x + y - \lambda z = 0} \cr } $$ has a non-trivial solution for :

For non-trivial solution, we have $$\left| {\matrix{ 1 & \lambda & { - 1} \cr \lambda & { - 1} & { - 1} \cr 1 & 1 & { - \lambda } \cr } } \right| = 0$$ $$ \Rightarrow 1(\lambda + 1) - \lambda ( - {\lambda ^2} + 1) - 1(\lambda + 1) = 0$$ $$ \Rightarrow \lambda ({\lambda ^2} - 1) = 0$$ $$ \Rightarrow \lambda = - 1,0,1$$

If $$A = \left[ {\matrix{ {5a} & { - b} \cr 3 & 2 \cr } } \right]$$ and $$A$$ adj $$A=A$$ $${A^T},$$ then $$5a+b$$ is equal to :

$$A\left( {Adj\,\,A} \right) = A\,{A^T}$$ $$ \Rightarrow {A^{ - 1}}A\left( {adj\,\,A} \right) = {A^{ - 1}}A\,{A^T}$$ $$Adj\,\,A = {A^T}$$ $$ \Rightarrow \left[ {\matrix{ 2 & b \cr { - 3} & {5a} \cr } } \right] = \left[ {\matrix{ {5a} & 3 \cr { - b} & 2 \cr } } \right]$$ $$ \Rightarrow a = {2 \over 5}\,\,$$ and $$\,\,b = 3$$ $$ \Rightarrow 5a + b = 5$$

If $$\left| {\matrix{ {x - 4} & {2x} & {2x} \cr {2x} & {x - 4} & {2x} \cr {2x} & {2x} & {x - 4} \cr } } \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}$$ then the o

$$\left| {\matrix{ {x - 4} & {2x} & {2x} \cr {2x} & {x - 4} & {2x} \cr {2x} & {2x} & {x - 4} \cr } } \right|$$ Applying c1 $$ \to $$ c1 + c2 + c3 $$ = \,\,\,\,\left| {\matrix{ {5x - 4} & {2x} & {2x} \cr {5x - 4} & {x - 4} & {2x} \cr {5x - 4} & {2x} & {x - 4} \cr } } \right|$$ Taking common (5x $$-$$ 4) from c1 $$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 1 & {x - 4} & {2x} \cr 1 & {2x} & {x - 4} \

Matrices and Determinants — Page 3 | JEE Mathematics

Q21

The system of linear equations

\begin{array}{ll}{x + \lambda y - z = 0} \\ {\lambda x - y - z = 0} \\ {x + y - \lambda z = 0} \end{array}

has a non-trivial solution for :

A infinitely many values of

\lambda .

B exactly one value of

\lambda .

C exactly two values of

\lambda .

D exactly three values of

\lambda .

Correct Answer

Option D

Solution

For non-trivial solution, we have

\left| \begin{array}{lll}1 & \lambda & { - 1} \\ \lambda & { - 1} & { - 1} \\ 1 & 1 & { - \lambda } \end{array} \right| = 0

\Rightarrow 1(\lambda + 1) - \lambda ( - {\lambda ^2} + 1) - 1(\lambda + 1) = 0

\Rightarrow \lambda ({\lambda ^2} - 1) = 0

\Rightarrow \lambda = - 1,0,1

Q22

If

A = \left[ \begin{array}{ll}{5a} & { - b} \\ 3 & 2 \end{array} \right]

and

A

adj

A=A

{A^T},

then

5a+b

is equal to :

A

4

B

13

C

-1

D

5

Correct Answer

Option D

Solution

A\left( {Adj\,\,A} \right) = A\,{A^T}

\Rightarrow {A^{ - 1}}A\left( {adj\,\,A} \right) = {A^{ - 1}}A\,{A^T}

Adj\,\,A = {A^T}

\Rightarrow \left[ \begin{array}{ll}2 & b \\ { - 3} & {5a} \end{array} \right] = \left[ \begin{array}{ll}{5a} & 3 \\ { - b} & 2 \end{array} \right]

\Rightarrow a = {2 \over 5}\,\,

and

\,\,b = 3

\Rightarrow 5a + b = 5

Q23

For two 3 × 3 matrices A and B, let A + B = 2BT and 3A + 2B = I3, where BT is the transpose of B and I3 is 3 × 3 identity matrix. Then :

A 5A + 10B = 2I3

B 10A + 5B = 3I3

C B + 2A = I3

D 3A + 6B = 2I3

Correct Answer

Option B

Solution

Given, A + B = 2BT .......(1) $\Rightarrow$ (A + B)T = (2BT)T $\Rightarrow$ AT + BT = 2B $\Rightarrow$ B =

{{{A^T} + {B^T}} \over 2}

Now put this in equation (1) So, A +

{{{A^T} + {B^T}} \over 2}

= 2BT $\Rightarrow$ 2A + AT = 3BT $\Rightarrow$ A =

{{3{B^T} - {A^T}} \over 2}

Also, 3A + 2B = I3 .......(2) $\Rightarrow$

3\left( {{{3{B^T} - {A^T}} \over 2}} \right) + 2\left( {{{{A^T} + {B^T}} \over 2}} \right)

= I3 $\Rightarrow$ 11BT - AT = 2I3 $\Rightarrow$ (11BT - AT)T = (2I3)T $\Rightarrow$ 11B - A = 2I3 ........(

3) Multiply (3) by 3 and then adding (2) and (3) we get, 35B = 7I3 $\Rightarrow$ B =

{{{I_3}} \over 5}

From (3), 11

{{{I_3}} \over 5}

- A = 2I3 $\Rightarrow$ A =

{{{I_3}} \over 5}

$\therefore$ 5A = 5B = I3 $\Rightarrow$ 10A + 5B = 3I3

Q24

The number of real values of

\lambda

for which the system of linear equations 2x + 4y

-

\lambda

z = 0 4x +

\lambda

y + 2z = 0

\lambda

x + 2y + 2z = 0 has infinitely many solutions, is :

A 0

B 1

C 2

D 3

Correct Answer

Option B

Solution

The system of equations can be written in the matrix form as

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

The system has infinite solutions; thus, we get

\left| \begin{array}{lll}2 & 4 & { - \lambda } \\ 4 & \lambda & 2 \\ \lambda & 2 & 2 \end{array} \right| = 0

\Rightarrow 0 = 2(2\lambda - 4) - 4(8 - 2\lambda ) - \lambda (8 - {\lambda ^2})

\Rightarrow 4\lambda - 8 - 32 + 8\lambda - 8\lambda + {\lambda ^3} = 0

\Rightarrow {\lambda ^3} + 4\lambda - 40 = 0

We can solve this by graphical method:

y = {x^3} + 4x - 40

For x = 0, y = $-$ 40: If we take y = $-$ 40, then we have

- 40 = {x^3} + 4x - 40

\Rightarrow {x^3} + 4x = 0

\Rightarrow x({x^2} + 4) = 0

\Rightarrow x = 0,{x^2} + 4 = 0

\Rightarrow x = \pm \,2i

The given equation of line intersects x only at one point; therefore, the real value of $\lambda$ is only one.

Q25

If the system of equations x+y+z=2 2x+4y–z=6 3x+2y+

\lambda

z=

\mu

has infinitely many solutions, then

A 2

\lambda

-

\mu

= 5

B

\lambda

- 2

\mu

= -5

C 2

\lambda

+

\mu

= 14

D

\lambda

+ 2

\mu

= 14

Correct Answer

Option C

Solution

D = 0\,\left| \begin{array}{lll}1 & 1 & 1 \\ 2 & 4 & { - 1} \\ 3 & 2 & \lambda \end{array} \right| = 0

$\Rightarrow$

(4\lambda + 2) - 1(2\lambda + 3) + 1(4 - 12) = 0

$\Rightarrow$

4\lambda + 2

- 2\lambda - 3

$-$ 8

= 0

$\Rightarrow$

2\lambda = 9 \Rightarrow \lambda = {9 \over 2}

{D_x} = \left| \begin{array}{lll}2 & 1 & 1 \\ 6 & 4 & { - 1} \\ \mu & 2 & { - 9/2} \end{array} \right| = 0

\Rightarrow \mu = 5

By checking all the options we find (C) is correct

2\lambda + \mu = 14

Q26

Suppose A is any 3

\times

3 non-singular matrix and ( A

-

3I) (A

-

5I) = O where I = I3 and O = O3. If

\alpha

A +

\beta

A-1 = 4I, then

\alpha

+

\beta

is equal to :

A 8

B 7

C 13

D 12

Correct Answer

Option A

Solution

Given, ( A $-$ 3I) (A $-$ 5I) = O $\Rightarrow$ A2 - 8A + 15I = O Multiplying both sides by A- 1, we get, A- 1A.A - 8A- 1A + 15A- 1I = A- 1O $\Rightarrow$ A - 8I + 15A- 1 = O $\Rightarrow$ A + 15A- 1 = 8I $\Rightarrow$

{A \over 2} + {{15{A^{ - 1}}} \over 2} = 4I

Comparing with the equation $\alpha$ A + $\beta$ A-1 = 4I, we get $\alpha$ =

{1 \over 2}

and $\beta$ =

{15 \over 2}

$\therefore$ $\alpha$ + $\beta$ =

{1 \over 2}

+

{15 \over 2}

=

{16 \over 2}

= 8

Q27

Let A be any 3

\times

3 invertible matrix. Then which one of the following is not always true ?

A adj (A) =

\left| \right.

A

\left| \right.

.A

-

1

B adj (adj(A)) =

\left| \right.

A

\left| \right.

.A

C adj (adj(A)) =

\left| \right.

A

\left| \right.

2.(adj(A))

-

1

D adj (adj(A)) =

\left| \, \right.

A

\left| \, \right.

.(adj(A))

-

1

Correct Answer

Option D

Solution

We know, the formula A-1 =

{{adj\left( A \right)} \over {\left| A \right|}}

$\therefore$ adj (A) =

\left| \right.

A

\left| \right.

.A $-$ 1 So, Option (A) is true. We know, the formula adj (adj (A)) =

{\left| A \right|^{n - 2}}.A

Now if we put n = 3 as given that A is a 3 $\times$ 3 matrix, we get adj (adj (A)) =

{\left| A \right|^{3 - 2}}.A

=

\left| A \right|.A

So, Option (B) is also true. We know, the formula adj (adj (A)) =

{\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}

Now if we put n = 3 as given that A is a 3 $\times$ 3 matrix, we get adj (adj (A)) =

{\left| A \right|^{3 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}

=

{\left| A \right|^{2}}{\left( {adj\left( A \right)} \right)^{ - 1}}

So, Option (C) is also true. Now in this formula adj (adj (A)) =

{\left| A \right|^{n - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}

if we put n = 2, we get adj (adj (A)) =

{\left| A \right|^{2 - 1}}{\left( {adj\left( A \right)} \right)^{ - 1}}

=

{\left| A \right|}{\left( {adj\left( A \right)} \right)^{ - 1}}

But as A is a 3 $\times$ 3 matrix so we can not take n = 2, so we can say for a 3 $\times$ 3 matrix option (D) is not true.

So, Option (D) is false.

Q28

If S is the set of distinct values of 'b' for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is :

A an empty set

B an infinite set

C a finite set containing two or more elements

D a singleton

Correct Answer

Option D

Solution

\left| \begin{array}{lll}1 & 1 & 1 \\ 1 & a & 1 \\ a & b & 1 \end{array} \right| = 0

$\Rightarrow$ 1 [a – b] – 1 [1 – a] + 1 [b – a2] = 0 $\Rightarrow$ (a - 1)2 = 0 $\Rightarrow$ a = 1 For a = 1, the equations become x + y + z = 1 x + y + z = 1 x + by + z = 0 These equations give no solution for b = 1 $\Rightarrow$ S is singleton set.

Q29

If the system of linear equations x + ky + 3z = 0 3x + ky - 2z = 0 2x + 4y - 3z = 0 has a non-zero solution (x, y, z), then

{{xz} \over {{y^2}}}

is equal to

A 30

B -10

C 10

D -30

Correct Answer

Option C

Solution

System of equations has non-zero solution when determinant of coefficient = 0. So, in this questions,

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

\Rightarrow \,\,\,\,

( $-$ 3K + 8) $-$ K ( $-$ 9 + 4) + 3(12 $-$ 2K) = 0

\Rightarrow \,\,\,\,

$-$ 3K + 8 + 9K $-$ 4K + 36 $-$ 6K = 0

\Rightarrow \,\,\,\,

$-$ 4K + 44 = 0

\Rightarrow \,\,\,\,

K = 11 Now the equations become x + 11y + 3z = 0 . . . (1) 3x + 11y $-$ 2z = 0 . . . (2) 2x + 4y $-$ 3z = 0 . . .

(3) By adding equation (1) and (3) we get, 3x + 15y = 0

\Rightarrow \,\,\,\,

x = $-$ 5y Putting x = $-$ 5y in equation (1) we get $-$ 5y + 11y + 3z = 0

\Rightarrow \,\,\,\,

6y + 3z = 0

\Rightarrow \,\,\,\,

z = $-$ 2y

\therefore\,\,\,\,

{{xz} \over {{y^2}}}

= {{\left( { - 5y} \right)\left( { - 2y} \right)} \over {{y^2}}}

= {{10{y^2}} \over {{y^2}}}

= 10

Q30

If

\left| \begin{array}{lll}{x - 4} & {2x} & {2x} \\ {2x} & {x - 4} & {2x} \\ {2x} & {2x} & {x - 4} \end{array} \right| = \left( {A + Bx} \right){\left( {x - A} \right)^2}

then the ordered pair (A, B) is equal to :

A (4, 5)

B (-4, -5)

C (-4, 3)

D (-4, 5)

Correct Answer

Option D

Solution

\left| \begin{array}{lll}{x - 4} & {2x} & {2x} \\ {2x} & {x - 4} & {2x} \\ {2x} & {2x} & {x - 4} \end{array} \right|

Applying c1 $\to$ c1 + c2 + c3

= \,\,\,\,\left| \begin{array}{lll}{5x - 4} & {2x} & {2x} \\ {5x - 4} & {x - 4} & {2x} \\ {5x - 4} & {2x} & {x - 4} \end{array} \right|

Taking common (5x $-$ 4) from c1

= \,\,\,\,\left( {5x - 4} \right)\left| \begin{array}{lll}1 & {2x} & {2x} \\ 1 & {x - 4} & {2x} \\ 1 & {2x} & {x - 4} \end{array} \right|

Apply R2 $\to$ R2 $-$ R1 and R3 $\to$ R3 $-$ R1

= \,\,\,\,\left( {5x - 4} \right)\left| \begin{array}{lll}1 & {2x} & {2x} \\ 0 & { - \left( {x + 4} \right)} & 0 \\ 0 & 0 & { - \left( {x + 4} \right)} \end{array} \right|

= \,\,\,\,\left( {5x - 4} \right){\left( {x + 4} \right)^2}

So, (A + Bx) (x $-$ A)2 = (5x $-$ 4) (x + 4)2 By comparing both sides we get, A = $-$ 4 and B = 5