Algebra Ex 3.16-10th Std Maths-Book Back Question And Answer
1. In the
write (i) The number of elements
(ii) The order of the matrix
(iii) Write the elements a22, a23, a24, a34, a43, a44.
Answer:
(i) The number of elements is 16
(ii) The order of the matrix is 4 × 4
(iii) Elements corresponds to
Question 2.
If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
The possible orders of the matrix having 18 elements are
The possible orders of the matrix having 6 elements are
Question 3.
Construct a 3 × 3 matrix whose elements are given by
(i) aij = |i – 2j|
Answer:
aij = |i – 2j|
The general 3 × 3 matrices is
a11 = |1 – 2(1)| = |1 – 2| = | – 1| = 1
a12 = |1 – 2(2)| = |1 – 4| = | – 3| = 3
a13 = |1 – 2(3)| = |1 – 6| = | – 5| = 5
a21 = |2 – 2(1)| = |2 – 2| = 0 = 0
a22 = |2 – 2(2)| = |2 – 4| = | – 2| = 2
a23 = |2 – 2(3)| = |2 – 6| = | – 4| = 4
a31 = |3 – 2(1)| = |3 – 2| = | 1 | = 1
a32 = |3 – 2(2)| = |3 – 4| = | – 1 | = 1
a33 = |3 – 2(3)| = |3 – 6| = | – 3 | = 3
The required matrix
(ii) aij = (i+j)33
Answer:
a11 = (1+1)33 = 233 = 83
a12 = (1+2)33 = 273 = 9
a13 = (1+3)33 = 643 = 643
a21 = (2+1)33 = 273 = 9
a22 = (2+2)33 = 643 = 643
a23 = (2+3)33 = 1253 = 1253
a31 = (3+1)33 = 643 = 643
a32 = (3+2)33 = 1253 = 1253
a33 = (3+3)33 = 2163 = 72
The required matrix
Question 4.
If 
then find the tranpose of A.
Answer:
transpose of A = (AT)
Question 5
If then find the tranpose of – A
Answer:
Transpose of – A = (-AT) = 
Question 6.
If A =
then verify (AT)T = A
Answer:
Hence it is verified
Question 7.
Find the values of x, y and z from the following equations
(i)
Answer:
Since the given matrices are equal then all the corresponding elements are equal.
y = 12, z = 3, x = 3
The value of x = 3, y = 12 and z = 3
(ii)
Answer:
x + y = 6 ……(1)
5 + z = 5
z = 5 – 5 = 0
xy = 8
y = 8x
Substitute the value of y = 8x in (1)
x + 8x = 6
x2 + 8 = 6x
x2 – 6x + 8 = 0
(x – 4) (x – 2) = 0
∴ x – 4 = 0 or x – 2 = 0
x = 4 or x = 2
y = 84 = 2 or y = 82 = 4
∴ The value of x, y and z are 4, 2, 0 (or) 2, 4, 0
(iii)
Solution:
x + y + z = 9 ……….(1)
x + z = 5 ……….(2)
y + z = 7 ……….(3)
Substitute the value of y = 4 in (3)
y + z = 7
4 + z = 7
z = 7 – 4
= 3
Substitute the value of z = 3 in (2)
x + 3 = 5
x = 5 – 3
= 2
∴ The value of x = 2 , y = 4 and z = 3
