Page 16 - Exemplar Solution - Updates, Questions & Discussions on Exemplar Solution at Shiksha

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

The matrix $[\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}]$ is a

(A) Diagonal matrix

(B) Symmetric matrix

(C) Skew symmetric matrix

(D) Scalar matrix

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t A = [\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}] \\ A^{'} = [\begin{matrix} 0 & 5 & - 8 \\ - 5 & 0 & - 12 \\ 8 & 12 & 0 \end{matrix}] \\ \Rightarrow A^{'} = - [\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}] = - A \\ A^{'} = - A, s o A i s a s k e w s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

Upvote

Exemplar Solution Matrices

The matrix $[\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}]$ is a

(A) Identity matrix

(B) Symmetric matrix

(C) Skew symmetric matrix

(D) None of these

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}] \\ A^{'} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}] = A \\ A^{'} = A, s o A i s a s y m m e t r i c m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Upvote

Exemplar Solution Matrices

If matrix $A = [a_{i j}]_{2 \times 2}$ , where $a_{i j} = 1$ if $i \neq j$ if $i = j$ ,then $A^{2}$ is equal to

(A) I

(B) A

(C) 0

(D) None of these

9 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = {[a_{i j}]}_{2 \times 2} \\ L e t A = {[\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}]}_{2 \times 2} \\ a_{11} = 0 [? i = j] \\ a_{12} = 1 [? i \neq j] \\ a_{21} = 1 [? i \neq j] \\ a_{22} = 0 [? i = j] \\ ∴ A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , then $A^{2}$ is equal to

(A) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

(B) $[\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}]$

(C) $[\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}]$

(D) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] \\ A^{2} = A . A = [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 0 + 1 & 0 + 0 \\ 0 + 0 & 1 + 0 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $A$ and $B$ are two matrices of the order $3 \times m$ and $3 \times n$ , respectively, and $m = n$ , then the order of the matrix $(5 A - 2 B)$ is

(A) $m \times 3$

(B) $3 \times 3$

(C) $m \times n$

(D) $3 \times n$

5 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} A s w e k n o w t h a t t h e a d d i t i o n a n d s u b t r a c t i o n o f t w o m a t r i c e s i s o n l y p o s s i b l e w h e n t h e y h a v e \\ s a m e o r d e r . I t i s a l s o g i v e n t h a t m = n . \\ ∴ O r d e r o f (5 A - 2 B) i s 3 \times n . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $A = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}], B = \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (x π) \end{matrix}],$ then $A - B$ is equal to:

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : A = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}] \\ a n d B = \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (π x) \end{matrix}] \\ A - B = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) \end{matrix}] - \frac{1}{π} [\begin{matrix} - c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) & - t a n^{- 1} (π x) \end{matrix}] \\ = \frac{1}{π} [\begin{matrix} s i n^{- 1} (x π) + c o s^{- 1} (x π) & t a n^{- 1} (\frac{x}{π}) - t a n^{- 1} (\frac{x}{π}) \\ s i n^{- 1} (\frac{x}{π}) - s i n^{- 1} (\frac{x}{π}) & c o t^{- 1} (π x) + t a n^{- 1} (π x) \end{matrix}] \\ = \frac{1}{π} [\begin{matrix} \frac{π}{2} & 0 \\ 0 & \frac{π}{2} \end{matrix}] [\begin{array}{l} ? s i n^{- 1} x + c o s^{- 1} x = \frac{π}{2} \\ t a n^{- 1} x + c o t^{- 1} x = \frac{π}{2} \end{array}] \\ = \frac{1}{π} \times \frac{π}{2} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = \frac{1}{2} I \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

If $[\begin{matrix} 2 x + y & 4 x \\ 5 x - 7 & 4 x \end{matrix}] = [\begin{matrix} 7 & 7 y - 13 \\ y & x + 6 \end{matrix}],$ then the value of $x + y$ is

(A) $x = 3, y = 1$

(B) $x = 2, y = 3$

(C) $x = 2, y = 4$

(D) $x = 3, y = 3$

2 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : [\begin{matrix} 2 x + y & 4 x \\ 5 x - 7 & 4 x \end{matrix}] = [\begin{matrix} 7 & 7 y - 13 \\ y & x + 6 \end{matrix}] \\ E q u a t i n g t h e c o r r e s p o n d i n g e l e m e n t s, w e g e t, \\ 2 x + y = 7 \dots (i) \\ a n d 4 x = 7 y - 13 \dots (i i) \\ f r o m e q n . (i i) 4 x - x = 6 \\ 3 x = 6 \\ ∴ x = 2 \\ f r o m e q n . (i) 2 \times 2 + y = 7 \\ 4 + y = 7 ∴ y = 7 - 4 = 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

Total number of possible matrices of order $3 \times 3$ with each entry 2 or 0 is

(A) 9

(B) 27

(C) 81

(D) 512

4 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f p o s s i b l e m a t r i c e s o f o r d e r 3 \times 3 w i t h e a c h e n t r y 0 o r 2 = 2^{3 \times 3} = 2^{9} = 512 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Three

The matrix $P = [\begin{matrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{matrix}]$ is a

(A) Square matrix

(B) Diagonal matrix

(C) Unit matrix

(D) None

3 Views|Posted a year ago

Asked by Payal Gupta

S

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t A = [\begin{matrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{matrix}] \\ H e r e, n u m b e r o f c o l u m n s a n d t h e n u m b e r o f r o w s a r e e q u a l i . e ., 3 . S o, A i s a s q u a r e m a t r i x . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Upvote

Exemplar Solution Maths NCERT Exemplar Solutions Class 11th Chapter Five

(i) For any two complex numbers <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>,</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> and any real numbers <math><mrow><mi>a</mi><mo>,</mo><mi>b</mi></mrow></math> , <math> <mrow> <mo>?</mo> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>?</mo> <mo>−</mo> <mo>?</mo> <msup> <mrow> <mi>b</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msup> <mrow> <mi>a</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <msub> <mrow> <mi>z</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>?</mo> <mo>=</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> <mo>.</mo> </mrow> </math> (ii) The value of <math><mrow><mroot><mrow><mo>−</mo><mn>2</mn><mn>5</mn></mrow><mrow></mrow></mroot><mo>×</mo><mroot><mrow><mo>−</mo><mn>9</mn></mrow><mrow></mrow></mroot></mrow></math> is . (iii) The number <math><mrow><mfrac><mrow><msup><mrow><mrow><mo>(</mo><mrow><mn>1</mn><mo>−</mo><mi>i</mi></mrow><mo>)</mo></mrow></mrow><mrow><mn>3</mn></mrow></msup></mrow><mrow><mn>1</mn><mo>−</mo><mi>i</mi></mrow></mfrac></mrow></math> is equal to . (iv) The sum of the series <math><mrow><mi>i</mi><mo>+</mo><msup><mrow><mi>i</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>+</mo><msup><mrow><mi>i</mi></mrow><mrow><mn>3</mn></mrow></msup><mo>+</mo><mo>…</mo></mrow></math> up to 1000 terms is . (v) Multiplicative inverse of <math><mrow><mn>1</mn><mo>+</mo><mi>i</mi></mrow></math> is . (vi) If <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></math> and <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> are complex numbers such that <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>1</mn></mrow></msub><mo>+</mo><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub></mrow></math> is a real number, then <math><mrow><msub><mrow><mi>z</mi></mrow><mrow><mn>2</mn></mrow></msub><mo>=</mo><mo>.</mo><mo>.</mo><mo>.</mo><mo>.</mo></mrow></math> (vii)  <math><mrow><mrow><mo>(</mo><mrow><mi>z</mi></mrow><mo>)</mo></mrow><mo>+</mo><mi>z</mi><mrow><mo>(</mo><mrow><mi>z</mi><mo>≠</mo><mn>0</mn></mrow><mo>)</mo></mrow></mrow></math> is . (viii) If  <math> <mrow> <mo>?</mo> <mi>z</mi> <mo>+</mo> <mn>4</mn> <mo>?</mo> <mo>≤</mo> <mn>3</mn> </mrow> </math> , then the greatest and least values of <math> <mrow> <mo>?</mo> <mi>z</mi> <mo>+</mo> <mn>1</mn> <mo>?</mo> </mrow> </math> are . and . (ix) If <math><mrow><mfrac><mrow><mi>z</mi><mo>−</mo><mn>2</mn></mrow><mrow><mi>z</mi><mo>+</mo><mn>6</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac><mo>,</mo></mrow></math> then the locus of <math><mrow><mi>z</mi></mrow></math> is . (x) If <math> <mrow> <mo>?</mo> <mi>z</mi> <mo>?</mo> <mo>=</mo> <mn>4</mn> </mrow> </math> and <math><mrow><mrow><mo>(</mo><mrow><mi>z</mi></mrow><mo>)</mo></mrow><mo>=</mo><mfrac><mrow><mn>5</mn><mi>π</mi></mrow><mrow><mn>6</mn></mrow></mfrac></mrow></math> , then <math><mrow><mi>z</mi><mo>=</mo><mo>…</mo><mo>…</mo><mo>…</mo><mo>…</mo></mrow></math>

4 Views|Posted a year ago

Asked by Payal Gupta

Exemplar Solution

Questions

Discussions

Active Users

Followers

The matrix $[\begin{matrix} 0 & - 5 & 8 \\ 5 & 0 & 12 \\ - 8 & - 12 & 0 \end{matrix}]$ is a

(A) Diagonal matrix

(B) Symmetric matrix

(C) Skew symmetric matrix

(D) Scalar matrix

The matrix $[\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 4 \end{matrix}]$ is a

(A) Identity matrix

(B) Symmetric matrix

(C) Skew symmetric matrix

(D) None of these

If matrix $A = [a_{i j}]_{2 \times 2}$ , where $a_{i j} = 1$ if $i \neq j$ if $i = j$ ,then $A^{2}$ is equal to

(A) I

(B) A

(C) 0

(D) None of these

If $A = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ , then $A^{2}$ is equal to

(A) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

(B) $[\begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix}]$

(C) $[\begin{matrix} 0 & 1 \\ 0 & 1 \end{matrix}]$

(D) $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

If $A$ and $B$ are two matrices of the order $3 \times m$ and $3 \times n$ , respectively, and $m = n$ , then the order of the matrix $(5 A - 2 B)$ is

(A) $m \times 3$

(B) $3 \times 3$

(C) $m \times n$

(D) $3 \times n$

If $[\begin{matrix} 2 x + y & 4 x \\ 5 x - 7 & 4 x \end{matrix}] = [\begin{matrix} 7 & 7 y - 13 \\ y & x + 6 \end{matrix}],$ then the value of $x + y$ is

(A) $x = 3, y = 1$

(B) $x = 2, y = 3$

(C) $x = 2, y = 4$

(D) $x = 3, y = 3$

Total number of possible matrices of order $3 \times 3$ with each entry 2 or 0 is

(A) 9

(B) 27

(C) 81

(D) 512

The matrix $P = [\begin{matrix} 0 & 0 & 4 \\ 0 & 4 & 0 \\ 4 & 0 & 0 \end{matrix}]$ is a

(A) Square matrix

(B) Diagonal matrix

(C) Unit matrix

(D) None