Matrices and Determinants

$\det A = \left| {\matrix{ { - 2} & {4 + d} & {\sin \theta - 2} \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & { - \sin \theta + 2 + 2d} \cr } } \right|$

(R₁ $ \to $ R₁ + R₃ $-$ 2R₂)

$ = \left| {\matrix{ 1 & 0 & 0 \cr 1 & {\sin \theta + 2} & d \cr 5 & {2\sin \theta - d} & {2 + 2d - \sin \theta } \cr } } \right|$

= (2 + sin $\theta $) ( 2 + 2d $-$ sin$\theta $) $-$ d(2sin$\theta $ $-$ d)

= 4 + 4d $-$ 2sin$\theta $ + 2sin$\theta $ + 2dsin$\theta $ $-$ sin²$\theta $ $-$ 2dsin$\theta $ + d²

d² + 4d + 4 $-$ sin²$\theta $

= (d + 2)² $-$ sin²$\theta $

For a given d, minimum value of

det(A) = (d + 2)² $-$ 1 = 8

$ \Rightarrow $  d = 1 or $-$ 5

x $-$ $4y + 7z = g$
      $3y$ $-$ $5z = h$
$-$$2x + 5y$ $-$ $9z = k$

$D = \left| {\matrix{ 1 & { - 4} & 7 \cr 0 & 3 & { - 5} \cr { - 2} & 5 & { - 9} \cr } } \right|$

$D = 1\left( { - 27 + 25} \right) - 2\left( {20 - 21} \right)$

$D = - 2 + 2 = 0$

If system is consistent then ${D_1} = {D_2} = {D_3} = 0$

$\left| {\matrix{ 1 & { - 4} & g \cr 0 & 3 & h \cr { - 2} & 5 & k \cr } } \right| = 0$

$1\left( {3k - 5h} \right) - 2\left( { - 4h - 3g} \right) = 0$

$3k - 5h + 8h + 6g = 0$

$6g + 3h + 3k = 0$

$2g + h + k = 0$

$A = \left[ {\matrix{ {{e^t}} & {{e^{ - t}}\cos t} & {{e^{ - t}}\sin t} \cr {{e^t}} & { - {e^{ - t}}\cos t - {e^{ - t}}\sin t} & { - {e^{ - t}}\sin t + {e^{ - t}}co{\mathop{\rm s}\nolimits} t} \cr {{e^t}} & {2{e^{ - t}}\sin t} & { - 2{e^{ - t}}\cos t} \cr } } \right]$

$\left| A \right| = {e^t}.\,{e^{ - t}}.{e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 1 & { - \cos t - \sin t} & { - \sin t + \cos t} \cr 1 & {2\sin t} & { - 2\cos t} \cr } } \right|$

Apply operations R₂ < R₂ $-$R₁, R₃ < R₃ $-$ R₁, R₁ < R₁

$\left| A \right| = {e^{ - t}}\left| {\matrix{ 1 & {\cos t} & {\sin t} \cr 0 & { - \sin t - 2\cos t} & { - 2\sin t + \cos t} \cr 0 & {2\sin t - \cos t} & { - 2\cos t - \sin t} \cr } } \right|$

Open the determinant by R₁

$\left| A \right| = 5{e^{ - t}}$

Invertible for all t $ \in $ R

$D = \left| {\matrix{ 1 & 1 & 1 \cr 2 & 3 & 2 \cr 2 & 3 & {{\alpha ^2} - 1} \cr } } \right|$

D = 3$a$² $-$ 3 $-$ 6 $-$ 2$a$² + 2 + 4 + 2$a$² $-$ 2 $-$ 4

D = ($a$² $-$ 3)

When D $ \ne $ 0 then system of equiation has unique solution.

$ \therefore $  3($a$² $-$ 3) $ \ne $ 0

$ \Rightarrow $  $\left| a \right|$ $ \ne \sqrt 3 $

When $3({a^2} - 3) = 0$

$ \Rightarrow $  $\left| a \right| = \sqrt 3 $ then D = 0

If D = 0 then two cases possible

(1)  System of equation has infinite many solution.

(2) System of equation has no solution and inconsistent.

Here D₁ = $\left| {\matrix{ 2 & 1 & 1 \cr 5 & 3 & 2 \cr {a + 1} & 3 & {{a^2} - 1} \cr } } \right| = {a^2} - a + 1$

D₂ = $\left| {\matrix{ 1 & 2 & 1 \cr 2 & 5 & 2 \cr 2 & {a + 1} & {{a^2} - 1} \cr } } \right| = {a^2} - 3$

D₃ = $\left| {\matrix{ 1 & 1 & 2 \cr 2 & 3 & 5 \cr 2 & 3 & {a + 1} \cr } } \right| = a - 4$

System of equation will have infinite solution if D₁ = D₂ = D₃ = 0.

And system of equation will have no solution if at last one of D₁, D₂, D₂ is non zero.

At $\left| a \right| = \sqrt 3 $ we get D = 0

But D₁ = 3 $ \pm $ $\sqrt 3 + 1$ $ \ne $ 0

and D₃ = $ \pm $ $\sqrt 3 - 4$ $ \ne $ 0

So, system of equations has no solution at $\left| a \right| = \sqrt 3 $ then system is in consistent

(A^$-$50) = (A^$-$1)⁵⁰

We know,

A^$-$1 = ${{adjA} \over {\left| A \right|}}$

$\left| A \right|$ = cos²$\theta $ + sin²$\theta $ = 1

cofactor of A = $\left[ {\matrix{ {\cos \theta } & { - \sin \theta } \cr {\sin \theta } & {\cos \theta } \cr } } \right]$

Adjoint of A = Transpose of cofactor matrix

$ \therefore $  Adj A = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

$ \therefore $  A^$-$1 = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

$ \therefore $  (A^$-$1)² = $\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]\left[ {\matrix{ {\cos \theta } & {\sin \theta } \cr { - \sin \theta } & {\cos \theta } \cr } } \right]$

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

Similarly,

(A^$-$1)³ = $\left[ {\matrix{ {\cos 3\theta } & {\sin 3\theta } \cr { - \sin 3\theta } & {\cos 3\theta } \cr } } \right]$

:

:

:

(A^$-$1)⁵⁰ = $\left[ {\matrix{ {\cos 50\theta } & {\sin 50\theta } \cr { - \sin 50\theta } & {\cos 50\theta } \cr } } \right]$

when $\theta $ = ${\pi \over {12}}$ then

${A^{ - 50}}$ = $\left[ {\matrix{ {\cos {{25\pi } \over 6}} & {\sin {{25\pi } \over 6}} \cr { - \sin {{25\pi } \over 6}} & {\cos {{25\pi } \over 6}} \cr } } \right]$

= $\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$

Note:

$\cos {{25\pi } \over 6} = \cos \left( {4\pi + {\pi \over 6}} \right) = \cos {\pi \over 6} = {{\sqrt 3 } \over 2}$

A = $\left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$

A² = A.A = $\left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right] \times \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$

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

A³ = A².A =  $\left[ {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right] \times \left[ {\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 1 & 1 & 1 \cr } } \right]$

=   $\left[ {\matrix{ 1 & 0 & 0 \cr 3 & 1 & 0 \cr 6 & 3 & 1 \cr } } \right]$

Similarly

A⁴ =   $\left[ {\matrix{ 1 & 0 & 0 \cr 4 & 1 & 0 \cr {10} & 4 & 1 \cr } } \right]$

From this we can say,

Aⁿ =   $\left[ {\matrix{ 1 & 0 & 0 \cr n & 1 & 0 \cr {{{n\left( {n + 1} \right)} \over 2}} & n & 1 \cr } } \right]$

$\therefore\,\,\,$ A²⁰ =   $\left[ {\matrix{ 1 & 0 & 0 \cr {20} & 1 & 0 \cr {210} & {20} & 1 \cr } } \right]$

$\therefore\,\,\,$ Sum of the first column

= 1 + 20 + 210

= 231

System of linear equation have no solution,

$\therefore\,\,\,$ determinant of coefficient = 0

$\left| {\matrix{ {k + 2} & {10} \cr k & {k + 3} \cr } } \right| = 0$

$ \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.

$\left| {\matrix{ {x - 4} & {2x} & {2x} \cr {2x} & {x - 4} & {2x} \cr {2x} & {2x} & {x - 4} \cr } } \right|$

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} \cr } } \right|$

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

$ = \,\,\,\,\left( {5x - 4} \right)\left| {\matrix{ 1 & {2x} & {2x} \cr 0 & { - \left( {x + 4} \right)} & 0 \cr 0 & 0 & { - \left( {x + 4} \right)} \cr } } \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

System of equations has non-zero solution when determinant of coefficient = 0.

So, in this questions,

$\left| {\matrix{ 1 & K & 3 \cr 3 & K & { - 2} \cr 2 & 4 & { - 3} \cr } } \right| = 0$

$ \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$

As the given system of equations has no solution then

$\Delta $ = 0 and at least one of $\Delta $₁, $\Delta $₂ and $\Delta $₂ should not be zero.

$ \therefore $ $\Delta $ = $\left| {\matrix{ 1 & a & 1 \cr 1 & 2 & 2 \cr 1 & 5 & 3 \cr } } \right| = 0$

$ \Rightarrow $ - $a$ - 1 = 0

$ \Rightarrow $ a = - 1

$\Delta $₂ = $\left| {\matrix{ 1 & 3 & 1 \cr 1 & 6 & 2 \cr 1 & b & 3 \cr } } \right| \ne 0$

$ \Rightarrow $ b $ \ne $ 0

Given,

( A $-$ 3I) (A $-$ 5I) = O

$ \Rightarrow $ A² - 8A + 15I = O

Multiplying both sides by A^{- 1}, we get,

A^{- 1}A.A - 8A^{- 1}A + 15A^{- 1}I = A^{- 1}O

$ \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

According to questions,

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

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

$ \Rightarrow $ A = $1 \over 3$$\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$ $\left[ {\matrix{ 3 & {-2} \cr 0 & 1 \cr } } \right]$

$ \Rightarrow $ A = $\left[ {\matrix{ \lambda & 0 \cr 0 & \lambda \cr } } \right]$ $\left[ {\matrix{ 1 & { - {2 \over 3}} \cr 0 & {{1 \over 3}} \cr } } \right]$

$ \Rightarrow $ A = $\left[ {\matrix{ \lambda & { - {2 \over 3}\lambda } \cr 0 & {{\lambda \over 3}} \cr } } \right]$

As $\left| {3A} \right|$ = 108

$ \Rightarrow $ 108 = $\left| {\matrix{ {3\lambda } & { - 2\lambda } \cr 0 & \lambda \cr } } \right|$

$ \Rightarrow $ 3$\lambda $² = 108

$ \Rightarrow $ $\lambda $² = 36

$ \Rightarrow $ $\lambda $ = $ \pm $6

When $\lambda $ = +6

then A = $\left[ {\matrix{ 6 & { - 4} \cr 0 & 2 \cr } } \right]$

$ \Rightarrow $ A² = $\left[ {\matrix{ {36} & { - 32} \cr 0 & 4 \cr } } \right]$

For $\lambda $ = -6

A = $\left[ {\matrix{ { - 6} & 4 \cr 0 & { - 2} \cr } } \right]$

$ \Rightarrow $ A² = $\left[ {\matrix{ {36} & { - 32} \cr 0 & 4 \cr } } \right]$

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

$ \therefore $ $\left| {\matrix{ 1 & 1 & 1 \cr 2 & 1 & { - 1} \cr 3 & 2 & k \cr } } \right|$ $ \ne $ 0

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

$ \Rightarrow $ k $ \ne $ 0

$ \therefore $ k $ \in $ R - {0}

Given, A + B = 2B^T .......(1)

$ \Rightarrow $ (A + B)^T = (2B^T)^T

$ \Rightarrow $ A^T + B^T = 2B

$ \Rightarrow $ B = ${{{A^T} + {B^T}} \over 2}$

Now put this in equation (1)

So, A + ${{{A^T} + {B^T}} \over 2}$ = 2B^T

$ \Rightarrow $2A + A^T = 3B^T

$ \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 $ 11B^T - A^T = 2I₃

$ \Rightarrow $ (11B^T - A^T)^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

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

$\left[ {\matrix{ 2 & 4 & { - \lambda } \cr 4 & \lambda & 2 \cr \lambda & 2 & 2 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$

The system has infinite solutions; thus, we get

$\left| {\matrix{ 2 & 4 & { - \lambda } \cr 4 & \lambda & 2 \cr \lambda & 2 & 2 \cr } } \right| = 0$

$ \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.

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.

Given,

        $\left| {\matrix{ 0 & {\cos x} & { - \sin x} \cr {\sin x} & 0 & {\cos x} \cr {\cos x} & {\sin x} & 0 \cr } } \right|$ = 0

$ \Rightarrow $$\,\,\,$ 0 (0 $-$ cosx sinx) $-$ cosx (0 $-$ cos²x) $-$ sinx(sin²x) = 0

$ \Rightarrow $$\,\,\,$ cos³x $-$ sin³ x = 0

$ \Rightarrow $$\,\,\,$ tan³x = 1

$ \Rightarrow $$\,\,\,$ tanx = 1

$ \therefore $ $\,\,\,$ $\sum\limits_{x\, \in \,\,S} {\,\tan \left( {{\pi \over 3} + x} \right)} $

= ${{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}$

= ${{\sqrt 3 + 1} \over {1 - \sqrt 3 }}$

= ${{\left( {\sqrt 3 + 1} \right)} \over {\left( {1 - \sqrt 3 } \right)}} \times {{1 + \sqrt 3 } \over {1 + \sqrt 3 }}$

= ${{1 + 3 + 2\sqrt 3 } \over { - 2}}$

= $-$ 2 $-$ $\sqrt 3 $

$\left| {\matrix{ 1 & 1 & 1 \cr 1 & a & 1 \cr a & b & 1 \cr } } \right| = 0$

$ \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.

We have, $A = \left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

$ \therefore $ A² = A.A = $\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

= $\left[ {\matrix{ {4 + 12} & { - 6 - 3} \cr { - 8 - 4} & {12 + 1} \cr } } \right]$

= $\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right]$

Now, 3A² + 12A

= $3\left[ {\matrix{ {16} & { - 9} \cr { - 12} & {13} \cr } } \right] + 12\left[ {\matrix{ 2 & { - 3} \cr { - 4} & 1 \cr } } \right]$

= $\left[ {\matrix{ {48} & { - 27} \cr { - 36} & {39} \cr } } \right] + \left[ {\matrix{ {24} & { - 36} \cr { - 48} & {12} \cr } } \right]$

= $\left[ {\matrix{ {72} & { - 63} \cr { - 84} & {51} \cr } } \right]$

$ \therefore $ adj(3A² + 12A) = $\left[ {\matrix{ {51} & {63} \cr {84} & {72} \cr } } \right]$

Given,

$A = \left[ {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right]$

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

$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right]$

A² $-$ 2A $-$ I

$ = \left[ {\matrix{ {13} & 3 \cr { - 9} & { - 2} \cr } } \right] - \left[ {\matrix{ { - 8} & { - 2} \cr 6 & 2 \cr } } \right] - \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$

$ = \left[ {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right]$

$\left| A \right| = \left| {\matrix{ { - 4} & { - 1} \cr 3 & 1 \cr } } \right|$ $=$ $-$ 4 + 3 $=$ $-$ 1

Now,

$\left| {{A^{2016}} - 2{A^{2015}} - {A^{2014}}} \right|$

$=$ ${\left| A \right|^{2014}}\left| {{A^2} - 2A - {\rm I}} \right|$

$ = {\left( { - 1} \right)^{2014}}\left| {\matrix{ {20} & 5 \cr { - 15} & { - 5} \cr } } \right|$

$=$ 1 $ \times $ ($-$ 100 + 75)

$=$ $-$ 25

Given,

A² $-$ 5A + 7I = 0

$ \Rightarrow $   A² $-$ 5A = $-$ 7I

$ \Rightarrow $   AAA^$-$1 $-$ 5AA^$-$1 = $-$ 7IA^$-$1

$ \Rightarrow $   AI $-$ 5I = $-$ 7A^$-$1

$ \Rightarrow $   A $-$ 5I = $-$ 7A^$-$1

$ \Rightarrow $   A^$-$1 = ${1 \over 7}$(5I $-$ A)

Hence, statement 1 is true.

Now A³ $-$ 2A² $-$ 3A + I

=   A(A²) $-$ 2A² $-$ 3A + I

=   A(5A $-$ 7I) $-$ 2A² $-$ 3A + I

=   5A² $-$ 7A $-$ 2A² $-$ 3A + I

=   3A² $-$ 10A + I

=   3(5A $-$ 7I) $-$ 10A + I

=   15A $-$ 21A $-$ 10A + I

=   5A $-$ 20I

=   5(A $-$ 4I)

So, statement 2 is also correct.

P = $\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & {{1 \over 2}} \cr { - {1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr } } \right]$

$ \therefore $   P^T = $\left[ {\matrix{ {{{\sqrt 3 } \over 2}} & { - {1 \over 2}} \cr {{1 \over 2}} & {{{\sqrt 3 } \over 2}} \cr  } } \right]$

As    PP^T = P^TP = I

given, Q = PAP^T

$ \therefore $   P^TQ = P^TP AP^T

$ \Rightarrow $   P^TQ = IAP^T = AP^T [ as    P^TP = I]

Now,

P^T  Q²⁰¹⁵  P

=  P^TQ   .   Q²⁰¹⁴  .  P

=  AP^T  Q²⁰¹⁴  P

=  AP^T  .  Q  .  Q²⁰¹³  .  P

=  A²P^T  .  Q²⁰¹³  .  P

.

.

.
=  A²⁰¹⁴  .  P^TQP

=  A²⁰¹⁴   .   AP^TP

=  A²⁰¹⁵

As   A = $\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]$

$ \therefore $  A² =   $\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]$   =  $\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$

A³  =   $\left[ {\matrix{ 1 & 2 \cr 0 & 1 \cr } } \right]$ $\left[ {\matrix{ 1 & 1 \cr 0 & 1 \cr } } \right]$  =  $\left[ {\matrix{ 1 & 3 \cr 0 & 1 \cr } } \right]$

A²⁰¹⁵  =  $\left[ {\matrix{ 1 & {2015} \cr 0 & 1 \cr } } \right]$

Given,

$\left| {\matrix{ {\cos x} & {\sin x} & {\sin x} \cr {\sin x} & {\cos x} & {\sin x} \cr {\sin x} & {\sin x} & {\cos x} \cr } } \right| = 0$

R₁  $ \to $  R₁  $-$  R₃

R₁  $ \to $  R₂  $-$  R₃

$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & {\sin x - \cos x} \cr {\sin x} & {\sin x} & { \cos x} \cr } } \right| = 0$

C₃  $ \to $  C₃  +  C₂

$\left| {\matrix{ {\cos x - \sin x} & 0 & {\sin x - \cos x} \cr 0 & {\cos x - \sin x} & 0 \cr {\sin x} & {\sin x} & {\sin x + \cos x} \cr } } \right| = 0$

Expanding using first column,

(cosx $-$ sinx)(cos $-$ sinx) (sinx + cos x)

      + sinx (cosx $-$ sinx) (sinx $-$ cosx) = 0

$ \Rightarrow $   (cosx $-$ sinx)² (sinx + cosx)

     $+$ sinx (cosx $-$ sinx)² = 0

$ \Rightarrow $    (cosx $-$ sinx)² (sinx + cosx $+$ sinx) = 0

$ \Rightarrow $    (2sinx + cosx )(cosx $-$ sinx)² = 0

$ \therefore $   cosx = -2sinx   or  cosx   =  sinx

$ \Rightarrow $ tanx = $ - {1 \over 2}$ or tanx = 1

$ \therefore $ x = $ - {\tan ^{ - 1}}\left( {{1 \over 2}} \right)$, ${\pi \over 4}$

$ \therefore $   Number of solutions   =  2

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$

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

$ \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$

$\left[ {\matrix{ 1 & 2 & 2 \cr 2 & 1 & { - 2} \cr a & 2 & b \cr } } \right]\left[ {\matrix{ 1 & 2 & a \cr 2 & 1 & 2 \cr 2 & { - 2} & b \cr } } \right] = \left[ {\matrix{ 9 & 0 & 0 \cr 0 & 9 & 0 \cr 0 & 0 & 9 \cr } } \right]$

$ \Rightarrow \left[ {\matrix{ {1 + 4 + 4} & {2 + 2 - 4} & {a + 4 + 2b} \cr {2 + 2 - 4} & {4 + 1 + 4} & {2a + 2 - 2b} \cr {a + 4 + 2b} & {2a + 2 - 2b} & {{a^2} + 4 + {b^2}} \cr } } \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {\matrix{ 9 & 0 & 0 \cr 0 & 9 & 0 \cr 0 & 0 & 9 \cr } } \right]$

$ \Rightarrow a + 4 + 2b = 0$ $ \Rightarrow a + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$2a + 2 - 2b = 0 \Rightarrow 2a - 2b = - 2$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a - b = - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

On solving $(i)$ and $(ii)$ we get

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1 + b + 2b = - 4\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$b=-1$ and $a=-2$

$\left( {a,b} \right) = \left( { - 2, - 1} \right)$

$\left. {\matrix{ {2{x_1} - 2{x_2} + {x^3} = \lambda {x_1}} \cr {2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}} \cr {\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} = \lambda {x_3}} \cr } } \right\}$

$\eqalign{ & \Rightarrow \,\,\,\,\,\,\,\left( {2 - \lambda } \right){x_1} - 2{x_2} + {x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,2{x_1} - \left( {3 + \lambda } \right){x_2} + 2{x_3} = 0 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - {x_1} + 2{x_2} - \lambda {x_3} = 0 \cr} $

For non-trivial solution, $\Delta = 0$

i.e. $\,\,\,\left| {\matrix{ {2 - \lambda } & { - 2} & 1 \cr 2 & { - \left( {3 + \lambda } \right)} & 2 \cr { - 1} & 2 & { - \lambda } \cr } } \right| = 0$

$ \Rightarrow \left( {2 - \lambda } \right)\left[ {\lambda \left( {3 + \lambda } \right) - 4} \right] + $

$\,\,\,\,\,\,\,\,\,2\left[ { - 2\lambda + 2} \right] + 1\left[ {4 - \left( {3 + \lambda } \right)} \right] = 0$

$ \Rightarrow {\lambda ^3} + {\lambda ^2} - 5\lambda + 3 = 0$

$ \Rightarrow \lambda = 1,1,3$

Hence, $\lambda $ has $2$ values.

$BB' = B\left( {{A^{ - 1}}A'} \right)' = B\left( {A'} \right)'\left( {{A^{ - 1}}} \right)' = BA\left( {{A^{ - 1}}} \right)'$

$ = \left( {{A^{ - 1}}A'} \right)\left( {A\left( {{A^{ - 1}}} \right)'} \right)$

$ = {A^{ - 1}}A.A'.\left( {{A^{ - 1}}} \right)'\,\,\,\,\,\,$ $\left\{ {} \right.$ as $\,\,\,\,\,\,$ $AA' = A'A$ $\left. \, \right\}$

$ = I\left( {{A^{ - 1}}A} \right)'$

$ = I.I = {I^2} = I$

Consider

$\left| {\matrix{ 3 & {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} \cr {1 + f\left( 1 \right)} & {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} \cr {1 + f\left( 2 \right)} & {1 + f\left( 3 \right)} & {1 + f\left( 4 \right)} \cr } } \right|$

$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{ {1 + 1 + 1} & {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} \cr {1 + \alpha + \beta } & {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} \cr {1 + {\alpha ^2} + {\beta ^2}} & {1 + {\alpha ^3} + {\beta ^3}} & {1 + {\alpha ^4} + {\beta ^4}} \cr } } \right|$

$\,\,\,\,\,\,\,\,\,\,\, = \left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right| \times \left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right|$

$\,\,\,\,\,\,\,\,\,\,\, = {\left| {\matrix{ 1 & 1 & 1 \cr 1 & \alpha & \beta \cr 1 & {{\alpha ^2}} & {{\beta ^2}} \cr } } \right|^2}$

$\,\,\,\,\,\,\,\,\,\,\, = {\left[ {\left( {1 - \alpha } \right)\left( {1 - \beta } \right)\left( {\alpha - \beta } \right)} \right]^2}$

So, $k=1$

From the given system, we have

${{k + 1} \over k} = {8 \over {k + 3}} \ne {{4k} \over {3k - 1}}$

( as System has no solution)

$ \Rightarrow {k^2} + 4k + 3 = 8k$

$ \Rightarrow k = 1,3$

If $k = 1$ then ${8 \over {1 + 3}} \ne {{4.1} \over 2}$ which is false

And if $k = 3$

Then ${8 \over 6} \ne {{4.3} \over {9 - 1}}$ which is true, therefore $k=3$

Hence for only one value of $k.$ System has no solution.

$\left| P \right| = 1\left( {12 - 12} \right) - \alpha \left( {4 - 6} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,3\left( {4 - 6} \right) = 2\alpha - 6$

Now, $adj\,\,A = P\,$ $\,\,\,\,\,\,\,\, \Rightarrow \left| {adj\,A} \right| = \left| P \right|$

$ \Rightarrow {\left| A \right|^2} = \left| P \right|$

$ \Rightarrow \left| P \right| = 16$

$ \Rightarrow 2\alpha - 6 = 16$

$ \Rightarrow \alpha = 11$

Given

${P^3} = {q^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

${P^2}Q = {Q^2}p\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Subtracting $(1)$ and $(2)$, we get

${P^3} - {P^2}Q = {Q^3} - {Q^2}P$

$ \Rightarrow {P^2}\left( {P - Q} \right) + {Q^2}\left( {P - Q} \right) = 0$

$ \Rightarrow \left( {{P^2} + {Q^2}} \right)\left( {P - Q} \right) = 0$

$ \Rightarrow \left| {{p^2} + {Q^2}} \right| = 0$

as $P \ne Q$

Let $A{u_1} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right)\,\,\,\,\,\,A{u_2} = \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right)$

Then, $A{u_1} + A{u_2} = \left( {\matrix{ 1 \cr 0 \cr 0 \cr } } \right) + \left( {\matrix{ 0 \cr 1 \cr 0 \cr } } \right)$

$ \Rightarrow A\left( {{u_1} + {u_2}} \right) = \left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Also, $A = \left( {\matrix{ 1 & 0 & 0 \cr 2 & 1 & 0 \cr 3 & 2 & 1 \cr } } \right)$

$ \Rightarrow \left| A \right| = 1\left( 1 \right) - 0\left( 2 \right) + 0\left( {4 - 3} \right) = 1$

We know,

${A^{ - 1}} = {1 \over {\left| A \right|}}\,adjA \Rightarrow {A^{ - 1}} = adj\left( A \right)$

( as $\left| A \right| = 1$ )

Now, from equation $(1)$, we have

${u_1} + {u_2} = {A^{ - 1}}\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$

$ = \left[ {\matrix{ 1 & 0 & 0 \cr { - 2} & 1 & 0 \cr 1 & { - 2} & 1 \cr } } \right]\left( {\matrix{ 1 \cr 1 \cr 0 \cr } } \right)$

$ = \left[ {\matrix{ 1 \cr { - 1} \cr { - 1} \cr } } \right]$

$\Delta = 0 \Rightarrow \left| {\matrix{ 4 & k & 2 \cr k & 4 & 1 \cr 2 & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow 4\left( {4 - 2} \right) - k\left( {k - 2} \right) + $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\left( {2k - 8} \right) = 0$

$ \Rightarrow 8 - {k^2} + 2k + 4k - 16 = 0$

$ \Rightarrow {k^2} - 6k + 8 = 0$

$ \Rightarrow \left( {k - 4} \right)\left( {k - 2} \right) = 0,k = 4,2$

$\therefore$ $A' = A,B' = B$

Now $\,\,\,\left( {A\left( {BA} \right)} \right)' = \left( {BA} \right)'A'$

$ = \left( {A'B'} \right)A' = \left( {AB} \right)A = A\left( {BA} \right)$

Similarly $\left( {\left( {AB} \right)A} \right)' = \left( {AB} \right)A$

So, $A\left( {BA} \right)\,\,\,\,$ and $A\left( {BA} \right)\,\,\,\,$ are symmetric matrices.

Again $\left( {AB} \right)' = B'A' = BA$

Now if $BA=AB$, then $AB$ is symmetric matrix.

$\left[ {\matrix{ 1 & {...} & {...} \cr {...} & 1 & {...} \cr {...} & {...} & 1 \cr } } \right]\,\,$ are $6$ non-singular matrices because $6$

blanks will be filled by $5$ zeros and $1$ one.

Similarly, $\left[ {\matrix{ {...} & {...} & 1 \cr {...} & 1 & {...} \cr 1 & {...} & {...} \cr } } \right]\,\,$ are $6$ non-singular matrices.

So, required cases are more than $7,$ non-singular $3 \times 3$ matrices.

Let $A = \left( {\matrix{ a & b \cr c & d \cr } } \right)$ where $a,b,c,d$ $ \ne 0$

${A^2} = \left( {\matrix{ a & b \cr c & d \cr } } \right)\left( {\matrix{ a & b \cr c & d \cr } } \right)$

$ \Rightarrow {A^2} = \left( {\matrix{ {{a^2} + bc} & {ab + bd} \cr {ac + cd} & {bc + {d^2}} \cr } } \right)$

$ \Rightarrow {a^2} + bc = 1,\,bc + {d^2} = 1$

$ab + bd = ac + cd = 0$

$c \ne 0\,\,\,\,\,b \ne 0$

$ \Rightarrow a + d = 0 \Rightarrow Tr\left( A \right) = 0$

$\left| A \right| = ad - bc = - {a^2} - bc = - 1$

$D = \left| {\matrix{ 1 & 2 & 1 \cr 2 & 3 & 1 \cr 3 & 5 & 2 \cr } } \right| = 0$

${D_1}\left| {\matrix{ 3 & 2 & 1 \cr 3 & 3 & 1 \cr 1 & 5 & 2 \cr } } \right| \ne 0$

$ \Rightarrow $ Given system, does not have any solution.

$ \Rightarrow $ No solution

We know that $\left| {adj\left( {adj\,\,A} \right)} \right| = {\left| {Adj\,\,A} \right|^{2 - 1}}$b

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\left| A \right|^{2 - 1}} = \left| A \right|$

$\therefore$ $\,\,\,\,\,\,$ Both the statements are true and statement $-2$ is a correct explanation for statement - $1.$

$\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {b + 1} & {c - 1} \cr {a - 1} & {b - 1} & {c + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}a} & {{{\left( { - 1} \right)}^{n + 1}}b} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + \left| {\matrix{ {a + 1} & {a - 1} & {{{\left( { - 1} \right)}^{n + 2}}a} \cr {b + 1} & {b - 1} & {{{\left( { - 1} \right)}^{n + 1}}b} \cr {c - 1} & {c + 1} & {{{\left( { - 1} \right)}^n}c} \cr } } \right| = 0$

(Taking transpose of second determinant)

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_1} \Leftrightarrow {C_3}$

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| - \left| {\matrix{ {{{\left( { - 1} \right)}^{n + 2}}a} & {a - 1} & {a + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}\left( { - b} \right)} & {b - 1} & {b + 1} \cr {{{\left( { - 1} \right)}^{n + 2}}c} & {c + 1} & {c - 1} \cr } } \right| = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} \Leftrightarrow {C_3}$

$ \Rightarrow \left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| + {\left( 1 \right)^{n + 2}}\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & {a + 1} & {a - 1} \cr { - b} & {b + 1} & {b - 1} \cr c & {c - 1} & {c + 1} \cr } } \right| = 0$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{C_2} - {C_1},{C_3} - {C_1}$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ a & 1 & { - 1} \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$

${R_1} + {R_3}$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left| {\matrix{ {a + c} & 0 & 0 \cr { - b} & {2b + 1} & {2b - 1} \cr c & { - 1} & 1 \cr } } \right| = 0$

$ \Rightarrow \left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right]\left( {a + c} \right)\left( {2b + 1 + 2b - 1} \right) = 0$

$ \Rightarrow 4b\left( {a + c} \right)\left[ {1 + {{\left( { - 1} \right)}^{n + 2}}} \right] = 0$

$ \Rightarrow 1 + {\left( { - 1} \right)^{n + 2}} = 0$ $\,\,\,\,\,$ as $\,\,\,\,\,b\left( {a + c} \right) \ne 0$

$ \Rightarrow n$ should be an odd integer.

The given equations are

$\matrix{ { - x + cy + bz = 0} \cr {cx - y + az = 0} \cr {bx + ay - z = 0} \cr } $

As $x,y,z$ are not all zero

$\therefore$ The above system should not have unique (zero) solution

$ \Rightarrow \Delta = 0 \Rightarrow \left| {\matrix{ { - 1} & c & b \cr c & { - 1} & a \cr b & a & { - 1} \cr } } \right| = 0$

$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$

$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$

$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$

Let $A = \left[ {\matrix{ a & b \cr c & d \cr } } \right]$ $\,\,\,$ then ${A^2} = 1$

$ \Rightarrow {a^2} + bc = 1\,\,\,\,ab + bd = 0$

$ac + cd = 0\,\,\,\,bc + {d^2} = 1$

From these four relations,

${a^2} + bc = bc + {d^2} \Rightarrow {a^2} = {d^2}$

and $\,\,b\left( {a + d} \right) = 0 = c\left( {a + d} \right)$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = - d$

We can take $a = 1,b = 0,c = 0,d = - 1$

as one possible set of values, then

$A = \left[ {\matrix{ 1 & 0 \cr 0 & { - 1} \cr } } \right]$

Clearly $A \ne I\,\,\,$ and $\,\,\,\,A \ne - I\,\,$ and $\,\,\,A = - 1$

$\therefore$ $\,\,\,\,\,$ Statement $1$ is true.

Also if $A \ne I\,\,\,\,\,tr\left( A \right) = 0$

$\therefore$ $\,\,\,\,\,$ Statement $2$ is false.

As all entries of square matrix $A$ are integers, therefore all co-factors should also be integers.

If det $A = \pm 1\,\,$ then ${A^{ - 1}}\,\,$ exists. Also all entries of ${A^{ - 1}}$ are integers.

$\left| {{A^2}} \right| = 25 \Rightarrow {\left| A \right|^2} = 25$

$ \Rightarrow {\left( {25\alpha } \right)^2} = 25 \Rightarrow \left| \alpha \right| = {1 \over 5}$

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 $y$

${A^2} - {B^2} = \left( {A - B} \right)\left( {A + B} \right)$

${A^2} - {B^2} = {A^2} + AB - BA - {B^2}$

$ \Rightarrow AB = BA$

$A = \left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right]\,\,\,\,B = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]$

$AB = \left[ {\matrix{ a & {2b} \cr {3a} & {4b} \cr } } \right]$

$BA = \left[ {\matrix{ a & 0 \cr 0 & b \cr } } \right]\left[ {\matrix{ 1 & 2 \cr 3 & 4 \cr } } \right] = \left[ {\matrix{ a & {2a} \cr {3b} & {4b} \cr } } \right]$

Hence, $AB=BA$ only when $a=b$

$\therefore$ There can be infinitely many $B's$

for which $AB=BA$

$ax + y + z = \alpha - 1$

$x + \alpha \,y + z = \alpha - 1;$

$x + y + z\alpha = \alpha - 1$

$\Delta = \left| {\matrix{ \alpha & 1 & 1 \cr 1 & \alpha & 1 \cr 1 & 1 & \alpha \cr } } \right|$

$ = \alpha \left( {{\alpha ^2} - 1} \right) - 1\left( {\alpha - 1} \right) + 1\left( {1 - \alpha } \right)$

$ = \alpha \left( {\alpha - 1} \right)\left( {\alpha + 1} \right) - 1\left( {\alpha - 1} \right) - 1\left( {\alpha - 1} \right)$

For infinite solutions, $\Delta = 0$

$ \Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 1 - 1} \right] = 0$

$ \Rightarrow \left( {\alpha - 1} \right)\left[ {{\alpha ^2} + \alpha - 2} \right] = 0$

$ \Rightarrow \alpha = - 2,1;$

But $\alpha \ne 1.\,\,\,$ $\therefore$ $\,\,\alpha = - 2$

As $\,\,\,\,{a_1},{a_2},{a_3},.........$ are in $G.P.$

$\therefore$ Using ${a_n} = a{r^{n - 1}},\,\,\,$ we get the given determinant,

as $\,\,\,\,\,\,\,\left| {\matrix{ {\log a{r^{n - 1}}} & {\log a{r^n}} & {\log a{r^{n + 1}}} \cr {\log a{r^{n + 2}}} & {\log a{r^{n + 3}}} & {\log a{r^{n + 4}}} \cr {\log a{r^{n + 5}}} & {\log a{r^{n + 6}}} & {\log a{r^{n + 7}}} \cr } } \right|$

Operating ${C_3} - {C_2}$ and ${C_2} - {C_1}$ and using

$\log m - \log n = \log {m \over n}\,\,\,\,$ we get

$ = \left| {\matrix{ {\log a{r^{n - 1}}} & {\log r} & {\log r} \cr {\log a{r^{n + 2}}} & {\log r} & {\log r} \cr {\log a{r^{n + 5}}} & {\log r} & {\log r} \cr } } \right| $

$=0$ (two columns being identical)

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