Free Matrices 03 Practice Test - 12th Grade - Commerce

Question 1

If $A = [\begin{matrix} c o s α & s i n α - s i n α & c o s α \end{matrix}]$ , then $A^{2} =$

[\begin{matrix} c o s 2 α & s i n 2 α s i n 2 α & c o s 2 α \end{matrix}]

[\begin{matrix} c o s 2 α & - s i n 2 α s i n 2 α & c o s 2 α \end{matrix}]

[\begin{matrix} c o s 2 α & s i n 2 α - s i n 2 α & c o s 2 α \end{matrix}]

[\begin{matrix} - c o s 2 α & s i n 2 α - s i n 2 α & - c o s 2 α \end{matrix}]

SOLUTION

Solution : C

Since $A^{2} = A . A = [\begin{matrix} c o s α & s i n α - s i n α & c o s α \end{matrix}] [\begin{matrix} c o s α & s i n α - s i n α & c o s α \end{matrix}] = [\begin{matrix} c o s 2 α & s i n 2 α - s i n 2 α & c o s 2 α \end{matrix}] .$

Question 2

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 - 2 & 3 & - 1 3 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦$ and I is a unit matrix of $3^{r d}$ order, then
$(A^{2} + 9 I)$ equals

A. 2A

B. 4A

C. 6A

D. None of these

SOLUTION

Solution : D

$A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 - 2 & 3 & - 1 3 & 1 & 2 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow A . A = A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 6 & 11 & 7 - 11 & 4 & - 11 7 & 11 & 12 \end{matrix} ⎤ ⎥ ⎦,$
$I = ⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then, $A^{2} + 9 I = ⎡ ⎢ ⎣ \begin{matrix} 15 & 11 & 7 - 11 & 13 & - 11 7 & 11 & 21 \end{matrix} ⎤ ⎥ ⎦ .$

Question 3

If $A = [\begin{matrix} 4 & 2 - 1 & 1 \end{matrix}]$ and I is the identity matrix of order 2, then (A - 2I)(A - 3I) =

I

0

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

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

SOLUTION

Solution : B

$(A - 2 I) (A - 3 I) = [\begin{matrix} 2 & 2 - 1 & - 1 \end{matrix}] [\begin{matrix} 1 & 2 - 1 & - 2 \end{matrix}] = [\begin{matrix} 0 & 0 0 & 0 \end{matrix}] = 0$

Question 4

If $A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ , then

A^{3} + 3 A^{2} + A - 9 I_{3} = 0

A^{3} - 3 A^{2} + A + 9 I_{3} = 0

A^{3} + 3 A^{2} - A + 9 I_{3} = 0

A^{3} - 3 A^{2} - A + 9 I_{3} = 0

SOLUTION

Solution : D

$⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ A^{2} = A . A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ = ⎡ ⎢ ⎣ \begin{matrix} 4 & 3 & 0 - 3 & 2 & - 2 6 & 4 & 5 \end{matrix} ⎤ ⎥ ⎦ A . A^{2} = ⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 1 0 & 1 & - 1 3 & - 1 & 1 \end{matrix} ⎤ ⎥ ⎦ ⎡ ⎢ ⎣ \begin{matrix} 4 & 3 & 0 - 9 & - 2 & - 7 21 & 11 & 7 \end{matrix} ⎤ ⎥ ⎦ \Rightarrow A^{3} - 3 A^{2} - A + 9 I_{3} = 0$

Question 5

If 3X + 2Y = I and 2X - Y = O, where I and O are unit and null matrices of order 3 respectively, then

X = (\frac{1}{7}), Y = (\frac{2}{7})

X = (\frac{2}{7}), Y = (\frac{1}{7})

X = (\frac{1}{7}) I, Y = (\frac{2}{7}) I

X = (\frac{2}{7}) I, Y = (\frac{1}{7}) I

SOLUTION

Solution : C

$\begin{matrix} 3 X + 2 Y = I 2 X - Y = 0 \end{matrix} \Rightarrow \begin{matrix} 3 X + 2 Y = I 4 X - 2 Y = 0 \end{matrix} \Rightarrow \begin{matrix} 7 X = I X = \frac{1}{7} I \end{matrix}$
(Solving simultaneously)
Therefore from (i), $2 Y = I - \frac{3}{7} I = \frac{4}{7} I \Rightarrow Y = \frac{2}{7} I$

Question 6

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 7

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 8

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 9

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 10

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 Matrices 03 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

Related ExamsDefinite Integrals and Areas 03Relations and Functions 03Integrals 03

Related Exams
Definite Integrals and Areas 03
Relations and Functions 03
Integrals 03