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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If the coordinates of the vertices of an equilateral triangle with sides of length ‘a’ are $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ , then

${[\begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix}]}^{2} = \frac{3 a^{2}}{4}$

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If $A + B + C = 0$ , then prove that $[\begin{matrix} 1 & c o s C & c o s B \\ c o s C & 1 & c o s A \\ c o s B & c o s A & 1 \end{matrix}] = 0$

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L . H . S . = | \begin{matrix} 1 & c o s C & c o s B \\ c o s C & 1 & c o s A \\ c o s B & c o s A & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} \\ = 1 | \begin{matrix} 1 & c o s A \\ c o s A & 1 \end{matrix} | - c o s C | \begin{matrix} c o s C & c o s B \\ c o s A & 1 \end{matrix} | + c o s B | \begin{matrix} c o s C & c o s B \\ 1 & c o s A \end{matrix} | \\ = 1 (1 - c o s^{2} A) - c o s C (c o s C - c o s A c o s B) + c o s B (c o s A c o s C - c o s B) \\ = s i n^{2} A - c o s^{2} C + c o s A c o s B c o s C + c o s A c o s B c o s C - c o s^{2} B \\ = s i n^{2} A - c o s^{2} B - c o s^{2} C + 2 c o s A c o s B c o s C \\ = - c o s (A + B) . c o s (A - B) - c o s^{2} C + 2 c o s A c o s B c o s C [? s i n^{2} A - c o s^{2} B = - c o s (A + B) . c o s (A - B)] \\ = - c o s (- C) . c o s (A - B) + c o s C + (2 c o s A c o s B - c o s C) [? A + B + C = 0] \\ = - c o s C (c o s A c o s B + s i n A s i n B) + c o s C + (2 c o s A c o s B - c o s C) \\ = - c o s C (c o s A c o s B + s i n A s i n B - 2 c o s A c o s B + c o s C) \\ = - c o s C (- c o s A c o s B + s i n A s i n B + c o s C) \\ = c o s C (c o s A c o s B - s i n A s i n B - c o s C) \\ = c o s C [c o s (A + B) - c o s C] \\ = c o s C [c o s (- C) - c o s C] [? A + B = - C] \\ = c o s C [c o s C - c o s C] = c o s C . 0 = 0 R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}]$ = (a – 1)3

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L . H . S . = | \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = | \begin{matrix} a^{2} - 1 & a - 1 & 0 \\ 2 a - 2 & a - 1 & 0 \\ 3 & 3 & 1 \end{matrix} | = | \begin{matrix} (a + 1) (a - 1) & a - 1 & 0 \\ 2 (a - 1) & a - 1 & 0 \\ 3 & 3 & 1 \end{matrix} | \\ T a k i n g (a - 1) c o m m o n f r o m C_{1} a n d C_{2} \\ = (a - 1) (a - 1) | \begin{matrix} a + 1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{3} \\ = {(a - 1)}^{2} [1 | \begin{matrix} a + 1 & 1 \\ 2 & 1 \end{matrix} |] \\ = {(a - 1)}^{2} (a + 1 - 2) = {(a - 1)}^{2} (a - 1) = {(a - 1)}^{3} R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix}]$ = 4 $x y z$

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L . H . S . = | \begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix} | \\ C_{1} \to C_{1} - (C_{2} + C_{3}) \\ = | \begin{matrix} 0 & z & y \\ - 2 x & z + x & x \\ - 2 x & x & x + y \end{matrix} | \\ T a k i n g - 2 c o m m o n f r o m C_{1} \\ = - 2 | \begin{matrix} 0 & z & y \\ x & z + x & x \\ x & x & x + y \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} \\ = - 2 [x | - z y - z y |] = - 2 (- x y z) = 4 x y z R . H . S . \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{matrix}]$ =0

7 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L . H . S . = | \begin{matrix} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{matrix} | \\ T a k i n g R_{1} \to x R_{1}, R_{2} \to y R_{2}, R_{3} \to z R_{3} a n d d i v i d i n g t h e determinant b y x y z . \\ = \frac{1}{x y z} | \begin{matrix} x y^{2} z^{2} & x y z & x y + z x \\ y z^{2} x^{2} & y z x & y z + x y \\ z x^{2} y^{2} & z x y & z x + z y \end{matrix} | \\ T a k i n g x y z c o m m o n f r o m C_{1} a n d C_{2} \\ = \frac{x y z . x y z}{x y z} | \begin{matrix} y z & 1 & x y + z x \\ z x & 1 & y z + x y \\ x y & 1 & z x + z y \end{matrix} | \\ C_{3} \to C_{3} + C_{1} \\ = x y z | \begin{matrix} y z & 1 & x y + y z + z x \\ z x & 1 & x y + y z + z x \\ x y & 1 & x y + y z + z x \end{matrix} | \\ T a k i n g (x y + y z + z x) c o m m o n f r o m C_{3} \\ = (x y z) (x y + y z + z x) | \begin{matrix} y z & 1 & 1 \\ z x & 1 & 1 \\ x y & 1 & 1 \end{matrix} | \\ = (x y z) (x y + y z + z x) | \begin{matrix} y z & 1 & 1 \\ z x & 1 & 1 \\ x y & 1 & 1 \end{matrix} | = 0 [? C_{2} a n d C_{3} a r e i d e n t i c a l] \\ L . H . S . = R . H . S . H e n c e p r o v e d . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix}]$

8 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ = | \begin{matrix} a + b + c & a + b + c & a + b + c \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | \\ T a k i n g (a + b + c) c o m m o n f r o m R_{1} \\ = (a + b + c) | \begin{matrix} 1 & 1 & 1 \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix} | \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ = (a + b + c) | \begin{matrix} 0 & 0 & 1 \\ b + c + a & - (b + c + a) & 2 b \\ 0 & a + b + c & c - a - b \end{matrix} | \\ T a k i n g (b + c + a) c o m m o n f r o m C_{1} a n d C_{2} \\ = {(a + b + c)}^{3} | \begin{matrix} 0 & 0 & 1 \\ 1 & - 1 & 2 b \\ 0 & 1 & c - a - b \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = {(a + b + c)}^{3} [1 | \begin{matrix} 1 & - 1 \\ 0 & 1 \end{matrix} |] \\ = {(a + b + c)}^{3} . \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix}]$

11 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{matrix} 3 x + 4 & x & x \\ 3 x + 4 & x + 4 & x \\ 3 x + 4 & x & x + 4 \end{matrix} | \\ T a k i n g (3 x + 4) c o m m o n f r o m C_{1} \\ = (3 x + 4) | \begin{matrix} 1 & x & x \\ 1 & x + 4 & x \\ 1 & x & x + 4 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = (3 x + 4) | \begin{matrix} 0 & - 4 & 0 \\ 0 & 4 & - 4 \\ 1 & x & x + 4 \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} = (3 x + 4) [1 | \begin{matrix} - 4 & 0 \\ 4 & - 4 \end{matrix} |] \\ = (3 x + 4) (16 - 0) = 16 (3 x + 4) \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{matrix}]$

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Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{matrix} x + y + z & - x + y & - x + z \\ x + y + z & 3 y & z - y \\ x + y + z & y - z & 3 z \end{matrix} | \\ T a k i n g (x + y + z) c o m m o n f r o m C_{1} \\ = (x + y + z) | \begin{matrix} 1 & - x + y & - x + z \\ 1 & 3 y & z - y \\ 1 & y - z & 3 z \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = (x + y + z) | \begin{matrix} 0 & - x - 2 y & - x + y \\ 0 & 2 y + z & - y - 2 z \\ 0 & y - z & 3 z \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} = (x + y + z) [1 | \begin{matrix} - x - 2 y & - x + y \\ 2 y + z & - y - 2 z \end{matrix} |] \\ = (x + y + z) [(- x - 2 y) (- y - 2 z) - (2 y + z) (- x + y)] \\ = (x + y + z) (x y + 2 z x + 2 y^{2} + 4 y z + 2 x y - 2 y^{2} + z x - z y) \\ = (x + y + z) (3 x y + 3 z x + 3 y z) \\ = 3 (x + y + z) (x y + y z + z x) \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} y & z y^{2} & 0 \end{matrix}]$

7 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} z & z y^{2} & 0 \end{matrix} | \\ T a k i n g x,^{2} y^{2} a n d z^{2} c o m m o n f r o m C_{1}, C_{2} a n d C_{3} r e s p e c t i v e l y \\ = x^{2} y^{2} z^{2} | \begin{matrix} 0 & x & x \\ y & 0 & y \\ z & z & 0 \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} = x^{2} y^{2} z^{2} [0 | \begin{matrix} 0 & y \\ z & 0 \end{matrix} | - x | \begin{matrix} y & y \\ z & 0 \end{matrix} | + x | \begin{matrix} y & 0 \\ z & z \end{matrix} |] \\ = x^{2} y^{2} z^{2} [- x (0 - y z) + x (y z - 0)] \\ = x^{2} y^{2} z^{2} (x y z + x y z) = x^{2} y^{2} z^{2} (2 x y z) = 2 x^{3} y^{3} z^{3} \end{array}$

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Ncert Solutions Maths class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}]$

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix} | \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ = | \begin{matrix} a + x + y + z & y & z \\ a + x + y + z & a + y & z \\ a + x + y + z & y & a + z \end{matrix} | = (a + x + y + z) | \begin{matrix} 1 & y & z \\ 1 & a + y & z \\ 1 & y & a + z \end{matrix} | \\ (T a k i n g a + x + y + z c o m m o n) \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ = (a + x + y + z) | \begin{matrix} 0 & - a & 0 \\ 0 & a & - a \\ 1 & y & a + z \end{matrix} | \\ E x p a n d i n g a l o n g C_{1} = (a + x + y + z) | 1 (a^{2} - 0) | = a^{2} (a + x + y + z) \end{array}$

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Kindly Consider the following

If $A + B + C = 0$ , then prove that $[\begin{matrix} 1 & c o s C & c o s B \\ c o s C & 1 & c o s A \\ c o s B & c o s A & 1 \end{matrix}] = 0$

Kindly Consider the following

$[\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}]$ = (a – 1)3

Kindly Consider the following

$[\begin{matrix} y + z & z & y \\ z & z + x & x \\ y & x & x + y \end{matrix}]$ = 4 $x y z$

Kindly Consider the following

$[\begin{matrix} y^{2} z^{2} & y z & y + z \\ z^{2} x^{2} & z x & z + x \\ x^{2} y^{2} & x y & x + y \end{matrix}]$ =0

Kindly Consider the following

$[\begin{matrix} a - b - c & 2 a & 2 a \\ 2 b & b - c - a & 2 b \\ 2 c & 2 c & c - a - b \end{matrix}]$

Kindly Consider the following

$[\begin{matrix} x + 4 & x & x \\ x & x + 4 & x \\ x & x & x + 4 \end{matrix}]$

Kindly Consider the following

$[\begin{matrix} 3 x & - x + y & - x + z \\ x - y & 3 y & z - y \\ x - z & y - z & 3 z \end{matrix}]$

Kindly Consider the following

$[\begin{matrix} 0 & x y^{2} & x z^{2} \\ x^{2} y & 0 & y z^{2} \\ x^{2} y & z y^{2} & 0 \end{matrix}]$

Kindly Consider the following

$[\begin{matrix} a + x & y & z \\ x & a + y & z \\ x & y & a + z \end{matrix}]$