Skip to content
Related Articles

Related Articles

Improve Article

Class 12 RD Sharma Solutions – Chapter 5 Algebra of Matrices – Exercise 5.3 | Set 2

  • Last Updated : 21 Jul, 2021
Geek Week

Question 26. If\begin{bmatrix}1 & -1 & x\end{bmatrix}\begin{bmatrix}0 & 1 & -1\\2 & 1 & 3\\1 & 1 & 1\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix} = 0, find x.

Solution:

We have,

=>\begin{bmatrix}1 & -1 & x\end{bmatrix}\begin{bmatrix}0 & 1 & -1\\2 & 1 & 3\\1 & 1 & 1\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix} = 0

=>\begin{bmatrix}0-2+x & 1-1+x & -1-3+x\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=0

=>\begin{bmatrix}-2+x & x & -4+x\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=0



=>\begin{bmatrix}0+x-4+x\end{bmatrix}=0

=> 2x – 4 = 0

=> 2x = 4

=> x = 2

Therefore, the value of x is 2.

Question 27. If A =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix} and I =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} , then prove that A2 – A + 2I = 0.

Solution:

We have,

A =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}



A2 =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}

=\begin{bmatrix}9-8 & -6+4\\12-8 & -8+4\end{bmatrix}

=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}

L.H.S. = A2 – A + 2I

=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}-\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}+2\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}-\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}+\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

=\begin{bmatrix}1-3 & -2+2\\4-4 & -4+2\end{bmatrix}+\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

=\begin{bmatrix}-2 & 0\\0 & -2\end{bmatrix}+\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

=\begin{bmatrix}-2+2 & 0+0\\0+0 & -2+2\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}



= 0

= R.H.S.

Hence proved.

Question 28. If A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} and I =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} , then find λ so that A2 = 5A + λI.

Solution:

We have,

A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

A2 =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}

We are given,



=> A2 = 5A + λI

=>\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}=5\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}+λ\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=>\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}=\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}λ & 0\\0 & λ\end{bmatrix}

=>\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}=\begin{bmatrix}15+λ & 5+0\\-5+0 & 10+λ\end{bmatrix}

=>\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}=\begin{bmatrix}15+λ & 5\\-5 & 10+λ\end{bmatrix}

On comparing both sides, we get

=> 8 = 15 + λ

=> λ = –7

Therefore, the value of λ is –7.

Question 29. If A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} , show that A2 – 5A + 7I2 = 0.

Solution:



We have,

A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

A2 =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}

L.H.S. = A2 – 5A + 7I2

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-5\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}+7\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}

=\begin{bmatrix}8-15+7 & 5-5+0\\-5+5+0 & 3-10+7\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}



= 0

= R.H.S.

Hence proved.

Question 30. If A =\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix} , show that A2 – 2A + 3I2 = 0.

Solution:

We have,

A =\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix}

A2 =\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix}\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix}

=\begin{bmatrix}4-3 & 6+0\\-2+0 & -3+0\end{bmatrix}

=\begin{bmatrix}1 & 6\\-2 & -3\end{bmatrix}

L.H.S. = A2 – 2A + 3I2



=\begin{bmatrix}1 & 6\\-2 & -3\end{bmatrix}-2\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix}+3\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}1 & 6\\-2 & -3\end{bmatrix}-\begin{bmatrix}4 & 6\\-2 & 0\end{bmatrix}+\begin{bmatrix}3 & 0\\0 & 3\end{bmatrix}

=\begin{bmatrix}1-4+3 & 6-6+0\\-2+2+0 & -3+0+3\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 31. Show that the matrix A =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} satisfies the equation A3 – 4A2 + A = 0.

Solution:

We have,

A =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}



A2 =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}4+3 & 6+6\\2+2 & 3+4\end{bmatrix}

=\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}

A3 = A2. A

=\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}14+12 & 21+24\\8+7 & 12+14\end{bmatrix}

=\begin{bmatrix}26 & 45\\15 & 26\end{bmatrix}

L.H.S. = A3 – 4A2 + A

=\begin{bmatrix}26 & 45\\15 & 26\end{bmatrix}-4\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}+\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}26 & 45\\15 & 26\end{bmatrix}-\begin{bmatrix}28 & 48\\16 & 28\end{bmatrix}+\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}



=\begin{bmatrix}26-28+2 & 45-48+3\\15-16+1 & 26-28+2\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 32. Show that the matrix A =\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix} is root of the equation A2 – 12A – I = 0

Solution:

We have,

A =\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix}

A2 =\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix}\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix}

=\begin{bmatrix}25+36 & 15+21\\60+84 & 36+49\end{bmatrix}

=\begin{bmatrix}61 & 36\\144 & 85\end{bmatrix}

L.H.S. = A2 – 12A – I

=\begin{bmatrix}61 & 36\\144 & 85\end{bmatrix}-12\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix}-\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}61 & 36\\144 & 85\end{bmatrix}-\begin{bmatrix}60 & 36\\144 & 84\end{bmatrix}-\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}61-60-1 & 36-36-0\\144-144-0 & 85-84-1\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 33. If A =\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix} find A2 – 5A – 14I.

Solution:



We have,

A =\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}

A2 =\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}

=\begin{bmatrix}9+20 & -15-10\\-12-8 & 20+4\end{bmatrix}

=\begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}

A2 – 5A – 14I =\begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}-5\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}-14\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}-\begin{bmatrix}15 & -25\\-20 & 10\end{bmatrix}-\begin{bmatrix}14 & 0\\0 & 14\end{bmatrix}

=\begin{bmatrix}29-15-14 & -25+25+0\\-20+20+0 & 24-10-14\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

Question 34. If A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} , find A2 – 5A + 7I = 0. Use this to find A4.

Solution:



We have,

A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

A2 =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}

L.H.S. = A2 – 5A + 7I = 0

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-5\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}+7\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}

=\begin{bmatrix}8-15+7 & 5-5+0\\-5+5+0 & 3-10+7\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}



= 0

= R.H.S.

Hence proved.

Now we have A2 – 5A + 7I = 0

=> A2 = 5A – 7I

=> A4 = (5A – 7I) (5A – 7I)

=> A4 = 25A2 – 35AI – 35AI + 49I

=> A4 = 25A2 – 70AI + 49I

=> A4 = 25 (5A – 7I) – 70AI + 49I

=> A4 = 125A – 175I – 70A + 49I

=> A4 = 55A – 126I

=> A4 =55\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}-126\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=> A4 =\begin{bmatrix}165 & 55\\-55 & 110\end{bmatrix}-\begin{bmatrix}126 & 0\\0 & 126\end{bmatrix}

=> A4 =\begin{bmatrix}165-126 & 55-0\\-55-0 & 110-126\end{bmatrix}

=> A4 =\begin{bmatrix}39 & 55\\-55 & -16\end{bmatrix}

Question 35. If A =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix} , find k such that A2 = kA – 2I2.

Solution:

We have,

A =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}

A2 =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}

=\begin{bmatrix}9-8 & -6+4\\12-8 & -8+4\end{bmatrix}



=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}

We are given,

=> A2 = kA – 2I2

=>\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix} = k\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}-2\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=>\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix} = \begin{bmatrix}3k & -2k\\4k & -2k\end{bmatrix}-\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

=>\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix} = \begin{bmatrix}3k-2 & -2k-0\\4k-0 & -2k-2\end{bmatrix}

On comparing both sides, we get

=> 3k – 2 = 1

=> 3k = 3

=> k = 1

Therefore, the value of k is 1.

Question 36. If A =\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} , find k such that A2 – 8A + kI = 0.

Solution:

We have,

A =\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}

A2 =\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}

=\begin{bmatrix}1-0 & 0+0\\-1-7 & 0+49\end{bmatrix}

=\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix}

We are given,

=> A2 – 8A + kI = 0

=>\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix}-8\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}+k\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=0



=>\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix}-\begin{bmatrix}8 & 0\\-8 & 56\end{bmatrix}+\begin{bmatrix}k & 0\\0 & k\end{bmatrix}=0

=>\begin{bmatrix}1-8+k & 0-0+0\\-8+8+0 & 49-56+k\end{bmatrix} = 0

=>\begin{bmatrix}-7+k & 0\\0 & -7+k\end{bmatrix} = 0

On comparing both sides, we get

=> –k + 7 = 0

=> k = 7

Therefore, the value of k is 7.

Question 37. If A =\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix} and f(x) = x2 – 2x – 3, show that f(A) = 0.

Solution:

We have,

A =\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix} and f(x) = x2 – 2x – 3

A2 =\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}

=\begin{bmatrix}1+4 & 2+2\\2+2 & 4+1\end{bmatrix}

=\begin{bmatrix}5 & 4\\4 & 5\end{bmatrix}

L.H.S. = f(A) = A2 – 2A – 3I2

=\begin{bmatrix}5 & 4\\4 & 5\end{bmatrix}-2\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}+3\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}5 & 4\\4 & 5\end{bmatrix}-\begin{bmatrix}2 & 4\\4 & 2\end{bmatrix}+\begin{bmatrix}3 & 0\\0 & 3\end{bmatrix}

=\begin{bmatrix}5-2-3 & 4-4-0\\4-4-0 & 5-2-3\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.



Hence proved.

Question 38. If A =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} and I =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} , find λ, μ so that A2 = λA + μI.

Solution:

We have,

A =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

A2 =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}4+3 & 6+6\\2+2 & 3+4\end{bmatrix}

=\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}

We are given,

=> A2 = λA + μI

=>\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}=λ\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} + μ\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=>\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}=\begin{bmatrix}2λ & 3λ\\λ & 2λ\end{bmatrix} + \begin{bmatrix}μ & 0\\0 & μ\end{bmatrix}

=>\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}=\begin{bmatrix}2λ+μ & 3λ+0\\λ+0 & 2λ+μ\end{bmatrix}

=>\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}=\begin{bmatrix}2λ+μ & 3λ\\λ & 2λ+μ\end{bmatrix}

On comparing both sides, we get,

=> 2λ + μ = 7 and λ = 4

=> 2(4) + μ = 7

=> μ = 7 – 8

=> μ = –1

Therefore, the value of λ is 4 and μ is –1.

Question 39. Find the value of x for which the matrix product\begin{bmatrix}2 & 0 & 7\\0 & 1 & 0\\1 & -2 & 1\end{bmatrix}\begin{bmatrix}-x & 14x & 7x\\0 & 1 & 0\\x & -4x & -2x\end{bmatrix} equals an identity matrix.

Solution:



We have,

=>\begin{bmatrix}2 & 0 & 7\\0 & 1 & 0\\1 & -2 & 1\end{bmatrix}\begin{bmatrix}-x & 14x & 7x\\0 & 1 & 0\\x & -4x & -2x\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}-2x+0+7x & 28x+0-28x & 14x+0-14x\\0+0+0 & 0+1-0 & 0+0-0\\-x-0+x & 14x-2-4x & 7x-0-2x\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}5x & 0 & 0\\0 & 1 & 0\\0 & 10x-2 & 5x\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

On comparing both sides, we get,

=> 5x = 1

=> x = 1/5

Therefore, the value of x is 1/5.

Question 40. Solve the following matrix equations:

(i)\begin{bmatrix}x & 1\end{bmatrix}\begin{bmatrix}1 & 0\\-2 & -3\end{bmatrix}\begin{bmatrix}x \\5\end{bmatrix}=0

Solution:

We have,

=>\begin{bmatrix}x & 1\end{bmatrix}\begin{bmatrix}1 & 0\\-2 & -3\end{bmatrix}\begin{bmatrix}x \\5\end{bmatrix}=0

=>\begin{bmatrix}x-2 & 0-3\end{bmatrix}\begin{bmatrix}x \\5\end{bmatrix}=0

=>\begin{bmatrix}x-2 & -3\end{bmatrix}\begin{bmatrix}x \\5\end{bmatrix}=0

=>\begin{bmatrix}x^2-2x-15\end{bmatrix}=0

=> x2 – 2x – 15 = 0

=> x2 – 5x + 3x – 15 = 0

=> x (x – 5) + 3 (x – 5) = 0

=> (x – 5) (x + 3) = 0

=> x = 5 or –3



Therefore, the value of x is 5 or –3.

(ii)\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1&0&2\end{bmatrix}\begin{bmatrix}0 \\2\\x\end{bmatrix}=0

Solution:

We have,

=>\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1&0&2\end{bmatrix}\begin{bmatrix}0 \\2\\x\end{bmatrix}=0

=>\begin{bmatrix}1+4+1 & 2+0+0 & 0+2+2\end{bmatrix}\begin{bmatrix}0 \\2\\x\end{bmatrix}=0

=>\begin{bmatrix}6 & 2 & 4\end{bmatrix}\begin{bmatrix}0 \\2\\x\end{bmatrix}=0

=>\begin{bmatrix}0+4+4x\end{bmatrix}=0

=> 4 + 4x = 0

=> 4x = –4

=> x = –1

Therefore, the value of x is –1.

(iii)\begin{bmatrix}x & -5 & -1\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}\begin{bmatrix}x \\4\\1\end{bmatrix}=0

Solution:

We have,

=>\begin{bmatrix}x & -5 & -1\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}\begin{bmatrix}x \\4\\1\end{bmatrix}=0

=>\begin{bmatrix}x-0-2 & 0-10-0 & 2x-5-3\end{bmatrix}\begin{bmatrix}x \\4\\1\end{bmatrix}=0

=>\begin{bmatrix}x-2 & -10 & 2x-8\end{bmatrix}\begin{bmatrix}x \\4\\1\end{bmatrix}=0

=>\begin{bmatrix}x^2-2x-40+2x-8\end{bmatrix}=0

=> x2 – 48 = 0



=> x2 = 48

=> x = ±4√3

Therefore, the value of x is ±4√3.

(iv)\begin{bmatrix}2x & 3\end{bmatrix}\begin{bmatrix}1 & 2\\-3 & 0\end{bmatrix}\begin{bmatrix}x\\8 \end{bmatrix}=0

Solution:

We have,

=>\begin{bmatrix}2x & 3\end{bmatrix}\begin{bmatrix}1 & 2\\-3 & 0\end{bmatrix}\begin{bmatrix}x\\8 \end{bmatrix}=0

=>\begin{bmatrix}2x-9 & 4x\end{bmatrix}\begin{bmatrix}x\\8 \end{bmatrix}=0

=>\begin{bmatrix}x(2x-9)+32x\end{bmatrix}=0

=>\begin{bmatrix}2x^2-9x+32x\end{bmatrix}=0

=> 2x2 + 23x = 0

=> x (2x + 23) = 0

=> x = 0 or x = –23/2

Therefore, the value of x is 0 or –23/2.

Question 41. If A =\begin{bmatrix}1 & 2 & 0\\3 & -4 & 5\\0&-1&3\end{bmatrix} , compute A2 – 4A + 3I3.

Solution:

We have,

A =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}

A2 =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}

=\begin{bmatrix}1+6+0 & 2-8-0 & 0+10+0\\3-12+0 & 6+16-5 & 0-20+15\\0-3+0&0+4-3&0-5+9\end{bmatrix}

=\begin{bmatrix}7 & -6 & 10\\-9 & 17 & -5\\-3&1&4\end{bmatrix}

So, A2 – 4A + 3I3 =\begin{bmatrix}7 & -6 & 10\\-9 & 17 & -5\\-3&1&4\end{bmatrix}-4\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}+3\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0&0&1\end{bmatrix}

=\begin{bmatrix}7 & -6 & 10\\-9 & 17 & -5\\-3&1&4\end{bmatrix}-\begin{bmatrix}4 & 0 & 8\\0 & 8 & 4\\8&0&12\end{bmatrix}+\begin{bmatrix}3 & 0 & 0\\0 & 3 & 0\\0&0&3\end{bmatrix}

=\begin{bmatrix}7-4+3 & -6-8+0 & 10-0+0\\-9-12+0 & 17+16+3 & -5-20+0\\-3-0+0&1+4+0&4-12+3\end{bmatrix}

=\begin{bmatrix}6 & -14 & 10\\-21 & 36 & -25\\-3&5&-5\end{bmatrix}

Question 42. If f(x) = x2 – 2x, find f(A), where A =\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix} .

Solution:

We have,

A =\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix} and f(x) = x2 – 2x

A2 =\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix}\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix}

=\begin{bmatrix}0+4+0 & 0+5+4 & 0+0+6\\0+20+0 & 4+25+0 & 8+0+0\\0+8+0&0+10+6&0+0+9\end{bmatrix}

=\begin{bmatrix}4 & 9 & 6\\20 & 29 & 8\\8&16&9\end{bmatrix}



So, f(A) = A2 – 2A

=\begin{bmatrix}4 & 9 & 6\\20 & 29 & 8\\8&16&9\end{bmatrix}-2\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix}

=\begin{bmatrix}4 & 9 & 6\\20 & 29 & 8\\8&16&9\end{bmatrix}-\begin{bmatrix}0 & 2 & 4\\8 & 10 & 0\\0&4&6\end{bmatrix}

=\begin{bmatrix}4-0 & 9-2 & 6-4\\20-8 & 29-10 & 8-0\\8-0&16-4&9-6\end{bmatrix}

=\begin{bmatrix}4 & 7 & 2\\12 & 19 & 0\\8&12&3\end{bmatrix}

Question 43. If f(x) = x3 + 4x2 – x, find f(A) where A =\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix} .

Solution:

We have,

A =\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix} and f(x) = x3 + 4x2 – x

A2 =\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}

=\begin{bmatrix}0+2+2 & 0-3-2 & 0+0+0\\0-6+0 & 2+9-0 & 4-0+0\\0-2+0&1+3-0&2-0+0\end{bmatrix}

=\begin{bmatrix}4 & -5 & 0\\-6 & 11 & 4\\-2&4&2\end{bmatrix}

A3 = A2. A

=\begin{bmatrix}4 & -5 & 0\\-6 & 11 & 4\\-2&4&2\end{bmatrix}\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}

=\begin{bmatrix}0-10+0 & 4+15-0 & 8-0+0\\0+22+4 & -6-33-4 & -12+0+0\\0+8+2&-2-12-2&-4+0+0\end{bmatrix}

=\begin{bmatrix}-10 & 19 & 8\\26 & -43 & -12\\10&-16&-4\end{bmatrix}

Now, f(A) = A3 + 4A2 – A

=\begin{bmatrix}-10 & 19 & 8\\26 & -43 & -12\\10&-16&-4\end{bmatrix}+4\begin{bmatrix}4 & -5 & 0\\-6 & 11 & 4\\-2&4&2\end{bmatrix}-\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}

=\begin{bmatrix}-10 & 19 & 8\\26 & -43 & -12\\10&-16&-4\end{bmatrix}+\begin{bmatrix}16 & -20 & 0\\-24 & 44 & 16\\-8&16&8\end{bmatrix}-\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}

=\begin{bmatrix}-10+16-0 & 19-20-1 & 8+0-2\\26-24-2 & -43+44+3 & -12+16-0\\10-8-1&-16+16+1&-4+8+0\end{bmatrix}

=\begin{bmatrix}6 & -2 & 6\\0 & 4 & 4\\1&1&4\end{bmatrix}



Question 44. If A =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix} , then show that A is a root of the polynomial f(x) = x3 – 6x2 + 7x +2.

Solution:

We have,

A =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix} and f(x) = x3 – 6x2 + 7x +2.

A2 =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}

=\begin{bmatrix}1+0+4 & 0+0+0 & 2+0+6\\0+0+2 & 0+4+0 & 0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}

=\begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8&0&13\end{bmatrix}

A3 = A2. A

=\begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8&0&13\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}

=\begin{bmatrix}5+0+16 & 0+0+0 & 10+0+24\\2+0+10 & 0+8+0 & 4+4+15\\8+0+26&0+0+0&16+0+39\end{bmatrix}

=\begin{bmatrix}21 & 0 & 34\\12 & 8 & 23\\34&0&55\end{bmatrix}

In order to show that A is a root of above polynomial, we need to prove that f(A) = 0.

Now, f(A) = A3 – 6A2 + 7A + 2I

=\begin{bmatrix}21 & 0 & 34\\12 & 8 & 23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0&0&1\end{bmatrix}

=\begin{bmatrix}21 & 0 & 34\\12 & 8 & 23\\34&0&55\end{bmatrix}-\begin{bmatrix}30 & 0 & 48\\12 & 24 & 30\\48&0&78\end{bmatrix}+\begin{bmatrix}7 & 0 & 14\\0 & 14 & 7\\14&0&21\end{bmatrix}+\begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0&0&2\end{bmatrix}

=\begin{bmatrix}21-30+7+2 & 0-0+0+0 & 34-48+14+0\\12-12+0+0 & 8-24+14+2 & 23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= 0

Hence proved.

Question 45. If A =\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix} , prove that A2 – 4A – 5I = 0.

Solution:

We have,



A =\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix}

A2 =\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix}

=\begin{bmatrix}1+4+4 & 2+2+4 & 2+4+2\\2+2+4 & 4+1+4 & 4+2+2\\2+4+2 & 4+2+2 & 4+4+1\end{bmatrix}

=\begin{bmatrix}9 & 8 & 8\\8 & 9 & 8\\8 & 8 & 9\end{bmatrix}

Now, L.H.S. = A2 – 4A – 5I

=\begin{bmatrix}9 & 8 & 8\\8 & 9 & 8\\8 & 8 & 9\end{bmatrix}-4\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix}-5\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}9 & 8 & 8\\8 & 9 & 8\\8 & 8 & 9\end{bmatrix}-\begin{bmatrix}4 & 8 & 8\\8 & 4 & 8\\8 & 8 & 4\end{bmatrix}-\begin{bmatrix}5 & 0 & 0\\0 & 5 & 0\\0 & 0 & 5\end{bmatrix}

=\begin{bmatrix}9-4-5 & 8-8-0 & 8-8-0\\8-8-0 & 9-4-5 & 8-8-0\\8-8-0 & 8-8-0 & 9-4-5\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 46. If A =\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix} , show that A2 – 7A + 10I3 = 0.

Solution:

We have,

A =\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}

A2 =\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}

=\begin{bmatrix}9+2+0 & 6+8+0 & 0+0+0\\3+4+0 & 2+16+0 & 0+0+0\\0+0+0 & 0+0+0 & 0+0+25\end{bmatrix}

=\begin{bmatrix}11 & 14 & 0\\7 & 18 & 0\\0 & 0 & 25\end{bmatrix}

Now, L.H.S. = A2 – 7A + 10I3

=\begin{bmatrix}11 & 14 & 0\\7 & 18 & 0\\0 & 0 & 25\end{bmatrix}-7\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}+10\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}



=\begin{bmatrix}11 & 14 & 0\\7 & 18 & 0\\0 & 0 & 25\end{bmatrix}-\begin{bmatrix}21 & 14 & 0\\7 & 28 & 0\\0 & 0 & 35\end{bmatrix}+\begin{bmatrix}10 & 0 & 0\\0 & 10 & 0\\0 & 0 & 10\end{bmatrix}

=\begin{bmatrix}11-21+10 & 14-14+0 & 0-0+0\\7-7+0 & 18-28+10 & 0-0+0\\0-0+0 & 0-0+0 & 25-35+10\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 47. Without using the concept of inverse of a matrix, find the matrix\begin{bmatrix}x & y\\z & u\end{bmatrix} such that,\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix} \begin{bmatrix}x & y\\z & u\end{bmatrix} =\begin{bmatrix}-16 & -6\\7 & 2\end{bmatrix}

Solution:

We have,

=>\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix} \begin{bmatrix}x & y\\z & u\end{bmatrix} =\begin{bmatrix}-16 & -6\\7 & 2\end{bmatrix}

=>\begin{bmatrix}5x-7z & 5y-7u\\-2x+3z & -2y+3u\end{bmatrix}=\begin{bmatrix}-16 & -6\\7 & 2\end{bmatrix}

On comparing both sides, we get,

5x – 7z = –16

5y – 7u = –6

–2x + 3z = 7

–2y + 3u = 2

On solving the above equations, we get

=> x = 1, y = –4, z = 3 and u = –2.

So, we get\begin{bmatrix}x & y\\z & u\end{bmatrix} =\begin{bmatrix}1 & -4\\3 & -2\end{bmatrix} .

Question 48. Find the matrix A such that

(i)\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}A=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

Solution:

Let A =\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}

Given equation is,

=>\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}A=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}x+a & y+b & z+c\\a+0 & b+0 & c+0\end{bmatrix}=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}x+a & y+b & z+c\\a & b & c\end{bmatrix}=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

On comparing both sides, we get, a = 1, b = 0 and c = 1.

And x + 1 = 3 => x = 2

Also, y = 3 and

z + 1 = 5 => z = 4



So, we have A =\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}=\begin{bmatrix}2 & 3 & 4\\1 & 0 & 1\end{bmatrix}

(ii)A\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}7 & -8 & -9\\2 & 4 & 6\end{bmatrix}

Solution:

Let A =\begin{bmatrix}w & x\\y & z\end{bmatrix}

Given equation is,

=>A\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}7 & -8 & -9\\2 & 4 & 6\end{bmatrix}

=>\begin{bmatrix}w & x\\y & z\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}7 & -8 & -9\\2 & 4 & 6\end{bmatrix}

=>\begin{bmatrix}w+4x & 2w+5x & 3w+6x\\y+4z & 2y+5z & 3y+6z\end{bmatrix}=\begin{bmatrix}7 & -8 & -9\\2 & 4 & 6\end{bmatrix}

On comparing both sides, we get,

w + 4x = 7

2w + 5x = –6

y + 4z = 2

2y + 5z = 4

On solving the above equations, we get

=> x = –2, y = 2, w = 1 and z = 0.

So, we get A =\begin{bmatrix}w & x\\y & z\end{bmatrix}=\begin{bmatrix}1 & -2\\2 & 0\end{bmatrix}

(iii)\begin{bmatrix}4\\1\\3\end{bmatrix}A=\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}

Solution:

Let A =\begin{bmatrix}x & y & z\end{bmatrix}

Given equation is,

=>\begin{bmatrix}4\\1\\3\end{bmatrix}A=\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}

=>\begin{bmatrix}4\\1\\3\end{bmatrix}\begin{bmatrix}x & y & z\end{bmatrix}=\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}

=>\begin{bmatrix}4x & 4y & 4z\\x & y & z\\3x & 3y & 3z\end{bmatrix}=\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}

On comparing both sides, we get,

=> 4x = – 4, 4y = 8 and 4z = 4.

=> x = –1, y = 2 and z = 1.

So, we get A =\begin{bmatrix}-1 & 2 & 1\end{bmatrix}

(iv)A=\begin{bmatrix}2 & 1 & 3\end{bmatrix}\begin{bmatrix}-1 & 0 & -1\\-1 & 1 & 0\\0 & 1 &1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

Solution:

We have,



A=\begin{bmatrix}2 & 1 & 3\end{bmatrix}\begin{bmatrix}-1 & 0 & -1\\-1 & 1 & 0\\0 & 1 &1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

A =\begin{bmatrix}-2-1+0 & 0+1+3 & -2+0+3\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

A =\begin{bmatrix}-3 & 4 & 1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

A =\begin{bmatrix}-3+0-1\end{bmatrix}

A =\begin{bmatrix}-4\end{bmatrix}

(v)\begin{bmatrix}2 & -1\\1 & 0\\-3 & -4\end{bmatrix}A=\begin{bmatrix}-1 & -8 & -10\\1 & -2 & -5\\9 & 22 & 15\end{bmatrix}

Solution:

Let A =\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}

Given equation is,

=>\begin{bmatrix}2 & -1\\1 & 0\\-3 & -4\end{bmatrix}A=\begin{bmatrix}-1 & -8 & -10\\1 & -2 & -5\\9 & 22 & 15\end{bmatrix}

=>\begin{bmatrix}2 & -1\\1 & 0\\-3 & -4\end{bmatrix}\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}=\begin{bmatrix}-1 & -8 & -10\\1 & -2 & -5\\9 & 22 & 15\end{bmatrix}

=>\begin{bmatrix}2x-a & 2y-b & 2z-c\\x & y & z\\-3x+4a & -3y+4b & -3z+4c\end{bmatrix}=\begin{bmatrix}-1 & -8 & -10\\1 & -2 & -5\\9 & 22 & 15\end{bmatrix}

On comparing both sides, we get,

=> x = 1, y = –2 and z = –5

And also we have,

2x – a = –1

2y – b = –8

2z – c = –10

On solving these, we get,

=> a = 3, b = 4 and c = 0.

So, we get A =\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}=\begin{bmatrix}1 & -2 & -5\\3 & 4 & 0\end{bmatrix}

(vi)A\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\\11 & 10 & 9\end{bmatrix}

Solution:

Let A =\begin{bmatrix}x & a\\y & b\\z & c\end{bmatrix}

Given equation is,

=>A\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\\11 & 10 & 9\end{bmatrix}

=>\begin{bmatrix}x & a\\y & b\\z & c\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\\11 & 10 & 9\end{bmatrix}

=>\begin{bmatrix}x+4a & 2x+5a & 3x+6a\\y+4b & 2y+5b & 3y+6b\\z+4c & 2z+5c & 3z+6c\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\\11 & 10 & 9\end{bmatrix}

On comparing both sides, we get

x + 4a = –7 and 2x + 5a = –8



=> x = 1 and a = –2

y + 4b = 2 and 2y + 5b = 4

=> b = 0 and y = 2

z + 4c = 11 and 2z + 5c = 10

=> c = 4 and z = –5

So, we get A =\begin{bmatrix}1 & -2\\2 & 0\\-5 & 4\end{bmatrix}

Question 49. Find a 2 × 2 matrix A such thatA\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix} = 6I2.

Solution:

Let A =\begin{bmatrix}w & x\\y & z\end{bmatrix}

Given equation is,

=>A\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix} = 6I

=>\begin{bmatrix}w & x\\y & z\end{bmatrix}\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix}=6\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=>\begin{bmatrix}w & x\\y & z\end{bmatrix}\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix}=\begin{bmatrix}6 & 0\\0 & 6\end{bmatrix}

=>\begin{bmatrix}w+x & -2w+4x\\y+z & -2y+4z\end{bmatrix}=\begin{bmatrix}6 & 0\\0 & 6\end{bmatrix}

On comparing both sides, we get

w + x = 6 and –2w + 4x = 0

=> w = 4 and x = 2

y + z = 0 and –2y + 4z = 6

=> y = –1 and z = 1

So, we get A =\begin{bmatrix}w & x\\y & z\end{bmatrix}=\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}

Question 50. If A =\begin{bmatrix}0 & 0\\4 & 0\end{bmatrix} , find A16.

Solution:

We have,

A =\begin{bmatrix}0 & 0\\4 & 0\end{bmatrix}

A2 =\begin{bmatrix}0 & 0\\4 & 0\end{bmatrix}\begin{bmatrix}0 & 0\\4 & 0\end{bmatrix}

=\begin{bmatrix}0+0 & 0+0\\0+0 & 0+0\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

A16 = A2 A2 A2 A2

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

Question 51. If A =\begin{bmatrix}0 & -x\\x & 0\end{bmatrix} , B =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} and x2 = –1, then show that (A + B)2 = A2 + B2.

Solution:

We have,



A =\begin{bmatrix}0 & -x\\x & 0\end{bmatrix} , B =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} and x2 = –1

L.H.S. = (A + B)2

=\left(\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}+\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\right)^2

=\left(\begin{bmatrix}0+0 & -x+1\\x+1 & 0+0\end{bmatrix}\right)^2

=\left(\begin{bmatrix}0 & -x+1\\x+1 & 0\end{bmatrix}\right)^2

=\begin{bmatrix}0 & -x+1\\x+1 & 0\end{bmatrix}\begin{bmatrix}0 & -x+1\\x+1 & 0\end{bmatrix}

=\begin{bmatrix}0+(1-x)(1+x)& 0+0\\0+0 & (x+1)(1-x)+0\end{bmatrix}

=\begin{bmatrix}1-x^2 & 0\\0 & 1-x^2\end{bmatrix}

=\begin{bmatrix}1-(-1) & 0\\0 & 1-(-1)\end{bmatrix}

=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

R.H.S. = A2 + B2

=\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}+\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}

=\begin{bmatrix}0-x^2 & 0+0\\0+0 & -x^2+0\end{bmatrix}+\begin{bmatrix}0+1 & 0+0\\0+0 & 1+0\end{bmatrix}

=\begin{bmatrix}-x^2 & 0\\0 & -x^2\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}1-x^2 & 0\\0 & 1-x^2\end{bmatrix}

=\begin{bmatrix}1-(-1) & 0\\0 & 1-(-1)\end{bmatrix}

=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

= L.H.S.

Hence proved.

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :