If for the matrix, $$A = \left[ {\matrix{ 1 & { - \alpha } \cr \alpha & \beta \cr } } \right]$$, $$A{A^T} = {I_2}$$, then the value of $${\alpha ^4} + {\beta ^4}$$ is :

$$\left[ {\matrix{ 1 & { - \alpha } \cr \alpha & \beta \cr } } \right]\left[ {\matrix{ 1 & \alpha \cr { - \alpha } & \beta \cr } } \right] = \left[ {\matrix{ {1 + {\alpha ^2}} & {\alpha - \alpha \beta } \cr {\alpha - \alpha \beta } & {{\alpha ^2} + {\beta ^2}} \cr } } \right] = \left[ {\matrix{ 1 & 0 \cr 0 & 1 \cr } } \right]$$ 1 + $$\alpha$$2 = 1 $$\alpha$$2 = 0 $$\alpha$$2 + $$\beta$$2 = 1 $$\beta$$2 = 1 $$\alpha$$4 = 0 $$\beta$$4 = 1 $$\alpha$$4 + $$\beta$$4 =

The value of $$\left| {\matrix{ {(a + 1)(a + 2)} & {a + 2} & 1 \cr {(a + 2)(a + 3)} & {a + 3} & 1 \cr {(a + 3)(a + 4)} & {a + 4} & 1 \cr } } \right|$$ is :

Given, $$\Delta $$ = $$\left| {\matrix{ {(a + 1)(a + 2)} & {a + 2} & 1 \cr {(a + 2)(a + 3)} & {a + 3} & 1 \cr {(a + 3)(a + 4)} & {a + 4} & 1 \cr } } \right|$$ R2 $$ \to $$ R2 $$-$$ R1 and R3 $$ \to $$ R3 $$-$$ R1 $$\Delta$$ = $$\left| {\matrix{ {(a + 1)(a + 2)} & {a + 2} & 1 \cr {(a + 2)(a + 3 - a - 1)} & 1 & 0 \cr {{a^2} + 7a + 12 - {a^2} - 3a - 2} & 2 & 0 \cr } } \right|$$ $$ = \left| {\matrix{ {{a^2} + 3a + 2} & {a + 2} & 1 \cr {2(a + 2)

Matrices and Determinants — Page 7 | JEE Mathematics

Q61

The system of linear equations

\lambda

x + 2y + 2z = 5 2

\lambda

x + 3y + 5z = 8 4x +

\lambda

y + 6z = 10 has

A a unique solution when

\lambda

= –8

B no solution when

\lambda

= 2

C infinitely many solutions when

\lambda

= 2

D no solution when

\lambda

= 8

Correct Answer

Option B

Solution

\Delta

=

\left| \begin{array}{lll}\lambda & 2 & 2 \\ {2\lambda } & 3 & 5 \\ 4 & \lambda & 6 \end{array} \right|

= $\lambda$ ( 18 – 5 $\lambda$ ) – 2(12 $\lambda$ – 20) + 2(2 $\lambda$ 2 – 12) = 18 $\lambda$ – 5 $\lambda$ 2 – 24 $\lambda$ + 40 + 4 $\lambda$ 2 – 24 = – $\lambda$ 2 – 6 $\lambda$ + 16 = – ( $\lambda$ + 8)( $\lambda$ – 2) For no solutions $\lambda$ = 0 $\Rightarrow$ $\lambda$ = – 8, $\lambda$ = 2 when $\lambda$ = 2

\Delta

x =

\left| \begin{array}{lll}5 & 2 & 2 \\ 8 & 3 & 5 \\ {10} & 2 & 6 \end{array} \right|

= 5 (18 – 10) – 2 (48 – 50) + 2 (16 – 30) = 40 + 4 – 28 $\ne$ 0 So no solution for $\lambda$ = 2

Q62

If the matrices A =

\left[ \begin{array}{lll}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & { - 1} & 3 \end{array} \right]

, B = adjA and C = 3A, then

{{\left| {adjB} \right|} \over {\left| C \right|}}

is equal to :

A 8

B 2

C 72

D 16

Correct Answer

Option A

Solution

A =

\left[ \begin{array}{lll}1 & 1 & 2 \\ 1 & 3 & 4 \\ 1 & { - 1} & 3 \end{array} \right]

$\Rightarrow$ |A| = 6

{{\left| {adjB} \right|} \over {\left| C \right|}}

=

{{\left| {adj\left( {adjA} \right)} \right|} \over {\left| {3A} \right|}}

=

{{{{\left| A \right|}^4}} \over {{3^3}\left| A \right|}}

=

{{{{\left| A \right|}^3}} \over {{3^3}}}

=

{{{6^3}} \over {{3^3}}}

= 8

Q63

If for some

\alpha

and

\beta

in R, the intersection of the following three places x + 4y – 2z = 1 x + 7y – 5z = b x + 5y +

\alpha

z = 5 is a line in R3, then

\alpha

+

\beta

is equal to :

A -10

B 0

C 10

D 2

Correct Answer

Option C

Solution

For planes to intersect on a line there should be infinite solution of the given system of equations.

For infinite solutions

\Delta

=

\left| \begin{array}{lll}1 & 4 & { - 2} \\ 1 & 7 & { - 5} \\ 1 & 5 & \alpha \end{array} \right|

= 0 $\Rightarrow$ 1(7 $\alpha$ + 25) – 4( $\alpha$ + 5) – 2(5 – 7) = 0 $\Rightarrow$ 7 $\alpha$ + 25 – 4 $\alpha$ – 20 + 4 = 0 $\Rightarrow$ 3 $\alpha$ + 9 = 0 $\Rightarrow$ $\alpha$ = -3 Also

\Delta

z = 0 $\Rightarrow$

\left| \begin{array}{lll}1 & 4 & 1 \\ 1 & 7 & \beta \\ 1 & 5 & 5 \end{array} \right|

= 0 $\Rightarrow$ 1(35 – 5 $\beta$ ) – 4(5 – $\beta$ ) + 1(5 – 7) = 0 $\Rightarrow$ 35 - 5 $\beta$ - 20 + 4 $\beta$ - 2 = 0 $\Rightarrow$ $\beta$ = 13 $\therefore$ $\alpha$ + $\beta$ = -3 + 13 = 10

Q64

Let A be a 2

\times

2 real matrix with entries from {0, 1} and |A|

\ne

0. Consider the following two statements : (P) If A

\ne

I2 , then |A| = –1 (Q) If |A| = 1, then tr(A) = 2, where I2 denotes 2

\times

2 identity matrix and tr(A) denotes the sum of the diagonal entries of A. Then :

A (P) is true and (Q) is false

B Both (P) and (Q) are false

C Both (P) and (Q) are true

D (P) is false and (Q) is true

Correct Answer

Option D

Solution

Let A =

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

, where a, b, c, d

\in

{0, 1} $\Rightarrow$ |A| = ad – bc $\therefore$ ad = 0 or 1 and bc = 0 or 1 So possible values of |A| are 1, 0 or –1 (P) If A $\ne$ I2 then |A| is either 0 or –1 (Q) If |A| = 1 then ad = 1 and bc = 0 $\Rightarrow$ a = d = 1 $\Rightarrow$ Tr(A) = 2

Q65

If for the matrix,

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

,

A{A^T} = {I_2}

, then the value of

{\alpha ^4} + {\beta ^4}

is :

A 3

B 2

C 1

D 4

Correct Answer

Option C

Solution

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

1 + $\alpha$ 2 = 1 $\alpha$ 2 = 0 $\alpha$ 2 + $\beta$ 2 = 1 $\beta$ 2 = 1 $\alpha$ 4 = 0 $\beta$ 4 = 1 $\alpha$ 4 + $\beta$ 4 = 1

Q66

If

A = \left( \begin{array}{ll}2 & 2 \\ 9 & 4 \end{array} \right)

and

I = \left( \begin{array}{ll}1 & 0 \\ 0 & 1 \end{array} \right)

then 10A–1 is equal to :

A 6I – A

B 4I – A

C A – 6I

D A – 4I

Correct Answer

Option C

Solution

According to Cayley Hamilton equation |A – $\lambda$ I| = 0 $\Rightarrow$

\left| \begin{array}{ll}{2 - \lambda } & 2 \\ 9 & {4 - \lambda } \end{array} \right|

= 0 $\Rightarrow$ (2 – $\lambda$ )(4 – $\lambda$ ) – 18 = 0 $\Rightarrow$ 8 – 2 $\lambda$ – 4 $\lambda$ + $\lambda$ 2 – 18 = 0 $\Rightarrow$ $\lambda$ 2 – 6 $\lambda$ – 10 = 0 $\therefore$ A2 – 6A– 10 = 0 $\Rightarrow$ A–1(A2) – 6A–1A – 10A–1 = 0 $\Rightarrow$ A – 6I – 10A–1 = 0 $\Rightarrow$ 10A–1 = A – 6I

Q67

The system of linear equations 3x - 2y - kz = 10 2x - 4y - 2z = 6 x+2y - z = 5m is inconsistent if :

A k

\ne

3, m

\in

R

B k = 3, m

\ne

{4 \over 5}

C k = 3, m

=

{4 \over 5}

D k

\ne

3, m

\ne

{4 \over 5}

Correct Answer

Option B

Solution

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

3(4 + 4) + 2( - 2 + 2) - k(4 + 4) = 0

\Rightarrow k = 3

{\Delta _x} = \left| \begin{array}{lll}{10} & { - 2} & { - 3} \\ 6 & { - 4} & { - 2} \\ {5m} & 2 & { - 1} \end{array} \right| \ne 0

10(4 + 4) + 2( - 6 + 10m) - 3(12 + 20m) \ne 0

80 - 12 + 20m - 36 - 60m \ne 0

40m \ne 32 \Rightarrow m \ne {4 \over 5}

{\Delta _y} = \left| \begin{array}{lll}3 & {10} & { - 3} \\ 2 & 6 & { - 2} \\ 1 & {5m} & { - 1} \end{array} \right| \ne 0

3( - 6 + 10m) - 10( - 2 + 2) - 3(10m - 6) \ne 0

- 18 + 30m - 30m + 18 \ne 0 \Rightarrow 0

{\Delta _z} = \left| \begin{array}{lll}3 & { - 2} & {10} \\ 2 & { - 4} & 6 \\ 1 & 2 & {5m} \end{array} \right| \ne 0

3( - 20m - 12) + 2(10m - 6) + 10(4 + 4) - 40m + 32 \ne 0 \Rightarrow m \ne {4 \over 5}

Q68

For the system of linear equations:

x - 2y = 1,x - y + kz = - 2,ky + 4z = 6,k \in R

, consider the following statements : (A) The system has unique solution if

k \ne 2,k \ne - 2

. (B) The system has unique solution if k =

-

2 (C) The system has unique solution if k = 2 (D) The system has no solution if k = 2 (E) The system has infinite number of solutions if k

\ne

-

2. Which of the following statements are correct?

A (B) and (E) only

B (C) and (D) only

C (A) and (E) only

D (A) and (D) only

Correct Answer

Option D

Solution

x - 2y + 0.z = 1

x - y + kz = - 2

0.x + ky + 4z = 6

\Delta = \left| \begin{array}{lll}1 & { - 2} & 0 \\ 1 & { - 1} & k \\ 0 & k & 4 \end{array} \right| = 4 - {k^2}

For unique solution

4 - {k^2} \ne 0

$\Rightarrow$ k $\ne$ $\pm$ 2 For k = 2 :

x - 2y + 0.z = 1

x - y + 2z = - 2

0.x + 2y + 4z = 6

\Delta x = \left| \begin{array}{lll}1 & { - 2} & 0 \\ 2 & { - 1} & 2 \\ 6 & 2 & 4 \end{array} \right| = ( - 8) + 2[ - 20]

\Delta x = - 48 \ne 0

For k = 2,

\Delta x \ne 0

$\therefore$ For K = 2; The system has no solution.

Q69

The following system of linear equations 2x + 3y + 2z = 9 3x + 2y + 2z = 9 x

-

y + 4z = 8

A does not have any solution

B has a solution (

\alpha

,

\beta

,

\gamma

) satisfying

\alpha

+

\beta

2 +

\gamma

3 = 12

C has a unique solution

D has infinitely many solutions

Correct Answer

Option C

Solution

\Delta = \left| \begin{array}{lll}2 & 3 & 2 \\ 3 & 2 & 2 \\ 1 & { - 1} & 4 \end{array} \right| = - 20 \ne 0

$\therefore$ unique solution

{\Delta _x} = \left| \begin{array}{lll}9 & 3 & 2 \\ 9 & 2 & 2 \\ 8 & { - 1} & 4 \end{array} \right| = 0

{\Delta _y} = \left| \begin{array}{lll}2 & 9 & 2 \\ 3 & 9 & 2 \\ 1 & 8 & 4 \end{array} \right| = - 20

{\Delta _z} = \left| \begin{array}{lll}2 & 3 & 9 \\ 3 & 2 & 9 \\ 1 & { - 1} & 8 \end{array} \right| = - 40

$\therefore$

x = {{{\Delta _x}} \over \Delta } = 0

y = {{{\Delta _y}} \over \Delta } = 1

z = {{{\Delta _z}} \over \Delta } = 2

Unique solution : (0, 1, 2)

Q70

The value of

\left| \begin{array}{lll}{(a + 1)(a + 2)} & {a + 2} & 1 \\ {(a + 2)(a + 3)} & {a + 3} & 1 \\ {(a + 3)(a + 4)} & {a + 4} & 1 \end{array} \right|

is :

A

-

2

B 0

C (a + 2)(a + 3)(a + 4)

D (a + 1)(a + 2)(a + 3)

Correct Answer

Option A

Solution

Given,

\Delta

=

\left| \begin{array}{lll}{(a + 1)(a + 2)} & {a + 2} & 1 \\ {(a + 2)(a + 3)} & {a + 3} & 1 \\ {(a + 3)(a + 4)} & {a + 4} & 1 \end{array} \right|

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

\Delta

=

\left| \begin{array}{lll}{(a + 1)(a + 2)} & {a + 2} & 1 \\ {(a + 2)(a + 3 - a - 1)} & 1 & 0 \\ {{a^2} + 7a + 12 - {a^2} - 3a - 2} & 2 & 0 \end{array} \right|

= \left| \begin{array}{lll}{{a^2} + 3a + 2} & {a + 2} & 1 \\ {2(a + 2)} & 1 & 0 \\ {4a + 10} & 2 & 0 \end{array} \right|

= 4(a + 2) - 4a - 10

= 4a + 8 - 4a - 10 = - 2