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Class 12th Matrices

67. Show that the matrix B'AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

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Vishal Baghel

Contributor-Level 10

We have,

(E) (B'AB)' = [B' (AB]'

= (AB)' (B')'

= B'A'B.

When A is symmetric, A' = A

(B'AB)' = B'AB

ie, B'AB is symmetric.

And when A is skew-symmetric, A¹ = -A

(B'AB)' = -B'AB.

ie, B'AB is skew-symmetric.

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8 months ago

Class 12th Matrices

66. If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

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Vishal Baghel

Contributor-Level 10

Given, A and B are symmetric matrices.

(E) Then A' = A and B' = B.

Now, (AB - BA)' = (AB)' - (BA)'

= B'A' - A'B'

= BA - AB

= - (AB - BA).

Hence, AB BA is skew-symmetric matrix

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Class 12th Matrices

65. If A = $[\begin{matrix} 3 & - 4 \\ 1 & - 1 \end{matrix}]$ , then prove that Aⁿ = $[\begin{matrix} 1 + 2 n & - 4 n \\ n & 1 - 2 n \end{matrix}]$ , where n is any positive integer.

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Vishal Baghel

Contributor-Level 10

We have,

$= [\begin{matrix} 3 (1 + 2 k) + (- 4 k) * 1 & - 4 (1 + 2 k) + (- 4 k) (- 1) \\ 3 k + 1 * (1 - 2 k) & - 4 * k + (1 - 2 k) (- 1) \end{matrix}]$

$= [\begin{matrix} 3 + 6 k - 4 k & - 4 - 8 k + 4 k \\ 3 k + 1 - 2 k & - 4 k + 2 k - 1 \end{matrix}] = [\begin{matrix} 3 + 2 k & - 4 - 4 k \\ 1 + k & - 1 - 2 k \end{matrix}]$

$= [\begin{matrix} 1 + 2 + 2 k & - 4 (1 + k) \\ 1 + k & 1 - 2 - 2 x \end{matrix}]$

$= [\begin{matrix} 1 + 2 (1 + k) & - 4 (1 + k) \\ 1 + k & 1 - 2 (1 + k) \end{matrix}]$

The results also holds for n = k + 1. Hence, An $= [\begin{matrix} 1 + 2 n & - 4 n \\ n & 1 - 2 n \end{matrix}]$

Holds for all natural number n.

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Class 12th Matrices

64. If A = $[\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}]$ , prove that Aⁿ = $[\begin{matrix} 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \\ 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \\ 3^{n - 1} & 3^{n - 1} & 3^{n - 1} \end{matrix}]$ , n ∈ N.

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Vishal Baghel

Contributor-Level 10

We have,

(E) P (n) : If A = $[$
$\begin{array}{l} \end{array} [\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}]$ ]. then Aⁿ = $[\begin{array}{l} 3^{n - 1} 3^{n - 1} 3^{n - 1} \\ 3^{n - 1} 3^{n - 1} 3^{n - 1} \\ 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{array}]$ n ? N.

P (1) : A¹= $[\begin{array}{l} 3^{1 - 1} 3^{1 - 1} 3^{1 - 1} \\ 3^{1 - 1} 3^{1 - 1} 3^{1 - 1} \\ 3^{1 - 1} 3^{1 - 1} 3^{1 - 1} \end{array}]$ = $[\begin{array}{l} 3^{0} 3^{0} 3^{0} \\ 3^{0} 3^{0} 3^{0} \\ 3^{0} 3^{0} 3^{0} \end{array}]$ = $[$
$\begin{array}{l} \end{array} [\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}]$ ]

So, the result holds true for n = 1.

Let the result be for n = k. So,

P (k): A^u= $[\begin{array}{l} 3^{k - 1} 3^{k - 1} 3^{k - 1} \\ 3^{k - 1} 3^{k - 1} 3^{k - 1} \\ 3^{k - 1} 3^{k - 1} 3^{k - 1} \end{array}]$

Then P (k+1): A^{k + 1}= A^k. A = $[\begin{array}{l} 3^{k - 1} 3^{k - 1} 3^{k - 1} \\ 3^{k - 1} 3^{k - 1} 3^{k - 1} \\ 3^{k - 1} 3^{k - 1} 3^{k - 1} \end{array}]$ $[$
$\begin{array}{l} \end{array} [\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}]$ ]

= $[\begin{matrix} 3^{k - 1} + 3^{k - 1} + 3^{k - 1} & 3^{k - 1} + 3^{k - 1} + 3^{k - 1} & 3^{k - 1} + 3^{k - 1} + 3^{k - 1} \\ 3^{k - 1} + 3^{k - 1} + 3^{k - 1} & 3^{k - 1} + 3^{k - 1} + 3^{k - 1} & 3^{k - 1} + 3^{k - 1} + 3^{k - 1} \\ 3^{k - 1} + 3^{k - 1} + 3^{k - 1} & 3^{k - 1} + 3^{k - 1} + 3^{k - 1} & 3^{k - 1} + 3^{k - 1} + 3^{k - 1} \end{matrix}]$

= A^k^{+ 1} = $[\begin{array}{l} 3^{(k + 1) - 1} 3^{(k + 1) - 1} 3^{(k + 1) - 1} \\ 3^{(k + 1) - 1} 3^{(k + 1) - 1} 3^{(k + 1) - 1} \\ 3^{(k + 1) - 1} 3^{(k + 1) - 1} 3^{(k + 1) - 1} \end{array}]$

The result holds for n = k + 1. Hence,

Aⁿ = $[\begin{array}{l} 3^{n - 1} 3^{n - 1} 3^{n - 1} \\ 3^{n - 1} 3^{n - 1} 3^{n - 1} \\ 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{array}]$ holds for all natural number.

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8 months ago

Class 12th Matrices

63. Let A = $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ , show that (aI + bA)ⁿ = aⁿI + na^{n – 1} bA, where I is the identity matrix of order 2 and n N.

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Vishal Baghel

Contributor-Level 10

We shall prove the result by using principal of mathematical induction

we have,

P (n) :- If A = $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ , then (a I + b A)ⁿ = aⁿ I + naⁿ^{- 1}b A where I is identity matrix of order 2, n∈N

P (1): (a I + b A)¹ = a¹I + 1 ´a^{1 - 1}b A

= a I + a⁰bA

= a¹I + b A {Qx⁰ = 1}

So the result is true for n = 1.

Let the result be true for n = k. So,

P (k): (a I + b A)^u = a^uI + u. a^u^-1b A. _____ (1)

Now, we prove that the result holds for n = k + 1,

P (k + 1): (a I + b A)^{k + 1} = (a I + b A). (a I + b A)^k

= (a I + b A) (auI + k au 1b A){using eqn (1)}

= a.a^kI² + k. a a^u^{- 1}b IA + a^kb.AI + a zz^k¹b²k A²

= a^{k + 1}I² + k a^u^{- 1+1}b IA + a^k b AI + a^k^{- 1}b² k A² _____ (2

...more

We shall prove the result by using principal of mathematical induction

we have,

P (n) :- If A = $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ , then (a I + b A)ⁿ = aⁿ I + naⁿ^{- 1}b A where I is identity matrix of order 2, n∈N

P (1): (a I + b A)¹ = a¹I + 1 ´a^{1 - 1}b A

= a I + a⁰bA

= a¹I + b A {Qx⁰ = 1}

So the result is true for n = 1.

Let the result be true for n = k. So,

P (k): (a I + b A)^u = a^uI + u. a^u^-1b A. _____ (1)

Now, we prove that the result holds for n = k + 1,

P (k + 1): (a I + b A)^{k + 1} = (a I + b A). (a I + b A)^k

= (a I + b A) (auI + k au 1b A){using eqn (1)}

= a.a^kI² + k. a a^u^{- 1}b IA + a^kb.AI + a zz^k¹b²k A²

= a^{k + 1}I² + k a^u^{- 1+1}b IA + a^k b AI + a^k^{- 1}b² k A² _____ (2)

Now, I2 = $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ = $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ = I

IA = $[\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ = A

AI = $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ = A.

A² = A.A = $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$ = 0.

Putting these values in eqn (2). We get,

P (k + 1) = a^{k + 1} I + kab A + a^{k + 1}^{- 1}b A + 0.

= a^{k + 1} I + (k + 1) a^{(k + 1)}^-1b A.

The result is true for n = k + 1. Thus by principle of mathematical induction

(a I + b A)ⁿ= aⁿI + naⁿ^-1bA for A = $[\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$

holds true for all natural number n.

less

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Class 12th Matrices

62.Matrices A and B will be inverse of each other only if

(A) AB = BA (B) AB = BA = 0

(C) AB = 0, BA = I (D) AB = BA = I

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Vishal Baghel

Contributor-Level 10

Matrices A and B will be inverse of each other if.

(E) AB = BA = I.

Here option D is correct.

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Class 12th Matrices

Kindly Consider the following

61. $[\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}]$

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Vishal Baghel

Contributor-Level 10

Let A = $[\begin{matrix} 2 & 0 & - 1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix}]$

$\Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 2 & 0 & - 1 \\ 0 & 1 & 3 \end{matrix}] = [\begin{matrix} - 2 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 2 - 2 (1) & 0 - 2 (1) & - 1 - 2 (2) \\ 0 & 1 & 3 \end{matrix}] = [\begin{matrix} - 2 & 1 & 0 \\ 1 - 2 (- 2) & 0 - 2 (1) & 0 - 2 (0) \\ 0 & 0 & 1 \end{matrix}]$ A. (R2 R2 ----> 2R1)

$\Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 0 & - 2 & - 5 \\ 0 & 1 & 3 \end{matrix}] = [\begin{matrix} - 2 & 1 & 0 \\ 5 & - 2 & 0 \\ 0 & 0 & 1 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & - 2 & - 5 \end{matrix}] = [\begin{matrix} - 2 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & - 2 & 0 \end{matrix}]$ A. (R2 <-->R3)

$\Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 + 2 (0) & - 2 + 2 (1) & - 5 + 2 (3) \end{matrix}] = [\begin{matrix} - 2 & 1 & 0 \\ 0 & 0 & 1 \\ 5 + 2 (0) & - 2 + 2 (0) & 0 + 2 (1) \end{matrix}]$ A. (R3 ->R3 + 2R2)

$\Rightarrow [\begin{matrix} 1 & 1 & 2 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 2 & 1 & 0 \\ 0 & 0 & 1 \\ 5 & - 2 & 2 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 - 2 (0) & 1 - 2 (0) & 2 - 2 (1) \\ 0 - 3 (0) & 1 - 3 (0) & 3 - 3 (1) \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 2 - 2 (5) & 1 - 2 (- 2) & 0 - 2 (2) \\ 0 - 3 (5) & 0 - 3 (- 2) & 1 - 3 (2) \\ 5 & - 2 & 2 \end{matrix}]$ A. $(\begin{array}{l} R_{1} \to R_{1} - 2 R_{3} & R_{2} \to R_{2} - 3 R_{3} \end{array})$

$\Rightarrow [\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 12 & 5 & - 4 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 - 0 & 1 - 1 & 0 - 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} - 12 - (+ 15) & 5 - 6 & - 4 - (- 5) \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$ A. (R1 --->R1 - R2).

$\Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$ A.

∴ A1 = $[\begin{matrix} 3 & - 1 & 1 \\ - 15 & 6 & - 5 \\ 5 & - 2 & 2 \end{matrix}]$

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Class 12th Matrices

Kindly Consider the following

60. $[\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 5 \\ 2 & 5 & 0 \end{matrix}]$

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Vishal Baghel

Contributor-Level 10

Let A = $[\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 5 \\ 2 & 5 & 0 \end{matrix}]$

We write A = IA

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 5 \\ 2 & 5 & 0 \end{matrix}] = [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ A.

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ - 3 + 3 (1) & 0 + 3 (3) & - 5 + 3 (- 2) \\ 2 - 2 (1) & 5 - 2 (3) & 0 - 2 (- 2) \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 + 3 (1) & 1 + 0 & 0 + 0 \\ 0 - 2 (1) & 0 - 0 & 1 - 0 \end{matrix}]$ A. $(\begin{array}{l} R_{2} \to R_{2} + 3 R_{1} & R_{3} \to R_{3} - 2 R_{1} \end{array})$

$= [\begin{matrix} 1 & 3 & - 2 \\ 0 & 9 & - 11 \\ 0 & - 1 & . 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1 \end{matrix}]$ A.

$\to [\begin{matrix} 1 & 3 & - 2 \\ 0 & - 1 & 4 p \\ 0 & 9 & - 11 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ - 2 & 0 & 1 \\ 3 & 1 & 0 \end{matrix}]$ A. (R2 R3)

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - 4 \\ 0 & 9 & - 11 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & - 1 \\ 3 & 1 & 0 \end{matrix}]$ A. (R2 (-1) R2)

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - 4 \\ 0 - 0 & 9 - 9 (1) & - 11 - 9 (- 4) \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & - 1 \\ 3 - 9 (2) & 1 - 0 & 0 - 9 (- 1) \end{matrix}]$ A. (R3 R3--> 4R2)

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - 4 \\ 0 & 0 & 25 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & - 1 \\ - 15 & 1 & 9 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & - 1 \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{5} \end{matrix}]$ A. $(R_{3} \to \frac{1}{25} R_{3})$

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 + 0 & 1 + 0 & - 4 + 4 (1) \\ 0 & 0 & 1 \end{matrix}] =$ $= [\begin{matrix} 1 & 0 \\ 2 + 4 (\frac{3}{5}) & 0 + 4 (\frac{1}{25}) & - 1 + 4 * (\frac{9}{25}) \\ \frac{- 3}{25} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A. (R2 R2 + 4R3)

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 + 0 & 3 + 0 & - 2 + 2 (1) \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 + 2 (- \frac{3}{5}) & 0 + 2 (\frac{1}{25}) & 0 + 2 (\frac{9}{23}) \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A. (R1 R1 + 2R3)

$\Rightarrow [\begin{array}{l} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{matrix} \frac{- 1}{5} & \frac{2}{25} & \frac{18}{25} \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 + 0 & 3 - 3 (1) & 0 + 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] =$ $[\begin{matrix} \frac{- 1}{5} - 3 (\frac{- 2}{5}) & \frac{2}{25} - 3 (\frac{4}{25}) & \frac{18}{25} - 3 (\frac{11}{25}) \\ \frac{- 2}{5} & \frac{4}{5} & \frac{11}{25} \\ \frac{- 3}{5} & 25 & \frac{9}{25} \end{matrix}]$ A. (R1 R1 3R2)

$\Rightarrow [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{matrix} 1 & \frac{- 10}{25} & \frac{- 15}{25} \\ \frac{- 2}{5} & \frac{4}{25} & \frac{1}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & \frac{- 2}{5} & \frac{- 3}{5} \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A.

∴A^-1 = $[\begin{matrix} 1 & \frac{- 2}{5} & \frac{- 3}{5} \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$

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Class 12th Matrices

Kindly Consider the following

59. $[\begin{matrix} 2 & - 3 & 3 \\ 2 & 2 & 3 \\ 3 & - 2 & 2 \end{matrix}]$

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Vishal Baghel

Contributor-Level 10

Let A = $[\begin{matrix} 2 & - 3 & 3 \\ 2 & 2 & 3 \\ 3 & - 2 & 2 \end{matrix}]$

We write, A = IA

$\Rightarrow [\begin{matrix} 2 & - 3 & 3 \\ 2 & 2 & 3 \\ 3 & - 2 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 3 & - 2 & 2 \\ 2 & 2 & 3 \\ 2 & - 3 & 3 \end{matrix}] = [\begin{matrix} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}]$ A. $[R_{3} \leftrightarrow R_{1})$

$\Rightarrow [\begin{matrix} 3 - 2 & - 2 - 2 & 2 - 3 \\ 2 & 2 & 3 \\ 2 & - 3 & 3 \end{matrix}] [\begin{matrix} 0 - 0 & 0 - 1 & 1 - 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{matrix}]$ A. (R1 R1 -->R2)

$\Rightarrow [\begin{matrix} 1 & - 4 & - 1 \\ 2 & 2 & 3 \\ 2 & - 3 & 3 \end{matrix}] = [\begin{matrix} 0 & - 1 & 1 \\ 0 & 1 & 0 \\ 10 & 0 & 0 \end{matrix}]$ A.

$\Rightarrow | \begin{matrix} 1 & - 4 & - 1 \\ 2 - 2 (1) & 2 + 2 (- 4) & 3 - 2 (- 1) \\ 2 - 2 (1) & - 3 + 2 (- 4) & 3 - 2 (- 1) \end{matrix} | = [\begin{matrix} 0 & - 1 & 1 \\ 0 - 2 (0) & 1 - 2 (- 1) & 0 - 2 (1) \\ 1 - 2 (0) & 0 - 2 (- 1) & 0 - 2 (1) \end{matrix}]$ A.

$(\begin{array}{l} R_{2} \to R_{2} - 2 R_{1} \end{array}$

$\begin{array}{l} R_{3} \to R_{3} - 2 R_{1} \end{array})$

$\Rightarrow [\begin{matrix} 1 & - 4 & - 1 \\ 0 & 10 & 5 \\ 0 & 5 & 5 \end{matrix}] = [\begin{matrix} 0 & - 1 & 1 \\ 0 & 3 & - 2 \\ 1 & 2 & - 2 \end{matrix}]$ A.

$1 : \frac{x^{2}}{36} + \frac{y^{2}}{16} = 1$

$\Rightarrow [\begin{matrix} 1 & - 4 & - 1 \\ 0 - 0 & 10 - 5 & 5 - 5 \\ 0 & 5 & 5 \end{matrix}] = [\begin{matrix} 0 & - 1 & 1 \\ 0 - 1 & 3 - 2 & - 2 - (- 2) \\ 1 & 2 & - 2 \end{matrix}]$ A. (R2 R2 --->R3)

$\Rightarrow [\begin{matrix} 1 & - 4 & - 1 \\ 0 & 5 & 0 \\ 0 & 5 & 5 \end{matrix}] = [\begin{matrix} 0 & - 1 & 1 \\ - 1 & 1 & 0 \\ 1 & 2 & - 2 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 & - 4 & - 1 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \end{matrix}] = [\begin{matrix} 0 & - 1 & 1 \\ \frac{- 1}{5} & \frac{1}{5} & 0 \\ \frac{1}{5} & \frac{2}{5} & \frac{- 2}{5} \end{matrix}]$ A. $(\begin{matrix} R_{2} \to \frac{1}{5} R_{2} & R_{3} \to \frac{1}{5} R_{3} \end{matrix})$

$\Rightarrow [\begin{matrix} 1 & - 4 & - 1 \\ 0 & 1 & 0 \\ 0 - 0 & 1 - 1 & 1 - 0 \end{matrix}] = [\begin{matrix} 0 & - 1 & 1 \\ \frac{- 1}{5} & \frac{1}{5} & 0 \\ 1 / 5 - (\frac{- 1}{5}) & \frac{2}{5} - \frac{1}{5} & 0 (\frac{- 2}{5}) - 0 \end{matrix}]$ A. (R3 R3 -->R2)

$\Rightarrow [\begin{matrix} 1 & - 4 & - 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 & - 1 & 1 \\ \frac{- 1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & \frac{- 2}{5} \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 + 0 & - 4 + 0 & - 1 + 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 0 + \frac{2}{5} & - 1 + \frac{1}{5} & 1 + (\frac{- 2}{5}) \\ \frac{- 1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & \frac{- 2}{5} \end{matrix}]$ A. (R1 R1 + R3)

$\Rightarrow [\begin{matrix} 1 & - 4 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} \frac{2}{5} & \frac{- 4}{5} & \frac{3}{5} \\ \frac{- 1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & \frac{- 2}{5} \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 + 0 & - 4 + 4 (1) & 0 + 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] =$ $[\begin{matrix} \frac{2}{5} + 4 (- \frac{1}{5}) & - \frac{4}{5} + 4 (\frac{1}{5}) & \frac{3}{5} + 4 (0) \\ \frac{- 1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & \frac{- 2}{5} \end{matrix}]$ A. (R1 R1 + 4R2)

$\Rightarrow [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{matrix} \frac{- 2}{5} & 0 & \frac{3}{5} \\ \frac{- 1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & \frac{- 2}{5} \end{matrix}]$ A.

∴A^-1 = $[\begin{matrix} \frac{- 2}{5} & 0 & \frac{3}{5} \\ \frac{- 1}{5} & \frac{1}{5} & 0 \\ \frac{2}{5} & \frac{1}{5} & \frac{- 2}{5} \end{matrix}]$

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New answer posted

8 months ago

Class 12th Matrices

Kindly Consider the following

58. $[\begin{matrix} 2 & 1 \\ 4 & 2 \end{matrix}]$

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Vishal Baghel

Contributor-Level 10

Let A = $[\begin{matrix} 2 & 1 \\ 4 & 2 \end{matrix}]$

We write, A = IA.

$\Rightarrow [\begin{matrix} 2 & 1 \\ 4 & 2 \end{matrix}] = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 & \frac{1}{2} \\ 4 & 2 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 \\ 0 & 1 \end{matrix}]$ A. $(R_{1} \to \frac{1}{2} R_{1})$

$\to [\begin{matrix} 1 & \frac{1}{2} \\ 4 - 4 (1) & 2 - 4 (\frac{1}{2}) \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 \\ 0 - 4 (\frac{1}{2}) & 1 - 4 (\frac{0}{2}) \end{matrix}]$ A. (R2 R2 --->4R1)

$\Rightarrow [\begin{matrix} 1 & \frac{1}{2} \\ 0 & 0 \end{matrix}] = [\begin{matrix} \frac{1}{2} & 0 \\ - 2 & 1 \end{matrix}]$ A.

∴A^-1 does not exit

0 0

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