The number of distinct real roots of $$\left| {\matrix{ {\sin x} & {\cos x} & {\cos x} \cr {\cos x} & {\sin x} & {\cos x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0

$$\left| {\matrix{ {\sin x} & {\cos x} & {\cos x} \cr {\cos x} & {\sin x} & {\cos x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0, - {\pi \over 4} \le x \le {\pi \over 4}$$ Apply : $${R_1} \to {R_1} - {R_2}$$ & $${R_2} \to {R_2} - {R_3}$$ $$ \Rightarrow $$ $$\left| {\matrix{ {\sin x - \cos x} & {\cos x - \sin x} & 0 \cr 0 & {\sin x - \cos x} & {\cos x - \sin x} \cr {\cos x} & {\cos x} & {\sin x} \cr } } \right| = 0$$ $$ \Rightarrow $$

If $$P = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right]$$, then P50 is :

$$P = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right]$$ $${P^2} = \left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]$$ $${P^3} = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 \cr {{1 \over 2}} & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr {{3 \over 2}} & 1 \cr } } \right]$$ $${P^4}

Let $$A = \left( {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right)$$. Then A2025 $$-$$ A2020 is equal to :

$$A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow {A^2} = \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right]$$ $${A^3} = \left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right] \Rightarrow {A^4} = \left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right]$$ $${A^n} = \left[ {\matrix{ 1 & 0 & 0 \cr {n - 1} &am

Let $$A = \left( {\matrix{ {[x + 1]} & {[x + 2]} & {[x + 3]} \cr {[x]} & {[x + 3]} & {[x + 3]} \cr {[x]} & {[x + 2]} & {[x + 4]} \cr } } \right)$$, where [t] denotes the greate

$$\left| {\matrix{ {[x + 1]} & {[x + 2]} & {[x + 3]} \cr {[x]} & {[x + 3]} & {[x + 3]} \cr {[x]} & {[x + 2]} & {[x + 4]} \cr } } \right| = 192$$ R1 $$\to$$ R1 $$-$$ R3 & R2 $$\to$$ R2 $$-$$ R3 $$\left[ {\matrix{ 1 & 0 & { - 1} \cr 0 & 1 & { - 1} \cr {[x]} & {[x] + 2} & {[x] + 4} \cr } } \right] = 192$$ $$2[x] + 6 + [x] = 192 \Rightarrow [x] = 62$$

Matrices and Determinants — Page 9 | JEE Mathematics

Q81

Let A and B be two 3

\times

3 real matrices such that (A2

-

B2) is invertible matrix. If A5 = B5 and A3B2 = A2B3, then the value of the determinant of the matrix A3 + B3 is equal to :

A 2

B 4

C 1

D 0

Correct Answer

Option D

Solution

C = A2 $-$ B2; | C | $\ne$ 0 A2 = B5 and A3B2 = A2B2 Now, A5 $-$ A3B2 = B5 $-$ A2B3 $\Rightarrow$ A3 (A2 $-$ B2) + B3 (A2 $-$ B2) = 0 $\Rightarrow$ (A3 + B3(A2 $-$ B2) = 0 $\Rightarrow$ A3 + B3 = 0

\left( \because{\left| {{A^2} - {B^2} \ne 0} \right|} \right)

Q82

The number of distinct real roots of

\left| \begin{array}{lll}{\sin x} & {\cos x} & {\cos x} \\ {\cos x} & {\sin x} & {\cos x} \\ {\cos x} & {\cos x} & {\sin x} \end{array} \right| = 0

in the interval

- {\pi \over 4} \le x \le {\pi \over 4}

is :

A 4

B 1

C 2

D 3

Correct Answer

Option B

Solution

\left| \begin{array}{lll}{\sin x} & {\cos x} & {\cos x} \\ {\cos x} & {\sin x} & {\cos x} \\ {\cos x} & {\cos x} & {\sin x} \end{array} \right| = 0, - {\pi \over 4} \le x \le {\pi \over 4}

Apply :

{R_1} \to {R_1} - {R_2}

&

{R_2} \to {R_2} - {R_3}

$\Rightarrow$

\left| \begin{array}{lll}{\sin x - \cos x} & {\cos x - \sin x} & 0 \\ 0 & {\sin x - \cos x} & {\cos x - \sin x} \\ {\cos x} & {\cos x} & {\sin x} \end{array} \right| = 0

$\Rightarrow$

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

$\Rightarrow$

{(\sin x - \cos x)^2}(\sin x + 2\cos x) = 0

$\therefore$

x = {\pi \over 4}

Q83

If

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

, then P50 is :

A

\left[ \begin{array}{ll}1 & 0 \\ {25} & 1 \end{array} \right]

B

\left[ \begin{array}{ll}1 & {50} \\ 0 & 1 \end{array} \right]

C

\left[ \begin{array}{ll}1 & {25} \\ 0 & 1 \end{array} \right]

D

\left[ \begin{array}{ll}1 & 0 \\ {50} & 1 \end{array} \right]

Correct Answer

Option A

Solution

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

{P^2} = \left[ \begin{array}{ll}1 & 0 \\ {{1 \over 2}} & 1 \end{array} \right]\left[ \begin{array}{ll}1 & 0 \\ {{1 \over 2}} & 1 \end{array} \right] = \left[ \begin{array}{ll}1 & 0 \\ 1 & 1 \end{array} \right]

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

{P^4} = \left[ \begin{array}{ll}1 & 0 \\ 1 & 1 \end{array} \right]\left[ \begin{array}{ll}1 & 0 \\ 1 & 1 \end{array} \right] = \left[ \begin{array}{ll}1 & 0 \\ 2 & 1 \end{array} \right]

$\vdots$ $\therefore$

{P^{50}} = \left[ \begin{array}{ll}1 & 0 \\ {25} & 1 \end{array} \right]

Q84

The ordered pair (a, b), for which the system of linear equations 3x

-

2y + z = b 5x

-

8y + 9z = 3 2x + y + az =

-

1 has no solution, is :

A

\left( {3,{1 \over 3}} \right)

B

\left( { - 3,{1 \over 3}} \right)

C

\left( { - 3, - {1 \over 3}} \right)

D

\left( {3, - {1 \over 3}} \right)

Correct Answer

Option C

Solution

\left| \begin{array}{lll}3 & { - 2} & 1 \\ 5 & { - 8} & 9 \\ 2 & 1 & a \end{array} \right| = 0 \Rightarrow - 14a - 42 = 0 \Rightarrow a = - 3

Now 3 (equation (1)) $-$ (equation (2)) $-$ 2 (equation (3)) is

3(3x - 2y + z - b) - (5x - 8y + 9z - 3) - 2(2x + y + az + 1) = 0

\Rightarrow - 3b + 3 - 2 = 0 \Rightarrow b = {1 \over 3}

So for no solution

a = - 3

and

b \ne {1 \over 3}

Q85

Let

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

. If A

-

1 =

\alpha

I +

\beta

A,

\alpha

,

\beta

\in

R, I is a 2

\times

2 identity matrix then 4(

\alpha

-

\beta

) is equal to :

A 5

B

{8 \over 3}

C 2

D 4

Correct Answer

Option D

Solution

A = \left[ \begin{array}{ll}1 & 2 \\ { - 1} & 4 \end{array} \right],|A| = 6

{A^{ - 1}} = {{adjA} \over {|A|}} = {1 \over 6}\left[ \begin{array}{ll}4 & { - 2} \\ 1 & 1 \end{array} \right] = \left[ \begin{array}{ll}{{2 \over 3}} & { - {1 \over 3}} \\ {{1 \over 6}} & {{1 \over 6}} \end{array} \right]

\left[ \begin{array}{ll}{{2 \over 3}} & { - {1 \over 3}} \\ {{1 \over 6}} & {{1 \over 6}} \end{array} \right] = \left[ \begin{array}{ll}\alpha & 0 \\ 0 & \alpha \end{array} \right] + \left[ \begin{array}{ll}\beta & {2\beta } \\ { - \beta } & {4\beta } \end{array} \right]

\left. \begin{array}{ll}\alpha + \beta = {2 \over 3} \\ \beta = - {1 \over 6} \end{array} \right\} \Rightarrow \alpha = {2 \over 3} + {1 \over 6} = {5 \over 6}

$\therefore$

4(\alpha - \beta ) = 4(1) = 4

Q86

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

-

y + z = 3 x + y + az = b has no solution, then :

A

a = - {1 \over 3},b \ne {7 \over 3}

B

a \ne {1 \over 3},b = {7 \over 3}

C

a \ne - {1 \over 3},b = {7 \over 3}

D

a = {1 \over 3},b \ne {7 \over 3}

Correct Answer

Option D

Solution

Here

D = \left| \begin{array}{lll}2 & 1 & 1 \\ 1 & { - 1} & 1 \\ 1 & 1 & a \end{array} \right|\begin{array}{ll}{ = 2(a - 1) - 1(a - 1) + 1 + 1} \\ { = 1 - 3a} \end{array}

{D_3} = \left| \begin{array}{lll}2 & 1 & 5 \\ 1 & { - 1} & 3 \\ 1 & 1 & b \end{array} \right|\begin{array}{ll}{ = 2( - b - 3) - 1(b - 3) + 5(1 + 1)} \\ { = 7 - 3b} \end{array}

for

a = {1 \over 3},b \ne {7 \over 3}

, system has no solutions.

Q87

Let

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

. If the system of linear equations

(1 + {\cos ^2}\theta )x + {\sin ^2}\theta y + 4\sin 3\,\theta z = 0

{\cos ^2}\theta x + (1 + {\sin ^2}\theta )y + 4\sin 3\,\theta z = 0

{\cos ^2}\theta x + {\sin ^2}\theta y + (1 + 4\sin 3\,\theta )z = 0

has a non-trivial solution, then the value of

\theta

is :

A

{{4\pi } \over 9}

B

{{7\pi } \over {18}}

C

{\pi \over {18}}

D

{{5\pi } \over {18}}

Correct Answer

Option B

Solution

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

{C_1} \to {C_1} + {C_2}

\left| \begin{array}{lll}2 & {{{\sin }^2}\theta } & {4\sin 3\,\theta } \\ 2 & {1 + {{\sin }^2}\theta } & {4\sin 3\,\theta } \\ 1 & {{{\sin }^2}\theta } & {1 + 4\sin 3\,\theta } \end{array} \right| = 0

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

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

or

4\sin 3\theta = - 2

\sin 3\theta = - {1 \over 2}

\theta = {{7\pi } \over {18}}

Q88

Let

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

. Then A2025

-

A2020 is equal to :

A A6

-

A

B A5

C A5

-

A

D A6

Correct Answer

Option A

Solution

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

{A^3} = \left[ \begin{array}{lll}1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{array} \right] \Rightarrow {A^4} = \left[ \begin{array}{lll}1 & 0 & 0 \\ 3 & 1 & 1 \\ 1 & 0 & 0 \end{array} \right]

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

{A^{2025}} - {A^{2020}} = \left[ \begin{array}{lll}0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]

{A^6} - A = \left[ \begin{array}{lll}0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right]

Q89

Let

A = \left( \begin{array}{lll}{[x + 1]} & {[x + 2]} & {[x + 3]} \\ {[x]} & {[x + 3]} & {[x + 3]} \\ {[x]} & {[x + 2]} & {[x + 4]} \end{array} \right)

, where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval :

A [68, 69)

B [62, 63)

C [65, 66)

D [60, 61)

Correct Answer

Option B

Solution

\left| \begin{array}{lll}{[x + 1]} & {[x + 2]} & {[x + 3]} \\ {[x]} & {[x + 3]} & {[x + 3]} \\ {[x]} & {[x + 2]} & {[x + 4]} \end{array} \right| = 192

R1 $\to$ R1 $-$ R3 & R2 $\to$ R2 $-$ R3

\left[ \begin{array}{lll}1 & 0 & { - 1} \\ 0 & 1 & { - 1} \\ {[x]} & {[x] + 2} & {[x] + 4} \end{array} \right] = 192

2[x] + 6 + [x] = 192 \Rightarrow [x] = 62

Q90

Let A(a, 0), B(b, 2b + 1) and C(0, b), b

\ne

0, |b|

\ne

1, be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is :

A

{{ - 2b} \over {b + 1}}

B

{{2b} \over {b + 1}}

C

{{2{b^2}} \over {b + 1}}

D

{{ - 2{b^2}} \over {b + 1}}

Correct Answer

Option D

Solution

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

\Rightarrow \left| \begin{array}{lll}a & 0 & 1 \\ b & {2b + 1} & 1 \\ 0 & b & 1 \end{array} \right| = \pm \,2

\Rightarrow a(2b + 1 - b) - 0 + 1({b^2} - 0) = \pm \,2

\Rightarrow a = {{ \pm \,2 - {b^2}} \over {b + 1}}

$\therefore$

a = {{2 - {b^2}} \over {b + 1}}

and

a = {{ - 2 - {b^2}} \over {b + 1}}

Sum of possible values of 'a' is

= {{ - 2{b^2}} \over {a + 1}}