Ncert Solutions Maths class 12th

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New answer posted•a year ago

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

Kindly Consider the following

$[\begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix}]$

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Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

$\begin{array}{l} L e t Δ = | \begin{matrix} x^{2} - x + 1 & x - 1 \\ x + 1 & x + 1 \end{matrix} | \\ C_{1} \to C_{1} - C_{2} \\ = | \begin{matrix} x^{2} - 2 x + 2 & x - 1 \\ 0 & x + 1 \end{matrix} | \\ = (x + 1) (x^{2} - 2 x + 2) - 0 \\ = x^{3} - 2 x^{2} + 2 x + x^{2} - 2 x + 2 = x^{3} - x^{2} + 2 \end{array}$

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New answer posted•a year ago

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

If $x + y + z = 0$ , prove that $[\begin{matrix} x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a \end{matrix}] = x y z [\begin{matrix} a & b & c \\ c & a & b \\ b & c & a \end{matrix}] .$

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

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

Prove that $[\begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} \\ c a - b^{2} & a b - c^{2} & b c - a^{2} \\ a b - c^{2} & b c - a^{2} & c a - b^{2} \end{matrix}]$ is divisible by $a + b + c$ and find the quotient.

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

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

If $a + b + c \neq 0$ and $[\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}] = 0,$ then prove that $a = b = c$ .

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} G i v e n t h a t : a + b + c \neq 0 a n d [\begin{matrix} a & b & c \\ b & c & a \\ c & a & b \end{matrix}] = 0 \\ C_{1} \to C_{1} + C_{2} + C_{3} \\ \Rightarrow [\begin{matrix} a + b + c & b & c \\ a + b + c & c & a \\ a + b + c & a & b \end{matrix}] = 0 \\ \Rightarrow (a + b + c) | \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{matrix} | = 0 (T a k i n g a + b + c c o m m o n f r o m C_{1}) \\ \Rightarrow a + b + c \neq 0 ∴ | \begin{matrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{matrix} | = 0 \\ R_{1} \to R_{1} - R_{2} a n d R_{2} \to R_{2} - R_{3} \\ \Rightarrow | \begin{matrix} 0 & b - c & c - a \\ 0 & c - a & a - b \\ 1 & a & b \end{matrix} | = 0 \\ E x p a n d i n g a l o n g C_{1} \\ \Rightarrow 1 | \begin{matrix} b - c & c - a \\ c - a & a - b \end{matrix} | = 0 \\ \Rightarrow (b - c) (c - a) - {(c - a)}^{2} = 0 \\ \Rightarrow a b - b^{2} - a c + b c - c^{2} - a^{2} + 2 a c = 0 \\ \Rightarrow - a^{2} - b^{2} - c^{2} + a b + b c + a c = 0 \\ \Rightarrow a^{2} + b^{2} + c^{2} - a b - b c - a c = 0 \\ \Rightarrow &thin \end{array}$

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New answer posted•a year ago

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

Given $A = [- \begin{matrix} 2 & 2 & - 4 \\ 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}], a n d B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}],$ find $B A$ and use this to solve the system of equations:

$y + 2 z = 7, x - y = 3, 2 x + 3 y + 4 z = 17 .$

18 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

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New answer posted•a year ago

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

Using matrix method, solve the system of equations

$3 x + 2 y - 2 z = 3, x + 2 y + 3 z = 6, 2 x - y + z = 2 .$

22 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

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

If $A = [\begin{matrix} 1 & 2 & 0 \\ 2 & 1 & 2 \\ 0 & 1 & 1 \end{matrix}],$ find $A^{- 1}$ . Using $A^{- 1}$ , solve the system of linear equations

$x - 2 y = 10, 2 x - y - z = 8, - 2 y + z = 7 .$

50 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted•a year ago

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

If possible, using elementary row transformations, find the inverse of the following matrices:

(i) $[\begin{matrix} 2 & - 1 & 3 \\ - 5 & 3 & 1 \\ - 3 & 2 & 3 \end{matrix}]$

(ii) $[\begin{matrix} 2 & 3 & - 3 \\ - 1 & - 2 & 2 \\ 1 & 1 & - 1 \end{matrix}]$

(iii) $[\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}]$

29 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol.

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New answer posted•a year ago

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

Find $x$ , $y$ , and $z$ if: $A = [\begin{matrix} 0 & 2 y & z \\ x & y & - z \\ x & - y & z \end{matrix}]$ satisfies $A^{T} = A^{- 1}$ .

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answers Type Questions as classified in NCERT Exemplar

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New answer posted•a year ago

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

If $A B = B A$ for any two square matrices, prove by mathematical induction that ${(A B)}^{n} = A^{n} B^{n}$ .

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} L e t P (n) : {(A B)}^{n} = A^{n} B^{n} \\ S t e p 1 : P u t n = 1, P (1) : A B = A B w h i c h i s t r u e f o r n = 1 \\ S t e p 2 : P u t n = k, P (k) : {(A B)}^{k} = A^{k} B^{k} L e t i t b e t r u e f o r a n y k \in N \\ S t e p 3 : P u t n = k + 1, P (k + 1) : {(A B)}^{k + 1} = A^{k + 1} B^{k + 1} \\ L . H . S . {(A B)}^{k + 1} = {(A B)}^{k} . A B \\ = A^{k} B^{k} . A B [F r o m s t e p 2] \\ = A^{k + 1} A^{k + 1} R . H . S . \\ H e n c e, i f P (n) i s t r u e f o r P (k) t h e n i t i s t r u e f o r P (k + 1) . \end{array}$

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Ncert Solutions Maths class 12th

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