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a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

${(A^{3})}^{- 1} = {(A^{- 1})}^{3}$ , where $A$ is a square matrix and $? A ? \neq 0$ .

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Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l} Since {(A^{K})}^{- 1} = {(A^{- 1})}^{K} w h e r e K \in N \\ S o, {(A^{3})}^{- 1} = {(A^{- 1})}^{3} i s t r u e \end{array}$

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

If $f (x) = [\begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix}] = A + B x + C x^{2} + \dots$ then $A =$ ____.

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Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t [\begin{matrix} {(1 + x)}^{17} & {(1 + x)}^{19} & {(1 + x)}^{23} \\ {(1 + x)}^{23} & {(1 + x)}^{29} & {(1 + x)}^{34} \\ {(1 + x)}^{41} & {(1 + x)}^{43} & {(1 + x)}^{47} \end{matrix}] = A + B x + C x^{2} + \dots \\ T a k i n g {(1 + x)}^{17}, {(1 + x)}^{23} a n d {(1 + x)}^{41} c o m m o n f r o m R_{1}, R_{2} a n d R_{3} r e s p e c t i v e l y \\ {(1 + x)}^{17} . {(1 + x)}^{23} . {(1 + x)}^{41} | \begin{matrix} 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \\ 1 & {(1 + x)}^{6} & {(1 + x)}^{11} \\ 1 & {(1 + x)}^{2} & {(1 + x)}^{6} \end{matrix} | \\ \Rightarrow {(1 + x)}^{17} . {(1 + x)}^{23} . {(1 + x)}^{41} . 0 (R_{1} a n d R_{3} a r e i d e n t i c a l) \\ ∴ 0 = A + B x + C x^{2} + \dots \\ B y c o m p a r i n g t h e l i k e t e r m s, w e g e t A = 0 . \end{array}$

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

Kindly Consider the following

$[\begin{matrix} 0 & - x y z & x - z \\ - y & x & - y - z \\ z & x - z & y \end{matrix}]$ =____.

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Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

If $x = - 9$ is a root of $[\begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix}] = 0,$ then the other two roots are ____.

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Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e, | \begin{matrix} x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x \end{matrix} | = 0 \\ E x p a n d i n g a l o n g R_{1} \\ \Rightarrow x | \begin{matrix} x & 2 \\ 6 & x \end{matrix} | - 3 | \begin{matrix} 2 & 2 \\ 7 & x \end{matrix} | + 7 | \begin{matrix} 2 & x \\ 7 & 6 \end{matrix} | = 0 \\ \Rightarrow x (x^{2} - 12) - 3 (2 x - 14) + 7 (12 - 7 x) = 0 \\ \Rightarrow x^{3} - 12 x - 6 x + 42 + 84 - 49 x = 0 \\ \Rightarrow x^{3} - 67 x + 126 = 0 \dots (1) \\ T h e r o o t s o f t h e e q u a t i o n m a y b e t h e f a c t o r s o f 126 i . e ., 2 * 7 * 9 \\ 9 i s t h e g i v e n r o o t o f t h e determinant p u t x = 2 i n e q . (1) \\ {(2)}^{3} - 67 * 2 + 126 \Rightarrow 8 - 134 + 126 = 0 \\ H e n c e, x = 2 i s t h e o t h e r r o o t . \\ N o w, p u t x = 7 i n e q . (1) \\ {(7)}^{3} - 67 * 7 + 126 \Rightarrow 343 - 469 + 126 = 0 \\ H e n c e, x = 7 i s a l s o t h e o t h e r r o o t o f t h e determinant . \end{array}$

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

The sum of the products of elements of any row with the cofactors of corresponding elements is equal to ___.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e s u m o f t h e p r o d u c t s o f e l e m e n t s o f a n y r o w w i t h t h e c o - f a c t o r s o f c o r r e s p o n d i n g e l e m e n t s \\ i s e q u a l t o t h e v a l u e o f t h e determinant o f t h e g i v e n m a t r i x . \\ L e t Δ = | \begin{matrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{matrix} | \\ E x p a n d i n g a l o n g R_{1} \\ a_{11} | \begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix} | - a_{12} | \begin{matrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{matrix} | + a_{13} | \begin{matrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix} | \\ \Rightarrow a_{11} M_{11} + a_{12} M_{12} + a_{13} M_{13} (w h e r e M_{11}, M_{12} a n d M_{13} a r e t h e minors o f t h e c o r r e s p o n d i n g e l e m e n t s) \end{array}$

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

If $A$ is a matrix of order $3 * 3$ , then the number of minors in the determinant of $A$ is ____.

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Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e o r d e r o f a m a t r i x i s 3 * 3 \\ ∴ T o t a l n u m b e r o f e l e m e n t s = 3 * 3 = 9 \\ H e n c e, t h e n u m b e r o f minors i n t h e determinant i s 9 . \end{array}$

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

If $A$ is a matrix of order $3 * 3$ , then ${(A^{2})}^{- 1} =$ ____.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol: $F o r a n y s q u a r e m a t r i x A, {(A^{2})}^{- 1} = {(A^{- 1})}^{2} .$

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

If $c o s^{2} θ = 0$ , then ___. $[\begin{matrix} 0 & c o s θ & s i n θ \\ c o s θ & s i n θ & 0 \\ s i n θ & 0 & c o s θ \end{matrix}] =$ ___________

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Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks 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

If $x, y, z \in R$ , then the value of the determinant $[\begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix}]$ is equal to ___.

0 Follower 17 Views

Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e, | \begin{matrix} {(2^{x} + 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | \\ C_{1} \to C_{1} - C_{2} \\ \Rightarrow | \begin{matrix} {(2^{x} + 2^{- x})}^{2} - {(2^{x} - 2^{- x})}^{2} & {(2^{x} - 2^{- x})}^{2} & 1 \\ {(3^{x} + 3^{- x})}^{2} - {(3^{x} - 3^{- x})}^{2} & {(3^{x} - 3^{- x})}^{2} & 1 \\ {(4^{x} + 4^{- x})}^{2} - {(4^{x} - 4^{- x})}^{2} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | \\ \Rightarrow | \begin{matrix} 4 . 2^{x} . 2^{- x} & {(2^{x} - 2^{- x})}^{2} & 1 \\ 4 . 3^{x} 3^{- x} & {(3^{x} - 3^{- x})}^{2} & 1 \\ 4 . 4^{x} 4^{- x} & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | [a p p l y i n g {(a + b)}^{2} - {(a - b)}^{2} = 4 a b] \\ \Rightarrow | \begin{matrix} 4 & {(2^{x} - 2^{- x})}^{2} & 1 \\ 4 & {(3^{x} - 3^{- x})}^{2} & 1 \\ 4 & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | \Rightarrow | \begin{matrix} 1 & {(2^{x} - 2^{- x})}^{2} & 1 \\ 1 & {(3^{x} - 3^{- x})}^{2} & 1 \\ 1 & {(4^{x} - 4^{- x})}^{2} & 1 \end{matrix} | (T a k i n g 4 c o m m o n f r o m C_{1}) \\ \Rightarrow 4.0 = 0 (? C_{1} a n d C_{3} a r e i d e n t i c a l c o l u m n s) \end{array}$

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New answer posted

a year ago

Determinants Maths NCERT Exemplar Solutions Class 12th Chapter Four

If $A$ is an invertible matrix of order $3 * 3$ , then $? A^{- 1} ?$ = ___.

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a Fill in the blanks as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e k n o w t h a t f o r a n i n v e r t i b l e m a t r i x A o f a n y o r d e r, \\ | A^{- 1} | = \frac{1}{| A |} \end{array}$

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