If $$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$$$$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]$$....$$\left[ {\matrix

Given $$\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$$$$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]$$....$$\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & {78} \cr 0 & 1 \cr } } \right]$$ $$ \Rightarrow $$ $$\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right].....\left[ {\matrix{ 1 & {n - 1} \cr 0 & 1 \

If A is a symmetric matrix and B is a skew-symmetric matrix such that A + B = $$\left[ {\matrix{ 2 & 3 \cr 5 & { - 1} \cr } } \right]$$, then AB is equal to :

$$A + B = \left[ {\matrix{ 2 & 3 \cr 5 & { - 1} \cr } } \right] = P(say)$$ Now $$A = {{P + {P^T}} \over 2}\& B = {{P - {P^T}} \over 2}$$ So $$A = {1 \over 2}\left( {\left[ {\matrix{ 2 & 3 \cr 5 & { - 1} \cr } } \right] + \left[ {\matrix{ 2 & 5 \cr 3 & { - 1} \cr } } \right]} \right) = \left[ {\matrix{ 2 & 4 \cr 4 & { - 1} \cr } } \right]$$ $$B = {1 \over 2}\left( {\left[ {\matrix{ 2 & 3 \cr 5 & { - 1} \cr } } \right] - \left[ {\matrix{ 2 & 5 \cr 3

Matrices and Determinants — Page 4 | JEE Mathematics

Q31

The number of values of k for which the system of linear equations, (k + 2)x + 10y = k kx + (k +3)y = k -1 has no solution, is :

A 1

B 2

C 3

D infinitely many

Correct Answer

Option A

Solution

System of linear equation have no solution,

\therefore\,\,\,

determinant of coefficient = 0

\left| \begin{array}{ll}{k + 2} & {10} \\ k & {k + 3} \end{array} \right| = 0

$\Rightarrow$

\,\,\,\,

(k + 2) (k + 3) $-$ 10 K = 0 $\Rightarrow$

\,\,\,\,

k2 $-$ 5k + 6 = 0

\therefore\,\,\,\,

k = 2, 3 When, k = 2 then equations become, 4x + 10y = 2 and 2x + 5y = 1 It has in finite number of solutions.

When k = 3, equations becomes 5x + 10y = 3 3x + 6y = 2 Those equation has no solutions.

\therefore\,\,\,\,

When k = 3, then system of equations have no solutions.

Q32

Let S be the set of all real values of k for which the systemof linear equations x + y + z = 2 2x + y

-

z = 3 3x + 2y + kz = 4 has a unique solution. Then S is :

A an empty set

B equal to {0}

C equal to R

D equal to R

-

{0}

Correct Answer

Option D

Solution

As system of linear equations have unique solutions so, determinant of coefficient $\ne$ 0 $\therefore$

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

$\ne$ 0 $\Rightarrow$ k + 2 - (2k + 3) + 1 $\ne$ 0 $\Rightarrow$ k $\ne$ 0 $\therefore$ k

\in

R - {0}

Q33

The greatest value of c

\in

R for which the system of linear equations x – cy – cz = 0 cx – y + cz = 0 cx + cy – z = 0 has a non-trivial solution, is :

A -1

B 0

C 1/2

D 2

Correct Answer

Option C

Solution

If the system of equations has non-trivial solutions, then D = 0

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

$\Rightarrow$ (1 - c2) + c(-c - c2) - c(c2 + c) = 0 $\Rightarrow$ (1 + c)(1 - c - 2c2) = 0 $\Rightarrow$ (1 + c)2 (1 - 2c) = 0 $\Rightarrow$ c = -1 or

{1 \over 2}

$\therefore$ Greatest value of c is

{1 \over 2}

.

Q34

If

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

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

....

\left[ \begin{array}{ll}1 & {n - 1} \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{ll}1 & {78} \\ 0 & 1 \end{array} \right]

, then the inverse of

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

is

A

\left[ \begin{array}{ll}1 & { 0} \\ {12} & 1 \end{array} \right]

B

\left[ \begin{array}{ll}1 & { 0} \\ {13} & 1 \end{array} \right]

C

\left[ \begin{array}{ll}1 & { - 13} \\ 0 & 1 \end{array} \right]

D

\left[ \begin{array}{ll}1 & { - 12} \\ 0 & 1 \end{array} \right]

Correct Answer

Option C

Solution

Given

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

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

....

\left[ \begin{array}{ll}1 & {n - 1} \\ 0 & 1 \end{array} \right] = \left[ \begin{array}{ll}1 & {78} \\ 0 & 1 \end{array} \right]

$\Rightarrow$

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

=

\left[ \begin{array}{ll}1 & {78} \\ 0 & 1 \end{array} \right]

$\Rightarrow$

\left[ \begin{array}{ll}1 & 6 \\ 0 & 1 \end{array} \right].....\left[ \begin{array}{ll}1 & {n - 1} \\ 0 & 1 \end{array} \right]

=

\left[ \begin{array}{ll}1 & {78} \\ 0 & 1 \end{array} \right]

$\Rightarrow$

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

=

\left[ \begin{array}{ll}1 & {78} \\ 0 & 1 \end{array} \right]

. . . . $\Rightarrow$

\left[ \begin{array}{ll}1 & {1 + 2 + 3 + .... + \left( {n - 1} \right)} \\ 0 & 1 \end{array} \right]

=

\left[ \begin{array}{ll}1 & {78} \\ 0 & 1 \end{array} \right]

By comparing both sides we get, 1 + 2 + 3 + ........+ (n - 1) = 78 $\Rightarrow$

{{n\left( {n - 1} \right)} \over 2}

= 78 $\Rightarrow$ n = 13, - 12(not possible) $\therefore$ The inverse of

\left[ \begin{array}{ll}1 & 13 \\ 0 & 1 \end{array} \right]

=

\left[ \begin{array}{ll}1 & -13 \\ 0 & 1 \end{array} \right]

Q35

If the system of equations 2x + 3y – z = 0, x + ky – 2z = 0 and 2x – y + z = 0 has a non-trival solution (x, y, z), then

{x \over y} + {y \over z} + {z \over x} + k

is equal to :-

A -4

B

{3 \over 4}

C

{1 \over 2}

D

-{1 \over 4}

Correct Answer

Option C

Solution

Given 2x + 3y – z = 0, x + ky – 2z = 0 2x – y + z = 0 For non trivial solution

\Delta = 0 \Rightarrow \left| \begin{array}{lll}2 & 3 & { - 1} \\ 1 & k & { - 2} \\ 2 & { - 1} & 1 \end{array} \right| = 0

\Rightarrow k = {9 \over 2}

$\therefore$ Equations are 2x + 3y – z = 0 ...(i) 2x – y + z = 0 ...(ii) 2x + 9y – 4z = 0 ...(iii) By (i) – (ii) we get, 4y - 2z = 0 $\Rightarrow$ 2y = z .......(iv) $\Rightarrow$

{y \over z} = {1 \over 2}

From equation (i) and (iv) 2x + 3y - 2y = 0 $\Rightarrow$ 2x + y = 0 $\Rightarrow$

{x \over y} = - {1 \over 2}

$\Rightarrow$

{x \over y} \times {y \over z} = - {1 \over 2} \times {1 \over 2} = - {1 \over 4}

$\Rightarrow$

{z \over x} = - 4

$\therefore$

{x \over y} + {y \over z} + {z \over x} + k = {{ - 1} \over 2} + {1 \over 2} - 4 + {9 \over 2}

=

{1 \over 2}

Q36

If

{\Delta _1} = \left| \begin{array}{lll}x & {\sin \theta } & {\cos \theta } \\ { - \sin \theta } & { - x} & 1 \\ {\cos \theta } & 1 & x \end{array} \right|

and

{\Delta _2} = \left| \begin{array}{lll}x & {\sin 2\theta } & {\cos 2\theta } \\ { - \sin 2\theta } & { - x} & 1 \\ {\cos 2\theta } & 1 & x \end{array} \right|

,

x \ne 0

; then for all

\theta \in \left( {0,{\pi \over 2}} \right)

:

A

{\Delta _1} - {\Delta _2}

= x (cos 2

\theta

– cos 4

\theta

)

B

{\Delta _1} + {\Delta _2}

= - 2x3

C

{\Delta _1} + {\Delta _2}

= – 2(x3 + x –1)

D

{\Delta _1} - {\Delta _2}

= - 2x3

Correct Answer

Option B

Solution

{\Delta _1} = \left| \begin{array}{lll}x & {\sin \theta } & {\cos \theta } \\ { - \sin \theta } & { - x} & 1 \\ {\cos \theta } & 1 & x \end{array} \right|

= x(–x2 –1) – sin $\theta$ (–xsin $\theta$ – cos $\theta$ ) + cos $\theta$ (–sin $\theta$ + xcos $\theta$ ) = –x3 – x + xsin2 $\theta$ + sin $\theta$ cos $\theta$ – cos $\theta$ sin $\theta$ + xcos2 $\theta$ = –x3 – x + x = –x3 Similarly

{\Delta _2} = - {x^3}

{\Delta _1} + {\Delta _2} = - 2{x^3}

Q37

If the system of linear equations x + y + z = 5 x + 2y + 2z = 6 x + 3y +

\lambda

z =

\mu

, (

\lambda

,

\mu

\in

R), has infinitely many solutions, then the value of

\lambda

+

\mu

is :

A 10

B 9

C 7

D 12

Correct Answer

Option A

Solution

x + y + z = 5 x + 2y + 2z = 6 x + 3y + $\lambda$ z = $\mu$ have infinite solution

\Delta

= 0,

\Delta

x =

\Delta

y =

\Delta

z = 0

\Delta = \left| \begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 3 & \lambda \end{array} \right| = 0

\Rightarrow 1(2\lambda - 6) - 1(\lambda - 2) + 1(3 - 2) = 0

\Rightarrow 2\lambda - 6 - \lambda + 2 + 1 = 0

\Rightarrow \lambda = 3

Now,

\Delta x = \left| \begin{array}{lll}5 & 1 & 1 \\ 6 & 2 & 2 \\ \mu & 3 & 3 \end{array} \right| = 0

,

\Delta

y =

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

\Rightarrow \left| \begin{array}{lll}1 & 5 & 1 \\ 0 & 1 & 1 \\ 0 & {\mu - 5} & 2 \end{array} \right| = 0

\Rightarrow \mu = 7

\Delta z = \left| \begin{array}{lll}1 & 1 & 5 \\ 1 & 2 & 6 \\ 1 & 3 & \mu \end{array} \right| = \left| \begin{array}{lll}1 & 1 & 5 \\ 0 & { - 1} & { - 1} \\ 0 & 2 & {\mu - 5} \end{array} \right|

\Rightarrow 1(5 - \mu + 2) = 0

\Rightarrow \mu = 7

So,

\lambda + \mu = 10

Q38

Let

\lambda

be a real number for which the system of linear equations x + y + z = 6, 4x +

\lambda

y –

\lambda

z =

\lambda

– 2, 3x + 2y – 4z = – 5 has infinitely many solutions. Then

\lambda

is a root of the quadratic equation:

A

\lambda

2 +

\lambda

- 6 = 0

B

\lambda

2 -

\lambda

- 6 = 0

C

\lambda

2 - 3

\lambda

- 4 = 0

D

\lambda

2 + 3

\lambda

- 4 = 0

Correct Answer

Option B

Solution

\Delta = 0

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

On solving we get $\lambda$ = 3

Q39

If A is a symmetric matrix and B is a skew-symmetric matrix such that A + B =

\left[ \begin{array}{ll}2 & 3 \\ 5 & { - 1} \end{array} \right]

, then AB is equal to :

A

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

B

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

C

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

D

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

Correct Answer

Option D

Solution

A + B = \left[ \begin{array}{ll}2 & 3 \\ 5 & { - 1} \end{array} \right] = P(say)

Now

A = {{P + {P^T}} \over 2}\& B = {{P - {P^T}} \over 2}

So

A = {1 \over 2}\left( {\left[ \begin{array}{ll}2 & 3 \\ 5 & { - 1} \end{array} \right] + \left[ \begin{array}{ll}2 & 5 \\ 3 & { - 1} \end{array} \right]} \right) = \left[ \begin{array}{ll}2 & 4 \\ 4 & { - 1} \end{array} \right]

B = {1 \over 2}\left( {\left[ \begin{array}{ll}2 & 3 \\ 5 & { - 1} \end{array} \right] - \left[ \begin{array}{ll}2 & 5 \\ 3 & { - 1} \end{array} \right]} \right) = \left[ \begin{array}{ll}0 & { - 1} \\ 1 & 0 \end{array} \right]

So

AB = \left( {\left[ \begin{array}{ll}2 & 4 \\ 4 & { - 1} \end{array} \right]\left[ \begin{array}{ll}0 & { - 1} \\ 1 & 0 \end{array} \right]} \right) = \left[ \begin{array}{ll}4 & { - 2} \\ { - 1} & { - 4} \end{array} \right]

Q40

Let A be a symmetric matrix of order 2 with integer entries. If the sum of the diagonal elements of A2 is 1, then the possible number of such matrices is :

A 6

B 4

C 1

D 12

Correct Answer

Option B

Solution

Let

A = \left[ \begin{array}{ll}a & b \\ b & c \end{array} \right]

{A^2} = \left[ \begin{array}{ll}a & b \\ b & c \end{array} \right]\left[ \begin{array}{ll}a & b \\ b & c \end{array} \right] = \left[ \begin{array}{ll}{{a^2} + {b^2}} & {ab + bc} \\ {ab + bc} & {{c^2} + {b^2}} \end{array} \right]

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

a = 1,b = 0,c = 0

a = 0,b = 0,c = 1

a = - 1,b = 0,c = 0

c = - 1,b = 0,a = 0