If $$D = \left| {\matrix{ 1 & 1 & 1 \cr 1 & {1 + x} & 1 \cr 1 & 1 & {1 + y} \cr } } \right|$$ for $$x \ne 0,y \ne 0,$$ then $$D$$ is :

Given, $$D = \left| {\matrix{ 1 & 1 & 1 \cr 1 & {1 + x} & 1 \cr 1 & 1 & {1 + y} \cr } } \right|$$ Apply $$\,\,\,{R^2} \to {R_2} - {R_1}$$ $$\,\,\,\,$$ and $$\,\,\,\,$$ $$R \to {R_3} - {R_1}$$ $$\therefore$$ $$\,\,\,\,\,D = \left| {\matrix{ 1 & 1 & 1 \cr 0 & x & 0 \cr 0 & 0 & y \cr } } \right| = xy$$ Hence, $$D$$ is divisible by both $$x$$ and $$y$$

If $$A = \left[ {\matrix{ a & b \cr b & a \cr } } \right]$$ and $${A^2} = \left[ {\matrix{ \alpha & \beta \cr \beta & \alpha \cr } } \right]$$, then

$${A^2} = \left[ {\matrix{ \alpha & \beta \cr \beta & \alpha \cr } } \right] = \left[ {\matrix{ a & b \cr b & a \cr } } \right]\left[ {\matrix{ a & b \cr b & a \cr } } \right]$$ $$ = \left[ {\matrix{ {{a^2} + {b^2}} & {2ab} \cr {2ab} & {{a^2} + {b^2}} \cr } } \right]$$ $$\alpha = {a^2} + {b^2};\,\,\beta = 2ab$$

If $${a_1},{a_2},{a_3},.........,{a_n},......$$ are in G.P., then the value of the determinant $$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}}

$$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$ $$ = \left| {\matrix{ {\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \cr {\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \cr {\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\

If $${a^2} + {b^2} + {c^2} = - 2$$ and f$$\left( x \right) = \left| {\matrix{ {1 + {a^2}x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr {\left( {1 + {a^2}} \right)x} &am

Applying, $${C_1} \to {C_1} + {C_2} + {C_3}\,\,\,$$ we get $$f\left( x \right) = \left| {\matrix{ {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \cr {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \cr {1 + \left( {{a^2} + {b^2} + {c^2} + 2} \right)x} & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \cr } } \right|$$ $$ = \left| {\matrix{ 1 & {\left( {1 + {

Let $$A = \left( {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right)$$ and $$B = \left( {\matrix{ a & 0 \cr 0 & b \cr } } \right),a,b \in N.$$ Then

$$A = \left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right]\,\,\,\,B = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]$$ $$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$$ $$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$$ Hence, $$AB=BA$$ only when $$a=b$$ $$\therefore$$ There can be infinitely many $$B's$$ for which $$AB=

Matrices and Determinants Questions — JEE Mathematics

Q1

The system of equations

- kx + 3y - 14z = 25

- 15x + 4y - kz = 3

- 4x + y + 3z = 4

is consistent for all k in the set

A R

B R

-

{

-

11, 13}

C R

-

{13}

D R

-

{

-

11, 11}

Correct Answer

Option D

Solution

The system may be inconsistent if

\left| \begin{array}{lll}{ - k} & 3 & { - 14} \\ { - 15} & 4 & { - k} \\ { - 4} & 1 & 3 \end{array} \right| = 0 \Rightarrow k = \, \pm \,11

Hence if system is consistent then

k \in R - \{ 11, - 11\}

.

Q2

If

D = \left| \begin{array}{lll}1 & 1 & 1 \\ 1 & {1 + x} & 1 \\ 1 & 1 & {1 + y} \end{array} \right|

for

x \ne 0,y \ne 0,

then

D

is :

A divisible by

x

but not

y

B divisible by

y

but not

x

C divisible by neither

x

nor

y

D divisible by both

x

and

y

Correct Answer

Option D

Solution

Given,

D = \left| \begin{array}{lll}1 & 1 & 1 \\ 1 & {1 + x} & 1 \\ 1 & 1 & {1 + y} \end{array} \right|

Apply

\,\,\,{R^2} \to {R_2} - {R_1}

\,\,\,\,

and

\,\,\,\,

R \to {R_3} - {R_1}

$\therefore$

\,\,\,\,\,D = \left| \begin{array}{lll}1 & 1 & 1 \\ 0 & x & 0 \\ 0 & 0 & y \end{array} \right| = xy

Hence,

D

is divisible by both

x

and

y

Q3

If

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

and

{A^2} = \left[ \begin{array}{ll}\alpha & \beta \\ \beta & \alpha \end{array} \right]

, then

A

\alpha = 2ab,\,\beta = {a^2} + {b^2}

B

\alpha = {a^2} + {b^2},\,\beta = ab

C

\alpha = {a^2} + {b^2},\,\beta = 2ab

D

\alpha = {a^2} + {b^2},\,\beta = {a^2} - {b^2}

Correct Answer

Option C

Solution

{A^2} = \left[ \begin{array}{ll}\alpha & \beta \\ \beta & \alpha \end{array} \right] = \left[ \begin{array}{ll}a & b \\ b & a \end{array} \right]\left[ \begin{array}{ll}a & b \\ b & a \end{array} \right]

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

\alpha = {a^2} + {b^2};\,\,\beta = 2ab

Q4

If

1,

\omega ,{\omega ^2}

are the cube roots of unity, then

\Delta = \left| \begin{array}{lll}1 & {{\omega ^n}} & {{\omega ^{2n}}} \\ {{\omega ^n}} & {{\omega ^{2n}}} & 1 \\ {{\omega ^{2n}}} & 1 & {{\omega ^n}} \end{array} \right|

is equal to

A

{\omega ^2}

B

0

C

1

D

\omega

Correct Answer

Option B

Solution

\Delta = \left| \begin{array}{lll}1 & {{\omega ^n}} & {{\omega ^{2n}}} \\ {{\omega ^n}} & {{\omega ^{2n}}} & 1 \\ {{\omega ^{2n}}} & 1 & {{\omega ^n}} \end{array} \right|

= 1\left( {{\omega ^{3n}} - 1} \right) - {\omega ^n}\left( {{\omega ^{2n}} - {\omega ^{2n}}} \right) +

\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\omega ^{2n}}\left( {{\omega ^n} - {\omega ^{4n}}} \right)

= {\omega ^{3n}} - 1 - 0 + {\omega ^{3n}} - {\omega ^{6n}}

= 1 - 1 + 1 - 1 = 0

\left[ {} \right.

as

\,\,\,\,\,

{\omega ^{3n}} = 1

\left. {} \right]

Q5

If the system of linear equations

x + 2ay + az = 0;

x + 3by + bz = 0;\,\,x + 4cy + cz = 0;

has a non - zero solution, then

a, b, c

.

A satisfy

a+2b+3c=0

B are in A.P

C are in G.P

D are in H.P.

Correct Answer

Option D

Solution

For homogeneous system of equations to have non zero solution,

\Delta = 0

\left| \begin{array}{lll}1 & {2a} & a \\ 1 & {3b} & b \\ 1 & {4c} & c \end{array} \right| = 0\,{C_2} \to {C_2} - 2{C_3}

\left| \begin{array}{lll}1 & 0 & a \\ 1 & b & b \\ 1 & {2c} & c \end{array} \right| = 0\,\,{R_3} \to {R_3} - {R_2},{R_2} \to {R_2} - {R_1}

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

b\left( {c - b} \right) - \left( {b - a} \right)\left( {2c - b} \right) = 0

On simplification,

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

$\therefore$

a,b,c

are in Harmonic Progression.

Q6

If

{a_1},{a_2},{a_3},.........,{a_n},......

are in G.P., then the value of the determinant

\left| \begin{array}{lll}{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \\ {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \\ {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \end{array} \right|,

is

A

-2

B

1

C

2

D

0

Correct Answer

Option D

Solution

\left| \begin{array}{lll}{\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \\ {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \\ {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \end{array} \right|

= \left| \begin{array}{lll}{\log {a_1}r{}^{n - 1}} & {\log {a_1}r{}^n} & {\log {a_1}r{}^{n + 1}} \\ {\log {a_1}r{}^{n + 2}} & {\log {a_1}r{}^{n + 3}} & {\log {a_1}r{}^{n + 4}} \\ {\log {a_1}r{}^{n + 5}} & {\log {a_1}r{}^{n + 6}} & {\log {a_1}r{}^{n + 7}} \end{array} \right|

= \left| \begin{array}{lll}{\log {a_1} + \left( {n - 1} \right)\log r} & {\log {a_1} + n\log r} & {\log {a_1} + \left( {n + 1} \right)\log r} \\ {\log {a_1} + \left( {n + 2} \right)\log r} & {\log {a_1} + \left( {n + 3} \right)\log r} & {\log {a_1} + \left( {n + 4} \right)\log r} \\ {\log {a_1} + \left( {n + 5} \right)\log r} & {\log {a_1} + \left( {n + 6} \right)\log r} & {\log {a_1} + \left( {n + 7} \right)\log r} \end{array} \right|

= 0\left[ \, \right.

Apply

\,\,\,\,{c_2} \to {c_2} - {1 \over 2}{c_1} - {1 \over 2}{c_3}\,\left. \, \right]

Q7

If

{A^2} - A + 1 = 0

, then the inverse of

A

is :

A

A+I

B

A

C

A-I

D

I-A

Correct Answer

Option D

Solution

Given

{A^2} - A + I = 0

{A^{ - 1}}{A^2} - {A^{ - 1}}A + {A^{ - 1}}.I = {A^{ - 1}}.0

(Multiplying

\,\,\,{A^{ - 1}}

on both sides)

\Rightarrow A - 1 + {A^{ - 1}} = 0

or

{A^{ - 1}} = 1 - A

Q8

Let

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

. The only correct statement about the matrix

A

is

A

{A^2} = 1

B

A=(-1)I,

where

I

is a unit matrix

C

{A^{ - 1}}

does not exist

D

A

is a zero matrix

Correct Answer

Option A

Solution

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

clearly

\,\,\,A \ne 0.\,

Also

\,\,\left| A \right| = - 1 \ne 0

$\therefore$

{A^{ - 1}}\,\,

exists, further

\left( { - 1} \right)I = \left[ \begin{array}{lll}{ - 1} & 0 & 0 \\ 0 & { - 1} & 0 \\ 0 & 0 & { - 1} \end{array} \right] \ne A

Also

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

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

Q9

If

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

and f

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

then f

(x)

is a polynomial of degree :

A

1

B

0

C

3

D

2

Correct Answer

Option D

Solution

Applying,

{C_1} \to {C_1} + {C_2} + {C_3}\,\,\,

we get

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

= \left| \begin{array}{lll}1 & {\left( {1 + {b^2}} \right)x} & {\left( {1 + {c^2}} \right)x} \\ 1 & {1 + {b^2}x} & {\left( {1 + {c^2}x} \right)} \\ 1 & {\left( {1 + {b^2}} \right)x} & {1 + {c^2}x} \end{array} \right|

\left[ \, \right.

As given that

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

\left. {} \right]

$\therefore$

{a^2} + {b^2} + {c^2} + 2 = 0

Applying

{R_1} \to {R_1} - {R_2},\,\,\,{R_2} \to {R_2} - {R_3}

$\therefore$

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

f\left( x \right) = {\left( {x - 1} \right)^2}

Hence degree

=2.

Q10

Let

A = \left( \begin{array}{ll}1 & 2 \\ 3 & 4 \end{array} \right)

and

B = \left( \begin{array}{ll}a & 0 \\ 0 & b \end{array} \right),a,b \in N.

Then

A there cannot exist any

B

such that

AB=BA

B there exist more then one but finite number of

B'

s such that

AB=BA

C there exists exactly one

B

such that

AB=BA

D there exist infinitely many

B'

s such that

AB=BA

Correct Answer

Option D

Solution

A = \left[ \begin{array}{ll}1 & 2 \\ 3 & 4 \end{array} \right]\,\,\,\,B = \left[ \begin{array}{ll}a & 0 \\ 0 & b \end{array} \right]

AB = \left[ \begin{array}{ll}a & {2b} \\ {3a} & {4b} \end{array} \right]

BA = \left[ \begin{array}{ll}a & 0 \\ 0 & b \end{array} \right]\left[ \begin{array}{ll}1 & 2 \\ 3 & 4 \end{array} \right] = \left[ \begin{array}{ll}a & {2a} \\ {3b} & {4b} \end{array} \right]

Hence,

AB=BA

only when

a=b

$\therefore$ There can be infinitely many

B's

for which

AB=BA