Chapter 3 MATRICES EXERCISE 3.2
NCERT Solutions for Class 12 Maths – Chapter 3 – MATRICES EXERCISE 3.2
Answer 1. Matrix Operations
Given matrices:
A = [2 4] B = [1 3] C = [-2 5]
[3 2] [-2 5] [ 3 4]
(i) A + B:
A + B = [2+1 4+3] = [3 7]
[3-2 2+5] [1 7]
(ii) A - B:
A - B = [2-1 4-3] = [1 1]
[3+2 2-5] [5 -3]
(iii) 3A - C:
3A - C = 3*[2 4] - [-2 5] = [6 12] - [-2 5]
[3 2] [ 3 4] [9 6] [ 3 4]
= [6+2 12-5] = [8 7]
[9-3 6-4] [6 2]
(iv) AB:
AB = A * B = [2*1+4*(-2) 2*3+4*5] = [2-8 6+20] = [-6 26]
[3*1+2*(-2) 3*3+2*5] [3-4 9+10] [-1 19]
(v) BA:
BA = B * A = [1*2+3*3 1*4+3*2] = [2+9 4+6] = [11 10]
[-2*2+5*3 -2*4+5*2] [-4+15 -8+10] [11 2]
Answer 2. Matrix Addition
To compute the sum of the matrices:
[a b] [a b] [a+a b+b] [2a 2b]
[-b a] + [b a] = [-b+b a+a] = [0 2a]
So, the sum of the matrices is:
[2a 2b]
[0 2a]
(ii) Matrix Addition
To compute the sum of the matrices:
[a² + b² b² + c²] [2ab 2bc] [a² + b² + 2ab b² + c² + 2bc]
[a² + c² a² + b²] + [-2ac -2ab] = [a² + c² - 2ac a² + b² - 2ab]
So, the sum of the matrices is:
[a² + b² + 2ab b² + c² + 2bc]
[a² + c² - 2ac a² + b² - 2ab]
(iii) Matrix Addition
To compute the sum of the matrices:
[-1 4 -6] [12 7 6] [(-1)+12 4+7 (-6)+6]
[ 8 5 16] + [ 8 0 5] = [ 8+8 5+0 16+5 ]
[ 2 8 5] [ 3 2 4] [ 2+3 8+2 5+4 ]
So, the sum of the matrices is:
[11 11 0]
[16 5 21]
[ 5 10 9]
(iv) Matrix Addition
To compute the sum of the matrices:
[cos²(x) sin²(x)] [sin²(x) cos²(x)] [cos²(x) + sin²(x) sin²(x) + cos²(x)]
[sin²(x) cos²(x)] + [cos²(x) sin²(x)] = [sin²(x) + cos²(x) cos²(x) + sin²(x)]
Since cos²(x) + sin²(x) = 1 (from trigonometric identity), and sin²(x) + cos²(x) = 1 as well:
[1 1]
[1 1]
Answer 3.
(i)Matrix Multiplication
To compute the product of the matrices:
[a b] [a -b] [a*a + b*(-b) a*(-b) + b*a] [a² - b² -ab + ab]
[-b a] * [b a] = [-b*a + a*b (-b)*b + a*a] = [-ab + ab a² + b² ]
So, the product of the matrices is:
[a² - b² 0 ]
[0 a² + b²]
(ii) Matrix Multiplication
To compute the product of the matrices:
[1] [2 3 4]
[2] * =
[3]
= [(1*2) + (2*3) + (3*4)]
= [2 + 6 + 12]
= [20]
Since the first matrix is a column matrix and the second one is a row matrix, we perform dot product:
[20]
(iii) Matrix Multiplication
To compute the product of the matrices:
[1 -2] * [1 2 3]
[2 3] [2 3 1]
= [(1*1) + (-2*2) (1*2) + (-2*3) (1*3) + (-2*1)]
[(2*1) + (3*2) (2*2) + (3*3) (2*3) + (3*1) ]
= [1 + (-4) 2 + (-6) 3 + (-2) ]
[2 + 6 4 + 9 6 + 3 ]
So, the product of the matrices is:
[-3 -4 1]
[8 13 9]
(iv) Matrix Multiplication
To compute the product of the matrices:
[2 3 4] [1 -3 5] [(2*1) + (3*0) + (4*3) (2*(-3)) + (3*2) + (4*0) (2*5) + (3*4) + (4*5)]
[3 4 5] * [0 2 4] = [ (3*1) + (4*0) + (5*3) (3*(-3)) + (4*2) + (5*0) (3*5) + (4*4) + (5*5)]
[4 5 6] [3 0 5] [(4*1) + (5*0) + (6*3) (4*(-3)) + (5*2) + (6*0) (4*5) + (5*4) + (6*5)]
So, the product of the matrices is:
[2 + 0 + 12 -6 + 6 + 0 10 + 12 + 20]
[3 + 0 + 15 -9 + 8 + 0 15 + 16 + 25]
[4 + 0 + 18 -12 + 10 + 0 20 + 20 + 30]
Which simplifies to:
[14 0 42]
[18 -1 56]
[22 -2 70]
(v) Matrix Multiplication
To compute the product of the matrices:
[2 1] [1 0 1] [(2*1) + (1*(-1)) (2*0) + (1*2) (2*1) + (1*1)]
[3 2] * [-1 2 1] = [ (3*1) + (2*(-1)) (3*0) + (2*2) (3*1) + (2*1)]
[-1 1] [(-1*1) + (1*(-1)) (-1*0) + (1*2) (-1*1) + (1*1)]
So, the product of the matrices is:
[2 - 1 + 0 0 + 2 2 + 1]
[3 - 2 + 0 0 + 4 3 + 2]
[-1 + 1 + 0 0 + 2 -1 + 1]
Which simplifies to:
[1 2 3]
[1 4 5]
[0 2 0]
(vi) Matrix Multiplication
To compute the product of the matrices:
[3 -1 3] [2 -3] = [3*2 + (-1)*1 + 3*3 3*(-3) + (-1)*0 + 3*1]
[-1 0 2] * [1 0] = [-1*2 + 0*1 + 2*3 -1*(-3) + 0*0 + 2*1]
[3 1]
= [6 + (-1) + 9 -3 + 0 + 3]
[-2 + 0 + 6 3 + 0 + 2]
= [14 0]
[4 5]
Answer 4.
Matrix Operations
To compute the sum of matrices A and B:
[1 2 -3] [3 -1 2] [1+3 2+(-1) (-3)+2]
[5 0 2] + [4 2 5] = [5+4 0+2 2+5 ]
[1 -1 1] [2 0 3] [1+2 (-1)+0 1+3 ]
So, the sum of matrices A and B is:
[4 1 -1]
[9 2 7]
[3 -1 4 ]
To compute the difference of matrices B and C:
[3 -1 2] [4 1 2] [3-4 (-1)-1 2-2]
[4 2 5] - [0 3 2] = [4-0 2-3 5-2]
[2 0 3] [1 -2 3] [2-1 0-(-2) 3-3]
So, the difference of matrices B and C is:
[-1 -2 0]
[4 -1 3]
[1 2 0]
To verify that A + (B - C) = (A + B) - C:
First, compute A + (B - C):
(B - C) = [-1 -2 0]
[4 -1 3]
[1 2 0]
(A + (B - C)) = [1 2 -3] [-1 -2 0] [1+(-1) 2+(-2) (-3)+0]
[5 0 2] + [4 -1 3] = [5+4 0+(-1) 2+3]
[1 -1 1] [1 2 0] [1+1 (-1)+2 1+0 ]
= [0 0 -3]
[9 -1 5]
[2 1 1]
Next, compute (A + B) - C:
(A + B) = [4 1 -1]
[9 2 7]
[3 -1 4 ]
(A + B) - C = [4 1 -1] [4 1 -1] [4-4 1-1 (-1)-(-1)]
[9 2 7] - [0 3 2 ] = [9-0 2-3 7-2 ]
[3 -1 4 ] [1 -2 3 ] [3-1 (-1)-(-2) 4-3 ]
= [0 0 0]
[9 -1 5]
[2 1 1]
Since both matrices are equal, the verification is successful.
Answer 5.
(v) Matrix Operation
To compute the expression 3A - 5B:
First, compute 3A:
3A = 3 * [2/3 1 5/3] [3 * (2/3) 3 * 1 3 * (5/3)]
[1/3 2/3 4/3] = [3 * (1/3) 3 * (2/3) 3 * (4/3)]
[7/3 2 2/3] [3 * (7/3) 3 * 2 3 * (2/3)]
= [2 3 5]
[1 2 4]
[7 6 2]
Next, compute 5B:
5B = 5 * [2/5 3/5 1] [5 * (2/5) 5 * (3/5) 5 * 1]
[1/5 2/5 4/5] = [5 * (1/5) 5 * (2/5) 5 * (4/5)]
[7/5 6/5 2/5] [5 * (7/5) 5 * (6/5) 5 * (2/5)]
= [2 3 5]
[1 2 4]
[7 6 2]
Now, compute the expression 3A - 5B:
3A - 5B = [2 - 2 3 - 3 5 - 5]
[1 - 1 2 - 2 4 - 4]
[7 - 7 6 - 6 2 - 2]
= [0 0 0]
[0 0 0]
[0 0 0]
So, the result is:
[0 0 0]
[0 0 0]
[0 0 0]
Answer 6.
(vi) Matrix Operation
To simplify the expression:
cosθ * [cosθ sinθ] + sinθ * [sinθ -cosθ]
[-sinθ cosθ] [cosθ sinθ]
First, distribute the cosθ and sinθ:
cosθ * [cosθ sinθ] + sinθ * [sinθ -cosθ]
[-sinθ cosθ] [cosθ sinθ]
= [cos2θ cosθsinθ] + [sin2θ -cosθsinθ]
[-sinθcosθ cos2θ] [cosθsinθ sin2θ]
Now, combine the terms:
[cos2θ + sin2θ cosθsinθ - cosθsinθ]
[-sinθcosθ + cosθsinθ cos2θ + sin2θ]
Since cos2θ + sin2θ = 1 and cosθsinθ - cosθsinθ = 0, the expression simplifies to:
[1 0]
[0 1]
Answer 7.
(i) Matrix Operation
To find X and Y:
We are given two equations:
X + Y = [7 0]
[2 5]
X - Y = [3 0]
[0 3]
Let's add the two equations:
(X + Y) + (X - Y) = [7 0] + [3 0]
[2 5] [0 3]
This simplifies to:
2X = [10 0]
[2 8]
Dividing both sides by 2:
X = [5 0]
[1 4]
Now, let's subtract the second equation from the first:
(X + Y) - (X - Y) = [7 0] - [3 0]
[2 5] [0 3]
This simplifies to:
2Y = [4 0]
[2 2]
Dividing both sides by 2:
Y = [2 0]
[1 1]
So, X =
[5 0]
[1 4]
and Y =
[2 0]
[1 1]
(ii) Matrix Operation
To find X and Y:
We are given two equations:
2X + 3Y = [2 3]
[4 0]
3X + 2Y = [2 -2]
[-1 5]
Multiply the first equation by 3 and the second equation by 2 to make the coefficients of X the same:
6X + 9Y = [6 9]
[12 0]
6X + 4Y = [4 -4]
[-2 10]
Now, subtract the second equation from the first:
(6X + 9Y) - (6X + 4Y) = [6 9] - [4 -4]
[12 0] [-2 10]
This simplifies to:
5Y = [2 13]
[14 -10]
Dividing both sides by 5:
Y = [2/5 13/5]
[14/5 -2]
Now, substitute Y into the first equation to solve for X:
2X + 3 * [2/5 13/5] = [2 3]
[14/5 -2] [4 0]
2X + [6/5 39/5] = [2 3]
[42/5 -6] [4 0]
2X = [2 3] - [6/5 39/5]
[4 0] [42/5 -6]
2X = [4/5 -24/5]
[-22/5 6 ]
X = [2/5 -12/5]
[-11/5 3 ]
So, X =
[2/5 -12/5]
[-11/5 3 ]
and Y =
[2/5 13/5]
[14/5 -2]