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New answer posted

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Maths Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

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

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Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

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

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

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

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Four

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

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

Maths 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.

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

Maths 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 $a = b = c$ .

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

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

Maths 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:

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

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Vishal Baghel

Contributor-Level 10

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

Sol:

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New answer posted

a year ago

Maths Maths NCERT Exemplar Solutions Class 12th Chapter Four

Using matrix method, solve the system of equations

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

0 Follower 14 Views

Vishal Baghel

Contributor-Level 10

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

Sol:

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New answer posted

a year ago

Maths 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