Free Objective Test 02 Practice Test - 11th and 12th

Question 1

If $A = [\begin{matrix} a & b b & a \end{matrix}]$ and $A^{2} = [\begin{matrix} α & β β & α \end{matrix}]$ , then

α = a^{2} + b^{2}, β = a b

α = a^{2} + b^{2}, β = 2 a b

α = a^{2} + b^{2}, β = a^{2} - b^{2}

α = 2 a b, β = a^{2} + b^{2}

SOLUTION

Solution : B

$A^{2} = [\begin{matrix} α & β β & α \end{matrix}] = [\begin{matrix} a & b b & a \end{matrix}] [\begin{matrix} a & b b & a \end{matrix}]; α = α^{2} + b^{2}; β = 2 a b$

Question 2

If $A = ⎡ ⎣ \begin{matrix} 1 & t a n \frac{θ}{2} - t a n \frac{θ}{2} & 1 \end{matrix} ⎤ ⎦$ and AB = I, then B =

c o s^{2} \frac{θ}{2} . A

c o s^{2} \frac{θ}{2} . A^{T}

c o s^{2} \frac{θ}{2} . I

None of these

SOLUTION

Solution : B

$| A | = 1 + t a n^{2} \frac{θ}{2} = s e c^{2} \frac{θ}{2} A B = I \Rightarrow B - I A^{- 1} \frac{[\begin{matrix} 1 & 0 0 & 1 \end{matrix}] ⎡ ⎢ ⎣ \begin{matrix} 1 & - t a n \frac{θ}{2} t a n \frac{θ}{2} & 1 \end{matrix} ⎤ ⎥ ⎦}{s e c^{2} \frac{θ}{2}} = c o s^{2} \frac{θ}{2} . A^{T} .$

Question 3

Let,
$A = ⎡ ⎢ ⎣ \begin{matrix} 4 & 6 & - 1 3 & 0 & 2 1 & - 2 & 5 \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 2 & 4 0 & 1 - 1 & 2 \end{matrix} ⎤ ⎥ ⎦$
$and C = [\begin{matrix} 1 & 2 & 3 \end{matrix}]$
The expression which is not defined is:

B^{'} B

C A B

A + B^{'}

A^{2} + A

SOLUTION

Solution : C

We can see from the options that if we take transpose of B,
$B^{'}$ will be of 2 x 3 matrix which cannot be added to a 3 x 3 matrix, as for the addition the order should be the same.

Question 4

If A is $3 \times 4$ matrix and B is a matrix such that A'B and BA' are both defined. Then B is of the type

3 \times 4

3 \times 3

4 \times 4

4 \times 3

SOLUTION

Solution : A

$A_{3 \times 4} \Rightarrow A_{4 \times 3}^{'}$ Now A'B defined
$\Rightarrow B$ is $3 \times p$
Again $B_{3 \times p} A_{4 \times 3}^{'}$ defined $\Rightarrow p = 4$
$∴ B$ is $3 \times 4$ .

Question 5

If $A = ⎡ ⎢ ⎣ \begin{matrix} a & 0 & 0 0 & b & 0 0 & 0 & c \end{matrix} ⎤ ⎥ ⎦$ , then $A^{n} =$

⎡ ⎢ ⎣ \begin{matrix} n a & 0 & 0 0 & n b & 0 0 & 0 & n c \end{matrix} ⎤ ⎥ ⎦

⎡ ⎢ ⎣ \begin{matrix} a & 0 & 0 0 & b & 0 0 & 0 & c \end{matrix} ⎤ ⎥ ⎦

⎡ ⎢ ⎣ \begin{matrix} a^{n} & 0 & 0 0 & b^{n} & 0 0 & 0 & c^{n} \end{matrix} ⎤ ⎥ ⎦

N o n e o f t h e s e

SOLUTION

Solution : C

Since $A^{2} = A . A = ⎡ ⎢ ⎣ \begin{matrix} a & 0 & 0 0 & b & 0 0 & 0 & c \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} a & 0 & 0 0 & b & 0 0 & 0 & c \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} a^{2} & 0 & 0 0 & b^{2} & 0 0 & 0 & c^{2} \end{matrix} ⎤ ⎥ ⎦$
And $A^{3} = ⎡ ⎢ ⎣ \begin{matrix} a^{3} & 0 & 0 0 & b^{3} & 0 0 & 0 & c^{3} \end{matrix} ⎤ ⎥ ⎦, . . . . \Rightarrow A^{n} = A^{n - 1} . A = ⎡ ⎢ ⎣ \begin{matrix} a^{n - 1} & 0 & 0 0 & b^{n - 1} & 0 0 & 0 & c^{n - 1} \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} a & 0 & 0 0 & b & 0 0 & 0 & c \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} a^{n} & 0 & 0 0 & b^{n} & 0 0 & 0 & c^{n} \end{matrix} ⎤ ⎥ ⎦ .$
Note: Students should remember this question as a formula.

Question 6

For each real number x such that $- 1 < x < 1, l e t A (x) b e t h e m a t r i x (1 - x)^{- 1} [\begin{matrix} 1 & - x - x & 1 \end{matrix}] a n d z = \frac{x + y}{1 + x y} T h e n,$

A (z) = A (x) + A (y)

A (z) = A (x) + [A (y)]^{- 1}

A (z) = A (x) A (y)

A (z) = A (x) - A (y)

SOLUTION

Solution : C

$A (z) = A (\frac{x + y}{1 + x y}) = [\frac{1 + x y}{(1 - x) (1 - y)}] ⎡ ⎢ ⎣ \begin{matrix} 1 & - (\frac{x + y}{1 + x y}) - (\frac{x + y}{1 + x y}) & 1 \end{matrix} ⎤ ⎥ ⎦$
$∴ A (x) . A (y) = A (z)$

Question 7

$I f A = [\begin{matrix} c o s θ & s i n θ s i n θ & - c o s θ \end{matrix}], B = [\begin{matrix} 1 & 0 - 1 & 1 \end{matrix}], C = A B A^{T}, t h e n A^{T} C^{n} A e q u a l s t o (n ϵ Z^{+})$

[\begin{matrix} - n & 1 1 & 0 \end{matrix}]

[\begin{matrix} 1 & - n 0 & 1 \end{matrix}]

[\begin{matrix} 0 & 1 1 & - n \end{matrix}]

[\begin{matrix} 1 & 0 - n & 1 \end{matrix}]

SOLUTION

Solution : D

$A = [\begin{matrix} c o s θ & s i n θ s i n θ & - c o s θ \end{matrix}] A A^{T} = I (i) N o w, C = A B A^{T} \Rightarrow A^{T} C = B A^{T} (i i) N o w A^{T} C^{n} A = A^{T} C . C^{n - 1} A = B A^{T} C^{n - 1} A (f r o m (i i)) = B A^{T} C . C^{n - 2} A = B^{2} A^{T} C^{n - 2} A = . . . . . . . = B^{n - 1} A^{T} C A = B^{n - 1} B A^{T} A = B^{n} = [\begin{matrix} 1 & 0 - n & 1 \end{matrix}]$

Question 8

$A = [a_{i j}]_{n \times n}$ and $a_{i j} = i^{2} - j^{2}$ then A is necessarily

A. a unit matrix

B. symmetric matrix

C. skew symmetric matrix

D. zero matrix

SOLUTION

Solution : C

$a_{j i} = j^{2} - i^{2} = - (i^{2} - j^{2}) = - a_{i j}$

Question 9

If A is a non-diagonal involutory matrix, then

A. A - I = 0

B. A + I = 0

C. A - I is non zero singular

D. none of these

SOLUTION

Solution : C

$A^{2} = I \Rightarrow A^{2} - I = 0$
$\Rightarrow$ (A+I)(A-I)=0
$∴$ either $| A + I | = 0$ or
$| A - I | = 0$
If $| A - I | \neq 0$ , then $(A + I) (A - I) = 0 \Rightarrow A + I = 0$ which is not so
$∴ | A - I |$ and $A - I \neq 0$ .

Question 10

If A and B are two non singular matrices and both are symmetric and commute each other then

A. Both

A^{- 1} B

and

A^{- 1} B^{- 1}

are symmetric

A^{- 1} B

is symmetric but

A^{- 1} B^{- 1}

is not symmetric

A^{- 1} B^{- 1}

is symmetric but

A^{- 1} B

is not symmetric

D. Neither

A^{- 1} B

nor

A^{- 1} B^{- 1}

are symmetric

SOLUTION

Solution : A

AB =BA
Previous & past multiplying both sides by $A^{- 1}$ .
$A^{- 1} (A B) A^{- 1} = A^{- 1} (B A) A^{- 1} (A^{- 1} A) (B A^{- 1}) = A^{- 1} B (A A^{- 1}) \Rightarrow (B A^{- 1})^{1} = (A^{- 1} B)^{1} = (A^{- 1})^{1} B^{1} (r e v e r s a l l a w s) = A^{- 1} B (a s B = B^{1}) (A^{- 1})^{1} = A^{- 1} \Rightarrow A^{- 1} B$ is symmetric
Similarly for $A^{- 1} B^{- 1}$ .

Question 11

If A is a skew-symmetric matrix of order 3, then the matrix $A^{4}$ is

A. skew symmetric

B. symmetric

C. diagonal

D. none of those

SOLUTION

Solution : B

We are given that the matrix A is skew symmetric which implies that,
$A^{T} = - A$ --------(1)
$(A^{4})^{T} = (A . A . A . A .)^{T} = A^{T} A^{T} A^{T} A^{T}$ (By applying the property of transpose of a matrix)
$\Rightarrow$ (-A) (-A) (-A) (-A) (Using equation (1))
$= (- 1)^{4} A^{4} = A^{4}$
i.e, $(A^{4})^{T} = (A^{4})$
$A^{4}$ is symmetric.

Question 12

If A is a $3 \times 3$ non-singular matrix such that $A A^{T} = A^{T} A a n d B = A^{- 1} A^{T}, t h e n B B^{T}$ is equal to

A. I + B

B. I

B^{- 1}

{(B^{- 1})}^{T}

SOLUTION

Solution : B

Given, $A A^{T} = A^{T} A a n d B = A^{- 1} A^{T}$
We are asked for the value of the expression $B B^{T}$ . Since B is given in terms of A we will substitute for the values of $B^{T}$ and B.
$B B^{T} = (A^{- 1} A^{T}) (A^{- 1} A^{T})^{T}$
$= A^{- 1} A^{T} A (A^{- 1})^{T} [∴ {(A B)}^{T} = B^{T} A^{T}]$
$= A^{- 1} A A^{T} (A^{- 1})^{T}$
$= I A^{T} (A^{- 1})^{T} [∴ (A^{- 1} A) = I]$
$= (A^{- 1} A)^{T}$
$= I^{T} = I$

Question 13

All the elements in a matrix A are complex numbers with imaginary parts not equal to zero. If $A^{*}$ is the conjugate of the matrix A, $a_{i j}$ is the general element of matrix A, then what is the general element of the matrix, $\frac{A + A^{*}}{2} .$

2 l m (a_{i j})

l m (a_{i j})

R e (a_{i j})

\frac{R e (a_{i j})}{2}

SOLUTION

Solution : C

By taking complex conjugate of a matrix we reverse the sign of imaginary parts of all the elements in the original matrix. i.e., if the element in A is x + iy, then the corresponding element in $A^{*}$ is x - iy.
So when A and $A^{*}$ is added the imaginary parts cancel out and the sum becomes 2 times the real part of element in A.
i.e., since $(a_{i j})$ is general element in A, the general element in $A + A^{*}$ becomes $2 R e (a_{i j}) .$
$∴$ General element in $\frac{A + A^{*}}{2} = R e (a_{i j}) .$

Question 14

If $A = [\begin{matrix} 1 & - 1 1 & 1 \end{matrix}]$ then $A^{16}$ =

[\begin{matrix} 0 & 256 256 & 0 \end{matrix}]

[\begin{matrix} 256 & 0 0 & 256 \end{matrix}]

[\begin{matrix} - 16 & 0 0 & - 16 \end{matrix}]

[\begin{matrix} 0 & 16 16 & 0 \end{matrix}]

SOLUTION

Solution : B

$A^{2} = [\begin{matrix} 0 & - 2 2 & 0 \end{matrix}], A^{4} = [\begin{matrix} - 4 & 0 0 & - 4 \end{matrix}] A^{8} = [\begin{matrix} 16 & 0 0 & 16 \end{matrix}], A^{16} = [\begin{matrix} 256 & 0 0 & 256 \end{matrix}]$

Question 15

If A and B are two square matrices which are skew symmetric then $(A B)^{T}$ equals to

A. AB

B. BA

C. -AB

D. -BA

SOLUTION

Solution : B

Required Expression,
$(A B)^{T} = B^{T} . A^{T}$
= (-B) . (-A) (Since A and B are said to be skew symmetric)
= BA

$∴$ (b) is the right option.

Free Objective Test 02 Practice Test - 11th and 12th

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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