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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Find the value of $θ$ satisfying

$[\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0$

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Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t A = [\begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix}] = 0 \\ | A | = | \begin{matrix} 1 & 1 & s i n 3 θ \\ - 4 & 3 & c o s 2 θ \\ 7 & - 7 & - 2 \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2} \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 7 & 3 & c o s 2 θ \\ 14 & - 7 & - 2 \end{matrix} | = 0 \\ T a k i n g 7 c o m m o n f r o m C_{1} \\ \Rightarrow = 7 | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ \Rightarrow = | \begin{matrix} 0 & 1 & s i n 3 θ \\ - 1 & 3 & c o s 2 θ \\ 2 & - 7 & - 2 \end{matrix} | = 0 \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow 1 | \begin{matrix} 1 & s i n 3 θ \\ - 7 & - 2 \end{matrix} | + 2 | \begin{matrix} 1 & s i n 3 θ \\ 3 & c o s 2 θ \end{matrix} | = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 (c o s 2 θ - 3 s i n 3 θ) = 0 \\ \Rightarrow - 2 + 7 s i n 3 θ + 2 c o s 2 θ - 6 s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 c o s 2 θ + s i n 3 θ = 0 \\ \Rightarrow - 2 + 2 (1 - 2 s i n^{2} θ) + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 2 + 2 - 4 s i n^{2} θ + 3 s i n θ - 4 s i n^{3} θ = 0 \\ \Rightarrow - 4 s i n^{3} θ - 4 s i n^{2} θ + 3 s i n θ = 0 \\ \Rightarrow - s i n θ (4 s i n^{2} θ + 4 s i n θ - 3) = 0 \\ - s i n θ = 0 o r 4 s i n^{2} θ + 4 s i n θ - 3 = 0 \\ ∴ θ = n π o r 4 s i n^{2} θ + 6 s i n θ - 2 s i n θ - 3 = 0 w h e n n \in I \\ \Rightarrow 2 s i n θ (2 s i n θ + 3) - 1 (2 s i n θ + 3) = 0 \\ \Rightarrow (2 s i n θ + 3) (2 s i n θ - 1) = 0 \\ \Rightarrow &thinsp \end{array}$

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New answer posted

a year ago

Determinants 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

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

0 Follower 10 Views

Vishal Baghel

Contributor-Level 10

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

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

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

0 Follower 8 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Determinants 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

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Determinants 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 $x y z$

0 Follower 11 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Determinants 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

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

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

0 Follower 8 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

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

0 Follower 11 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

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

0 Follower 8 Views

Vishal Baghel

Contributor-Level 10

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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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

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

0 Follower 7 Views

Vishal Baghel

Contributor-Level 10

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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