Quantitative Aptitude PYQs

Concept:

Let z = x + iy be a complex number.

• Modulus of z = \(\left| {\rm{z}} \right| = {\rm{}}\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} = {\rm{}}\sqrt {{\rm{Re}}{{\left( {\rm{z}} \right)}^2} + {\rm{Im\;}}{{\left( {\rm{z}} \right)}^2}}\)
• arg (z) = arg (x + iy) = \({\tan ^{ - 1}}\left( {\frac{{\rm{y}}}{{\rm{x}}}} \right)\)

 

Calculation:

Given that,

\(\frac{{1 + 2{\rm{i}}}}{{1 - {{\left( {1 - {\rm{i}}} \right)}^2}}}\)

⇒ (1 - i)² = 1² + i² - 2.1.i      ...[∵ (a- b)² = a² + b²- 2ab]

⇒ (1 - i)² = 1 – 1 - 2i

⇒ (1 - i)² = - 2i

So,

⇒ 1 - (1 - i)² = 1 + 2i

Put this value in main equation

\(\Rightarrow \frac{{1 + 2{\rm{i}}}}{{1 + 2{\rm{i}}}} = 1\) = 1 + 0 i

So,

Modulus = \(\sqrt {{1^2} + {0^2}} \) = 1

Argument = \({\tan ^{ - 1}}\left( {\frac{0}{1}} \right) = 0\)

Concept:

• Find the coordinates of the points of intersection with the x axis, we must solve the equation f(x) = 0 or we can say that to finding roots of quadratic equation we must calculate the intersection point with x axis.
• The points where the graph of a polynomial cuts the x-axis are the zeros of the polynomial
• If the graph of a polynomial doesn't cut or touches the x- axis it means that the polynomial has no real root i.e it has complex roots.

 

Calculation:

Given that the graph of a quadratic polynomial lies entirely above x-axis so it never cuts the x axis so quadratic equation does not have any real roots or we can say it does have complex roots.

Concept:

Let us consider the standard form of a quadratic equation, ax² + bx + c =0 Where a, b and c are constants with a ≠ 0

Let α and β be the two roots of the above quadratic equation. 

The sum of the roots of a quadratic equation are: \({\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

The product of the roots is given by: \({\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

• \(\cot \left( {{\rm{a}} + {\rm{b}}} \right) = \frac{{\cot {\rm{\;a}}.\cot {\rm{\;b\;}} - 1}}{{\cot {\rm{\;a}} + \cot {\rm{\;b}}}}\)

 

Calculation:

Given that, cot α and cot β are the roots of the equation x² + bx + c = 0 so,

⇒ cot α + cot β = - b/1 = - b     ...1)

⇒ cot α. cot β = c/1 = c     ...2)

value of cot (α + β)

\(\Rightarrow \cot \left( {{\rm{\alpha }} + {\rm{\beta }}} \right) = \frac{{\cot {\rm{\;\alpha }}.\cot {\rm{\;\beta }} - 1}}{{\cot {\rm{\;\alpha }} + \cot {\rm{\;\beta }}}}\)     ...3) [\(\because\cot \left( {{\rm{a}} + {\rm{b}}} \right) = \frac{{\cot {\rm{a}}.\cot {\rm{b}} - 1}}{{\cot {\rm{a}} + \cot {\rm{b}}}}\)]

From equation 1, 2 and 3

\(\Rightarrow \cot \left( {{\rm{\alpha }} + {\rm{\beta }}} \right) = \frac{{{\rm{c}} - 1}}{{ - {\rm{b}}}}\)

\(\Rightarrow \cot \left( {{\rm{\alpha }} + {\rm{\beta }}} \right) = \frac{{1 - {\rm{c}}}}{{\rm{b}}}\)

Concept:

1. Let us consider the standard form of a quadratic equation, ax² + bx + c = 0 Where a, b and c are constants with a ≠ 0

Let α and β be the two roots of the above quadratic equation.

 

• The sum of the roots of a quadratic equation are:

\({\rm{\alpha }} + {\rm{\beta }} = - \frac{{\rm{b}}}{{\rm{a}}} = - \frac{{{\rm{coefficient\;of\;x}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

• The product of the roots is given by:

\({\rm{\alpha \beta }} = \frac{{\rm{c}}}{{\rm{a}}} = \frac{{{\rm{constant\;term}}}}{{{\rm{coefficient\;of\;}}{{\rm{x}}^2}}}\)

1. If a, b, c are in A.P ⇔ b = \(\frac{{{\rm{a\;}} + {\rm{\;c}}}}{2}\)

 

Calculation:

Given that,

Let’s take α, β are roots of ax² + x + c = 0 so,

\(\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = {\rm{}} - \frac{1}{{\rm{a}}}\)     ...1)

\(\Rightarrow {\rm{\alpha }}.{\rm{\beta }} = \frac{{\rm{c}}}{{\rm{a}}}\)     ...2)

Given

\(\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = \frac{1}{{{{\rm{\alpha }}^2}}} + \frac{1}{{{{\rm{\beta }}^2}}}\)

\(\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = \frac{{{{\rm{\alpha }}^2} + {{\rm{\beta }}^2}}}{{{{\rm{\alpha }}^2}.{{\rm{\beta }}^2}}}\)

\(\Rightarrow {\rm{\alpha }} + {\rm{\beta }} = \frac{{{{\left( {{\rm{\alpha }} + {\rm{\beta }}} \right)}^2} - 2{\rm{\alpha \beta }}}}{{{{\left( {{\rm{\alpha \beta }}} \right)}^2}}}\)     ...[∵ (a+ b)² = a² + b²+ 2ab]

Put all values,

\(\Rightarrow - \frac{1}{{\rm{a}}} = {\rm{}}\frac{{{{\left( { - \frac{1}{{\rm{a}}}} \right)}^2} - \frac{{2{\rm{c}}}}{{\rm{a}}}}}{{{{\left( {\frac{{\rm{c}}}{{\rm{a}}}} \right)}^2}}}\)

\(\Rightarrow - \frac{1}{{\rm{a}}} = {\rm{\;}}\frac{{1 - 2{\rm{ac}}}}{{{{\rm{c}}^2}}}\)

Now, Cross multiply

\({{\rm{c}}^2} = {\rm{}} - {\rm{a}} + 2{{\rm{a}}^2}{\rm{c}}\)

\(\Rightarrow {{\rm{c}}^2} + {\rm{a}} = 2{{\rm{a}}^2}{\rm{c}}\)

\( \Rightarrow {{\rm{a}}^2}{\rm{c}} = \frac{{{{\rm{c}}^2} + {\rm{a}}}}{2}\)     ...[∵ a, b, c are in A.P ⇔ b = (a + c)/2]

So we can say that a, ca², c² are in AP.

Concept:

• C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) ………….. C(n, n) = 2ⁿ
• C(n, 0) = C(n, n) = 1
• C(n, r) = \(^{\rm{n}}{{\rm{C}}_{\rm{r}}} = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n}} - {\rm{r}}} \right)!}}\)

 

Calculation:

Given that,

⇒

C(7, 0) + C(7, 1) + C(7, 1) + C(7, 2) + C(7, 2) … + C(7, 6) + C(7, 7)

⇒ C(7, 0) + 2.[ C(7, 1) + C(7, 2) + … + C(7, 6)] + C(7, 7)

Add and subtract C(7, 0),C(7, 7) so,

⇒

C(7, 0) + 2.[ C(7, 1) + C(7, 2) + … + C(7, 6)] + C(7, 0) + C(7, 7) + C(7, 0) + C(7, 7) - C(7, 0) - C(7, 7)

⇒ 2[ C(7, 0) + C(7, 1) + C(7, 2) + … + C(7, 6) + C(7, 7)] – C(7,0) – C(7, 7)

⇒

2[2⁷] - C(7,0) – C(7, 7)     ...[∵ C(n, 0) + C(n, 1) + C(n, 2) + C(n, 3) ………….. C(n, n) = 2ⁿ]

⇒

2 × 128 – 1 – 1

⇒ 254

Concept:

Combination: Selecting r objects from given n objects.

• The number of selections of r objects from the given n objects and arranging them is denoted by
• \(^{\rm{n}}{{\rm{C}}_{\rm{r}}} = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n}} - {\rm{r}}} \right)!}}\)

 

Calculation:

Given letters are A, E, I, T, Q, O, N, U

5 vowels and 3 consonant so,

First we arrange first and last letter of word

We have to choose only consonant so total ways are \(^3{{\rm{C}}_2}2!\) = 6

Now we have 6 places to choose and arrange 6 letters so,

\(^6{{\rm{C}}_6}6!\) = 720

So total 8 words letters are = 720 × 6 = 4320

Concept:

• Sum of AP = \({{\rm{s}}_{\rm{n}}} = \frac{{\rm{n}}}{2}\left( {{\rm{a}} + {\rm{l}}} \right)\) where a is first term and l is last term of AP.
• Nth term of AP = a_n = a + (n - 1)d where a first letter and d is difference between two consecutive terms.

 

Calculation:

Two digit odd numbers AP series = 11, 13, 15 ... 99

⇒ a = 11, a_n = 99 and d = 13 – 11 = 2

⇒ a_n = a + (n - 1)d

⇒ 99 = 11 + (n – 1) × 2

⇒ 88 = (n - 1) × 2

⇒ 44 = n – 1

⇒ n = 45

Now,

\(\Rightarrow {{\rm{s}}_{\rm{n}}} = \frac{{\rm{n}}}{2}\left( {{\rm{a}} + {\rm{l}}} \right)\)

\( \Rightarrow {{\rm{s}}_{\rm{n}}} = \frac{{45}}{2}\left( {11 + 99} \right)\)

\( \Rightarrow {{\rm{s}}_{\rm{n}}} = \frac{{110 \times 45}}{2}\)

⇒ s_n = 2475

Concept:

Covert Binary to Decimal:

For binary number with n digits;

D_n-1 …. D₃D₂D₁D₀

The decimal number is equal to the sum of binary digits (D_n) times their power of 2 (2ⁿ)

Decimal = D₀ × 2⁰ + D₁ × 2¹ + D₂ × 2² + ….

Calculation:

Given that,

(1p101)₂ + (10q1)₂ = (100r00)₂

 First we Change binary to decimal

⇒ (1p101)₂ = \(1 \times {2^4} + {\rm{p}} \times {2^3} + 1 \times {2^2} + 0 + 1 \times {2^0}\) = 21 + 8p     ...1)

⇒ (10q1)₂ = \(1 \times {2^3} + 0 + {\rm{q}} \times {2^1} + 1 \times {2^0}\) = 9 + 2q     ...2)

⇒ (100r00)₂ = \(1 \times {2^5} + 0 + 0 + {\rm{r}} \times {2^2} + 0 + 0\) = 4r + 32     ...3)

So,

⇒ 21 + 8p + 9 +2q = 4r + 32

⇒ 8p + 2q = 4r + 2

So from options p = 0, q = 1 and r = 0 will satisfy above equation.

Concept:

• If a set does not contain any element than it is called empty set.
• i = \(\sqrt { - 1}\)

 

Calculation:

Given that,

S = {x : x² + 1 = 0, x is real}

⇒ x² + 1 = 0

⇒ x² = - 1

⇒ x = ±√(-1)

⇒ x = ± i imaginary

But given that set only contain real value of x but after solving x² + 1 we will get only complex value so both are contradict to each other, So we can say that set does not have any value in it. it is an empty set.

Concept:

1. General term: General term in the expansion of (x + y) ⁿ is given by

• \({{\rm{T}}_{\left( {{\rm{r\;}} + {\rm{}}1} \right)}} = {\rm{}}{{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)

1. \(^{\rm{n}}{{\rm{C}}_{\rm{r}}} = \frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n}} - {\rm{r}}} \right)!}}\)
2. (n - r)! = (n - r) (n – r - 1)!

 

Calculation:

Given that, (x - y)ⁿ where n ≥ 5

And T₅ + T₆ = 0

\( \Rightarrow {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_4} \times {{\rm{x}}^{{\rm{n}} - 4}} \times {\left( { - {\rm{y}}} \right)^{4{\rm{\;}}}} + {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_5} \times {{\rm{x}}^{{\rm{n}} - 5}} \times {( - {\rm{y}})^5} = 0\)     ...[\({{\rm{T}}_{\left( {{\rm{r}} + 1} \right)}} = {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}} \times {{\rm{x}}^{{\rm{n}} - {\rm{r}}}} \times {{\rm{y}}^{\rm{r}}}\)]

\(\Rightarrow {{\rm{x}}^{{\rm{n}} - 4}}.{{\rm{y}}^4}\left[ {^{\rm{n}}{{\rm{C}}_4} - {{\rm{\;}}^{\rm{n}}}{{\rm{C}}_5} \times {{\rm{x}}^{ - 1}} \times {\rm{y}}} \right] = 0\)

\(\Rightarrow {}_{}^n{C_4} - {}_{}^n{C_5} \times {x^{ - 1}} \times y = 0\) = 0

\(\Rightarrow {}_{}^n{C_4} = {}_{}^n{C_5} \times {x^{ - 1}} \times y\)

\(\Rightarrow \frac{{{\rm{n}}!}}{{4!\left( {{\rm{n}} - {\rm{}}4} \right)!}} = {\rm{}}\frac{{{\rm{n}}!}}{{5!\left( {{\rm{n}} - {\rm{}}5} \right)!}}{\rm{}} \times \frac{{\rm{y}}}{{\rm{x}}}\)

\(\Rightarrow \frac{1}{{\left( {{\rm{n}} - 4} \right)\left( {{\rm{n}} - 5} \right)!}} = \frac{1}{{5 \times \left( {{\rm{n}} - 5} \right)!}} \times \frac{{\rm{y}}}{{\rm{x}}}\)     ...[∵ (n - r)! = (n - r) (n – r - 1)!]

\(\Rightarrow \frac{{\rm{x}}}{{\rm{y}}} = \frac{{{\rm{n}} - 4}}{5}\)

Concept:

1. If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}} \end{array}} \right]\) then determinant of A is given by: |A| = a₁₁ × a₂₂ – a₂₁ × a₁₂
2. |Aⁿ| = |A|ⁿ

Calculation:

Given that,

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm{\alpha }}&2\\ 2&{\rm{\alpha }} \end{array}} \right]\) and det{A³} = 125

⇒ |A| = α × α – 2 × 2

⇒ |A| = α² - 4     ...1)

Given |A³| = 125

⇒ |A|³ = 125     ...[∵ |Aⁿ| = |A|ⁿ]

⇒ |A| = (125)^1/3 = 5     ...2)

From equations 1 and 2

⇒ α² - 4 = 5

⇒ α² = 9

⇒ α = ± 3

Concept:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}&{{{\rm{a}}_{13}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}}&{{{\rm{a}}_{23}}}\\ {{{\rm{a}}_{31}}}&{{{\rm{a}}_{32}}}&{{{\rm{a}}_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}

Calculation:

\(\left| {\begin{array}{*{20}{c}} 0&{{\rm{x}} - {\rm{a}}}&{{\rm{x}} - {\rm{b}}}\\ {{\rm{x}} + {\rm{a}}}&0&{{\rm{x}} - {\rm{c}}}\\ {{\rm{x}} + {\rm{b}}}&{{\rm{x}} + {\rm{c}}}&0 \end{array}} \right| = 0\)

By solving it

⇒ 0 – (x – a) [0 - (x + b) (x - c)] + (x – b) [(x + a) (x + c) – 0] = 0

⇒ (x – a) (x + b) (x - c) + (x - b) (x + a) (x + c) = 0

So from options we can say that only one solution possible which is x = 0.

While putting x = 0

⇒ a. b. c – a. b. c = 0

Concept:

1. Transpose of a Matrix:

• The new matrix obtained by interchanging the rows and columns of the original matrix is called as the transpose of the matrix.
• It is denoted by A′or A^T

2. Multiplication of matrices:

• The number of columns of the 1^st matrix must equal the number of rows of the 2^nd matrix.
• The result will have the same number of rows as the 1^st matrix, and the same number of columns as the 2^nd matrix.
• To multiply an m × n matrix by an n × p matrix, the n must be the same, and the result is an m × p matrix.

 

Calculation:

Given: \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\alpha }}}&{\sin {\rm{\alpha }}}\\ { - \sin {\rm{\alpha }}}&{\cos {\rm{\alpha }}} \end{array}} \right]\)

\({{\rm{A}}^{\rm{T}}} = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\alpha }}}&{ - \sin {\rm{\alpha }}}\\ {\sin {\rm{\alpha }}}&{\cos {\rm{\alpha }}} \end{array}} \right]\)

Now,

\({\rm{A}}{{\rm{A}}^{\rm{T}}} = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\alpha }}}&{\sin {\rm{\alpha }}}\\ { - \sin {\rm{\alpha }}}&{\cos {\rm{\alpha }}} \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\alpha }}}&{ - \sin {\rm{\alpha }}}\\ {\sin {\rm{\alpha }}}&{\cos {\rm{\alpha }}} \end{array}} \right]\)

\( \Rightarrow {\rm{A}}{{\rm{A}}^{\rm{T}}} = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}{\rm{\alpha }} + {\rm{\;}}{{\sin }^2}{\rm{\alpha }}}&{ - \cos {\rm{\alpha }}\sin {\rm{\alpha }} + {\rm{\;}}\cos {\rm{\alpha }}\sin {\rm{\alpha }}}\\ { - \cos {\rm{\alpha }}\sin {\rm{\alpha }} + {\rm{\;}}\cos {\rm{\alpha }}\sin {\rm{\alpha }}}&{{{\sin }^2}{\rm{\alpha }} + {\rm{\;}}{{\cos }^2}{\rm{\alpha }}} \end{array}} \right]\)

\( \Rightarrow {\rm{A}}{{\rm{A}}^{\rm{T}}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right] = {{\rm{I}}_2}\)

So, AA^T is identity matrix.

Concept:

Let the system of equations be,

a₁x + b₁y + c₁z = d₁

a₂x + b₂y + c₂z = d₂

a₃x + b₃y + c₃z = d₃

\({\rm{\;}} \Rightarrow \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_1}}&{{{\rm{b}}_1}}&{{{\rm{c}}_1}}\\ {{{\rm{a}}_2}}&{{{\rm{b}}_2}}&{{{\rm{c}}_2}}\\ {{{\rm{a}}_3}}&{{{\rm{b}}_3}}&{{{\rm{c}}_3}} \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {{{\rm{d}}_1}}\\ {{{\rm{d}}_2}}\\ {{{\rm{d}}_3}} \end{array}} \right]\)

⇒ AX = B

∴ X = A^-1 B = \(\frac{{{\rm{adj\;}}\left( {\rm{A}} \right)}}{{\det {\rm{\;}}({\rm{A}})}}{\rm{\;B}}\)

• If det (A) ≠ 0, the system is consistent having unique solution.
• If det (A) = 0 and (adj A). B = 0, the system is consistent, with infinitely many solutions.
• If det (A) = 0 and (adj A). B ≠ 0, the system is inconsistent (no solution)

 

Calculation:

Given:

x + 2y + 3z = 1

2x + y + 3z = 2

5x + 5y + 9z = 4

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 1&2&3\\ 2&1&3\\ 5&5&9 \end{array}} \right]{\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1\\ 2\\ 4 \end{array}} \right] \Leftrightarrow {\rm{AX}} = {\rm{B}}\)

\({\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 2&1&3\\ 5&5&9 \end{array}} \right]\)

Det (A) = 1 (9 – 15) – 2 (18 – 15) + 3 (10 – 5)

Det (A) = -6 – 6 + 15 = 3 ≠ 0

∴ The system is consistent having a unique solution.

Concept:

Multiplication of matrices:

• The number of columns of the 1^st matrix must equal the number of rows of the 2^nd matrix.
• The result will have the same number of rows as the 1^st matrix, and the same number of columns as the 2^nd matrix.
• To multiply an m × n matrix by an n × p matrix, the n must be the same, and the result is an m × p matrix.

 

Calculation:

Given: AB = C

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} {{\rm{x}} + {\rm{y}}}&{\rm{y}}\\ {\rm{x}}&{{\rm{x}} - {\rm{y}}} \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} 3\\ { - 2} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 4\\ { - 2} \end{array}} \right]\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} {3{\rm{x}} + 3{\rm{y}} - 2{\rm{y}}}\\ {3{\rm{x}} - 2{\rm{x}} + 2{\rm{y}}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 4\\ { - 2} \end{array}} \right]\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} {3{\rm{x}} + {\rm{y}}}\\ {{\rm{x}} + 2{\rm{y}}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 4\\ { - 2} \end{array}} \right]\)

So, 3x + y = 4   ….

(1) x + 2y = -2       …. (2) Solving equation 1 and 2, we get

x = 2 and y = -2

Now,

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{\rm{x}} + {\rm{y}}}&{\rm{y}}\\ {\rm{x}}&{{\rm{x}} - {\rm{y}}} \end{array}} \right] \)

\(= {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {2 - 2}&{ - 2}\\ 2&{2 - \left( { - 2} \right)} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 0&{ - 2}\\ 2&4 \end{array}} \right]\)

\({{\rm{A}}^2} = {\rm{A}} \times {\rm{A}} \)

\(= {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 0&{ - 2}\\ 2&4 \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} 0&{ - 2}\\ 2&4 \end{array}} \right]\)

\( \Rightarrow {{\rm{A}}^2} = \left[ {\begin{array}{*{20}{c}} { - 4}&{ - 8}\\ 8&{12} \end{array}} \right]\)

Concept:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}&{{{\rm{a}}_{13}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}}&{{{\rm{a}}_{23}}}\\ {{{\rm{a}}_{31}}}&{{{\rm{a}}_{32}}}&{{{\rm{a}}_{33}}} \end{array}} \right]\) then determinant of A is given by:

• |A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}
• Elementary row or column transformations do not change the value of the determinant of a matrix.

 

Calculation:

Given:

\({\rm{Let\;\Delta }} = \left| {\begin{array}{*{20}{c}} 1&1&1\\ 1&{1 + {\rm{xyz}}}&1\\ 1&1&{1 + {\rm{xyz}}} \end{array}} \right|\)

Apply R₂ → R₂ – R₁ and R₃ → R₃ – R₁, we get

\(= \left| {\begin{array}{*{20}{c}} 1&1&1\\ 0&{{\rm{xyz}}}&0\\ 0&0&{{\rm{xyz}}} \end{array}} \right|\)

Expanding along R₁, we get

Δ = 1 (x²y²z² – 0) – 0 + 0

∴ Δ = x²y²z²

Concept:

If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {{{\rm{a}}_{11}}}&{{{\rm{a}}_{12}}}&{{{\rm{a}}_{13}}}\\ {{{\rm{a}}_{21}}}&{{{\rm{a}}_{22}}}&{{{\rm{a}}_{23}}}\\ {{{\rm{a}}_{31}}}&{{{\rm{a}}_{32}}}&{{{\rm{a}}_{33}}} \end{array}} \right]\) then determinant of A is given by:

|A| = a₁₁ × {(a₂₂ × a₃₃) – (a₂₃ × a₃₂)} - a₁₂ × {(a₂₁ × a₃₃) – (a₂₃ × a₃₁)} + a₁₃ × {(a₂₁ × a₃₂) – (a₂₂ × a₃₁)}

Calculation:

Given:

\(\left| {\begin{array}{*{20}{c}} {\rm{x}}&{\rm{y}}&0\\ 0&{\rm{x}}&{\rm{y}}\\ {\rm{y}}&0&{\rm{x}} \end{array}} \right| = 0\)

Expanding along R₁, we get

⇒ x (x² – 0) – y (0 – y²) + 0 = 0

⇒ x³ + y³ = 0

\(\Rightarrow {\rm{\;}}{{\rm{y}}^3}{\rm{\;}}\left( {\frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} + {\rm{\;}}1} \right){\rm{\;}} = {\rm{\;}}0{\rm{\;}}\)

\(\therefore \left( {\frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} + {\rm{\;}}1} \right) = {\rm{\;}}0{\rm{\;and\;}}{{\rm{y}}^3} \ne 0\)

\(\Rightarrow \frac{{{{\rm{x}}^3}}}{{{{\rm{y}}^3}}}{\rm{\;}} = {\rm{}} - 1\)

\(\Rightarrow \frac{{\rm{x}}}{{\rm{y}}} = {\left( { - 1} \right)^{\frac{1}{3}}}\)

Hence x/y is one of the cube roots of -1

Calculation:

Given: Set A of all matrices of order 3 × 3 with entries 0 or 1 only.

Let B be the subset of A consisting of all matrices whose determinant is 1.

⇒ B ∈ (0, 1) and |B| = 1

Assume B = \(\left| {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right|\)

Now, |B| = 1 (1 – 0) – 0 + 0

∴ |B| = 1

Let C be the subset of A consisting of all matrices whose determinant is -1.

⇒ C ∈ (0, 1) and |C| = -1

Assume C = \(\left| {\begin{array}{*{20}{c}} 1&0&0\\ 0&0&1\\ 0&1&0 \end{array}} \right|\)

Now, |C| = 1 (0 – 1) – 0 + 0

∴ |C| = -1

Here, number of elements in B and C are equal.

Concept:

• cos² θ – sin² θ = cos 2θ
• 2 sin θ cos θ = sin 2θ
• cos (x + y) = cos x cos y – sin x sin y
• sin (x + y) = sin x cos y + cos x sin y

Calculation:

Given:

\({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\)

\({{\rm{A}}^2} = {\rm{AA}} = \left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\)

\( \Rightarrow {{\rm{A}}^2} = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}{\rm{\theta }} - {\rm{\;}}{{\sin }^2}{\rm{\theta }}}&{\cos {\rm{\theta }}\sin {\rm{\theta }} + {\rm{\;}}\sin {\rm{\theta }}\cos {\rm{\theta }}}\\ { - \cos {\rm{\theta }}\sin {\rm{\theta }} - {\rm{\;}}\cos {\rm{\theta }}\sin {\rm{\theta }}}&{ - {{\sin }^2}{\rm{\theta }} + {\rm{\;}}{{\cos }^2}{\rm{\theta }}} \end{array}} \right]\)

\( \Rightarrow {{\rm{A}}^2} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos 2{\rm{\theta }}}&{\sin 2{\rm{\theta }}}\\ { - \sin 2{\rm{\theta }}}&{\cos 2{\rm{\theta }}} \end{array}} \right]\)

Now,

\(\Rightarrow {{\rm{A}}^3} = {{\rm{A}}^2}{\rm{A}} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos 2{\rm{\theta }}}&{\sin 2{\rm{\theta }}}\\ { - \sin 2{\rm{\theta }}}&{\cos 2{\rm{\theta }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\cos {\rm{\theta }}}&{\sin {\rm{\theta }}}\\ { - \sin {\rm{\theta }}}&{\cos {\rm{\theta }}} \end{array}} \right]\)

\(\Rightarrow {{\rm{A}}^3} = \left[ {\begin{array}{*{20}{c}} {\cos 2{\rm{\theta }}\cos {\rm{\theta }} - \sin 2{\rm{\theta }}\sin {\rm{\theta \;}}}&{\cos 2{\rm{\theta }}\sin {\rm{\theta }} + \sin 2{\rm{\theta }}\cos {\rm{\theta }}}\\ { - \sin 2{\rm{\theta }}\cos {\rm{\theta }} - {\rm{\;}}\cos 2{\rm{\theta }}\sin {\rm{\theta }}}&{ - \sin 2{\rm{\theta }}\sin {\rm{\theta }} + \cos 2{\rm{\theta }}\cos {\rm{\theta }}} \end{array}} \right]{\rm{\;}}\)

\(\Rightarrow {{\rm{A}}^3} = \left[ {\begin{array}{*{20}{c}} {\cos 2{\rm{\theta }}\cos {\rm{\theta }} - \sin 2{\rm{\theta }}\sin {\rm{\theta \;}}}&{\cos 2{\rm{\theta }}\sin {\rm{\theta }} + \sin 2{\rm{\theta }}\cos {\rm{\theta }}}\\ { - \left( {\cos 2{\rm{\theta }}\sin {\rm{\theta }} + \sin 2{\rm{\theta }}\cos {\rm{\theta }}} \right)}&{\cos 2{\rm{\theta }}\cos {\rm{\theta }} - \sin 2{\rm{\theta }}\sin {\rm{\theta }}} \end{array}} \right]\)

\(\Rightarrow {{\rm{A}}^3} = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos (2{\rm{\theta }} + {\rm{\theta }}){\rm{\;}}}&{\sin (2{\rm{\theta }} + {\rm{\theta }}){\rm{\;}}}\\ { - \sin (2{\rm{\theta }} + {\rm{\theta }})}&{\cos (2{\rm{\theta }} + {\rm{\theta }}){\rm{\;}}} \end{array}} \right] = {\rm{\;}}\left[ {\begin{array}{*{20}{c}} {\cos 3{\rm{\theta }}}&{\sin 3{\rm{\theta }}}\\ { - \sin 3{\rm{\theta }}}&{\cos 3{\rm{\theta }}} \end{array}} \right]{\rm{\;}}\)

∴ Option 1^st is correct answer.

Concept:

• Order of a matrix is determined by the number of rows and columns the matrix consists.
• To multiply an m × n matrix by an n × p matrix, the n must be the same, and the result is an m × p matrix.

Calculation:

We have to find the order of \(\left[ {{\rm{x\;\;y\;\;z}}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right]\)

Let \(\left[ {{\rm{x\;\;y\;\;z}}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{a}}&{\rm{h}}&{\rm{g}}\\ {\rm{h}}&{\rm{b}}&{\rm{f}}\\ {\rm{g}}&{\rm{f}}&{\rm{c}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\rm{x}}\\ {\rm{y}}\\ {\rm{z}} \end{array}} \right] = {\rm{ABC}}\)

Order of matrix A is (1 × 3)

Order of matrix B is (3 × 3)

Order of matrix C is (3 × 1)

Now, the Order of AB: A _{(1 × 3)} B _{(3 × 3)} is (1 × 3)

∴ Order of ABC = Order of (AB) _{(1 × 3)} C _{(3 × 1)} = (1 × 1)

So, Option 2^nd is the correct answer.