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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

The number of distinct real roots of $[\begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix}] = 0$ in the interval $- \frac{π}{4} \leq x \leq \frac{π}{4}$ is

(A) 0

(B) 2

(C) 1

(D) 3

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t | \begin{matrix} s i n x & c o s x & c o s x \\ c o s x & s i n x & c o s x \\ c o s x & c o s x & s i n x \end{matrix} | = 0 \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow | \begin{matrix} 2 c o s x + s i n x & c o s x & c o s x \\ 2 c o s x + s i n x & s i n x & c o s x \\ 2 c o s x + s i n x & c o s x & s i n x \end{matrix} | = 0 (T a k i n g 2 c o s x + s i n x c o m m o n f r o m C_{1}) \\ \Rightarrow (2 c o s x + s i n x) | \begin{matrix} 1 & c o s x & c o s x \\ 1 & s i n x & c o s x \\ 1 & c o s x & s i n x \end{matrix} | = 0 \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow (2 c o s x + s i n x) | \begin{matrix} 0 & c o s x - s i n x & 0 \\ 0 & s i n x - c o s x & c o s x - s i n x \\ 1 & c o s x & s i n x \end{matrix} | = 0 \\ \Rightarrow (2 c o s x + s i n x) [1 | \begin{matrix} c o s x - s i n x & 0 \\ s i n x - c o s x & c o s x - s i n x \end{matrix} |] \\ \Rightarrow (2 c o s x + s i n x) {(c o s x - s i n x)}^{2} = 0 \\ 2 c o s x + s i n x = 0 a n d {(c o s x - s i n x)}^{2} = 0 \\ 2 + t a n x = 0 a n d (c o s x - s i n x) = 0 \\ ∴ t a n x = - 2 a n d \Rightarrow t a n x = 1 \\ B u t - \frac{π}{4} \leq x \leq \frac{π}{4} a n d \Rightarrow t a n x = t a n \frac{π}{4} \\ ∴ x = \frac{π}{4} \in [- \frac{π}{4}, \frac{π}{4}] \\ S o, x &thins \end{array}$

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

The determinant $[\begin{matrix} b^{2} - a b & b - c & b c - a c \\ b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix}]$ equals

(A) $a b c (b - c) (c - a) (a - b)$

(B) $(b - c) (c - a) (a - b)$

(C) $(a + b + c) (b - c) (c - a) (a - b)$

(D) None of these

0 Follower 8 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t Δ = | \begin{matrix} b^{2} - a b & b - c & b c - a c \\ a b - a^{2} & a - b & b^{2} - a b \\ b c - a c & c - a & a b - a^{2} \end{matrix} | \\ \Rightarrow = | \begin{matrix} b (b - a) & b - c & c (b - a) \\ a (b - a) & a - b & b (b - a) \\ c (b - a) & c - a & a (b - a) \end{matrix} | (T a k i n g (b - a) c o m m o n f r o m C_{1} a n d C_{3}) \\ \Rightarrow = {(b - a)}^{2} | \begin{matrix} b & b - c & c \\ a & a - b & b \\ c & c - a & a \end{matrix} | \\ C_{1} \to C_{1} - C_{3} \\ \Rightarrow = {(a - b)}^{2} | \begin{matrix} b - c & b - c & c \\ a - b & a - b & b \\ c - a & c - a & a \end{matrix} | (C_{1} a n d C_{2} a r e i d e n t i c a l c o l u m n s .) \\ \Rightarrow = {(a - b)}^{2} . 0 = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

The area of a triangle with vertices $(- 3, 0)$ , $(3, 0)$ , and $(0, k)$ is 9 sq. units. The value of $k$ will be

(A) 9

(B) 3

(C) –9

(D) 6

0 Follower 21 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} A r e a o f t r i a n g l e w i t h v e r t i c e s (x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3}) w i l l b e : \\ Δ = \frac{1}{2} | \begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{matrix} | \Rightarrow Δ = \frac{1}{2} | \begin{matrix} - 3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{matrix} | \\ \Rightarrow = \frac{1}{2} [- 3 | \begin{matrix} 0 & 1 \\ k & 1 \end{matrix} | - 0 | \begin{matrix} 3 & 1 \\ 0 & 1 \end{matrix} | + 1 | \begin{matrix} 3 & 0 \\ 0 & k \end{matrix} |] \\ \Rightarrow = \frac{1}{2} [- 3 (- k) - 0 + 1 (3 k)] \\ \Rightarrow = \frac{1}{2} (3 k + 3 k) \Rightarrow = \frac{1}{2} (6 k) = 3 k \\ 3 k = 9 \Rightarrow k = 3 \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

The value of determinant $[\begin{matrix} a - b & b + c & a \\ b - a & c + a & b \\ c - a & a + b & c \end{matrix}]$ is

(A) $a^{3} + b^{3} + c^{3}$

(B) $3 b c$

(C) $a^{3} + b^{3} + c^{3} - 3 a b c$

(D) None of these

0 Follower 66 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

If $[\begin{matrix} 2 x & 5 \\ 8 & x \end{matrix}] = [\begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix}]$ then the value of $x$ is

(A) 3

(B) ±3

(C) ±6

(D) 6

0 Follower 10 Views

Vishal Baghel

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \\ \Rightarrow | \begin{matrix} 2 x & 5 \\ 8 & x \end{matrix} | = | \begin{matrix} 6 & - 2 \\ 7 & 3 \end{matrix} | \\ \Rightarrow 2 x^{2} - 40 = 18 + 14 \Rightarrow 2 x^{2} = 32 + 40 \\ \Rightarrow 2 x^{2} = 72 \Rightarrow x^{2} = 36 \\ ∴ x \pm 6 \\ H e n c e, t h e c o r r e c t x^{2} o p t i o n i s (c) . \end{array}$

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Find $A^{- 1}$ if $A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}]$ and show that $A^{- 1} = \frac{3 I - A^{2}}{2}$

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} H e r e, A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ | A | = 0 | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | - 1 | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | + 1 | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | \\ = 0 - 1 (0 - 1) + 1 (1 - 0) \\ = 1 + 1 = 2 \neq 0 (n o n - singular m a t r i x) \\ N o w, c o - f a c t o r s, \\ a_{11} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{12} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{13} = + | \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} | = 1 \\ a_{21} = - | \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} | = 1, a_{22} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1, a_{23} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1 \\ a_{31} = + | \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} | = 1, a_{32} = - | \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} | = 1, a_{33} = + | \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} | = - 1 \\ A d j (A) = {[\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}]}^{'} = [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ ∴ A^{- 1} = \frac{1}{| A |} A d j (A) = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ N o w, A^{2} = A . A = [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] [\begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix}] \\ = [\begin{matrix} 0 + 1 + 1 & 0 + 0 + 1 & 0 + 1 + 0 \\ 0 + 0 + 1 & 1 + 0 + 1 & 1 + 0 + 0 \\ 0 + 1 + 0 & 1 + 0 + 0 & 1 + 1 + 0 \end{matrix}] = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ H e n c e, A^{2} = [\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] \\ N o w, w e h a v e t o p r o v e t h a t A^{- 1} = \frac{A^{2} - 3 I}{2} \\ R . H . S . = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - 3 [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]}{2} \\ = \frac{[\begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}] - [\begin{matrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{matrix}]}{2} = \frac{1}{2} [\begin{matrix} - 1 & 1 & 1 \\ 1 & - 1 & 1 \\ 1 & 1 & - 1 \end{matrix}] \\ = A^{- 1} = L . H . S . \\ H e n c e, p r o v e d . \end{array}$

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Show that the $? A B C$ is an isosceles triangle if the determinant

$[\begin{matrix} 1 & 1 & 1 \\ 1 + c o s A & 1 + c o s B & 1 + c o s C \\ c o s^{2} A + c o s A & c o s^{2} B + c o s B & c o s^{2} C + c o s C \end{matrix}] = 0$

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

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

Show that the points $(a + 5, a - 4)$ , $(a - 2, a + 3)$ , and $(a, a)$ do not lie on a straight line for any value of $a$ .

0 Follower 12 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} I f t h e g i v e n points l i e o n a s t r a i g h t l i n e, t h e n t h e a r e a o f t h e t r i a n g l e f o r m e d b y j o i n i n g t h e \\ points p a i r w i s e i s z e r o . \\ S 0, | \begin{matrix} a + 5 & a - 4 & 1 \\ a - 2 & a + 3 & 1 \\ a & a & 1 \end{matrix} | \\ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} \\ \Rightarrow | \begin{matrix} 7 & - 7 & 0 \\ - 2 & 3 & 0 \\ a & a & 1 \end{matrix} | \\ E x p a n d i n g a l o n g C_{3} \\ \Rightarrow 1 . | \begin{matrix} 7 & - 7 \\ - 2 & 3 \end{matrix} | = 21 - 14 = 7 u n i t s \\ A s 7 \neq 0 . H e n c e, t h e t h r e e points d o n o t l i e o n a s t r a i g h t l i n e f o r a n y v a l u e o f a . \end{array}$

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If $a_{1}, a_{2}, a_{3}, . . ., a_{r}$ are in G.P., then prove that the determinant $[\begin{matrix} a_{r + 1} & a_{r + 5} & a_{r + 9} \\ a_{r + 7} & a_{r + 11} & a_{r + 15} \\ a_{r + 11} & a_{r + 17} & a_{r + 21} \end{matrix}]$ is

independent of $r$ .

0 Follower 10 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} I f a_{1}, a_{2}, a_{3}, \dots a_{r} b e t h e t e r m s o f G . P ., t h e n \\ a_{n} = A R^{n - 1} (w h e r e A i s t h e f i r s t t e r m a n d R i s t h e c o m m o n r a t i o o f t h e G . P .) \\ ∴ a_{r + 1} = A R^{r + 1 - 1} = A R^{r}; a_{r + 5} = A R^{r + 5 - 1} = A R^{r + 4} \\ a_{r + 9} = A R^{r + 9 - 1} = A R^{r + 8}; a_{r + 7} = A R^{r + 7 - 1} = A R^{r + 6} \\ a_{r + 11} = A R^{r + 11 - 1} = A R^{r + 10}; a_{r + 15} = A R^{r + 15 - 1} = A R^{r + 14} \\ a_{r + 17} = A R^{r + 17 - 1} = A R^{r + 16}; a_{r + 21} = A R^{r + 21 - 1} = A R^{r + 20} \\ ∴ T h e determinant b e c o m e s \\ | \begin{matrix} A R^{r} & A R^{r + 4} & A R^{r + 8} \\ A R^{r + 6} & A R^{r + 10} & A R^{r + 14} \\ A R^{r + 10} & A R^{r + 16} & A R^{r + 20} \end{matrix} | \\ T a k i n g A R^{r}, A R^{r + 6} a n d A R^{r + 10} c o m m o n f r o m R_{1,} R_{2,} a n d R_{3} r e s p e c t i v e l y . \\ A R^{r} . A R^{r + 6} . A R^{r + 10} | \begin{matrix} 1 & R^{4} & R^{8} \\ 1 & R^{4} & R^{8} \\ 1 & R^{6} & R^{10} \end{matrix} | \\ = A R^{r} . A R^{r + 6} . A R^{r + 10} | 0 | [? R_{1} a n d R_{2} a r e i d e n t i c a l r o w s] \\ = 0 \\ H e n c e, t h e g i v e n determinant i s i n d e p e n d e n t o f r . \end{array}$

0 0

New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

If $[\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] ? = 0,$ then find values of $x$ .

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

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

$\begin{array}{l} L e t A = [\begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix}] = 0 \\ | A | = | \begin{matrix} 4 - x & 4 + x & 4 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ R_{1} \to R_{1} + R_{2} + R_{3} \\ \Rightarrow | \begin{matrix} 12 + x & 12 + x & 12 + x \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ T a k i n g (12 + x) c o m m o n f r o m R_{1} \\ \Rightarrow (12 + x) | \begin{matrix} 1 & 1 & 1 \\ 4 + x & 4 - x & 4 + x \\ 4 + x & 4 + x & 4 - x \end{matrix} | = 0 \\ C_{1} \to C_{1} - C_{2}, C_{2} \to C_{2} - C_{3} \\ \Rightarrow (12 + x) | \begin{matrix} 0 & 0 & 1 \\ 2 x & - 2 x & 4 + x \\ 0 & 2 x & 4 - x \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow (12 + x) [1 . | \begin{matrix} 2 x & - 2 x \\ 0 & 2 x \end{matrix} |] = 0 \\ \Rightarrow (12 + x) (4 x^{2} - 0) = 0 \Rightarrow 12 + x = 0 o r 4 x^{2} = 0 \\ \Rightarrow x = - 12 o r x = 0 \end{array}$

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