Free Matrices 02 Practice Test - 12th Grade - Commerce

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 = [3 1 2]. The expression which is not defined is

A. B'B

B. CAB

C. 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 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 $A^{- 1} B^{- 1}$ .

Free Matrices 02 Practice Test - 12th Grade - Commerce

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

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