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VOOZH | about |
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VOOZH | about |
Matrices are a fundamental concept in linear algebra, used extensively in mathematics, physics, engineering, computer science, and various other fields. A matrix is a rectangular array of numbers, symbols, or expressions arranged in rows and columns. The numbers or elements inside a matrix are enclosed within square brackets and can be used to represent systems of linear equations, transformations, and many real-world data structures. Operations such as addition, subtraction, multiplication, and finding determinants and inverses of matrices help solve complex mathematical problems efficiently.
(i) A + B
(ii) A - B
(iii) 3A - C
(iv) AB
(v) BA
Solution:
(i)
(ii)
(iii)
(iv)
(v)
Solution:
(i)
(ii)
(iii)
(iv)
Solution:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Solution:
Now we have to show A + (B - C) = (A + B) - C
L.H.S = R.H.S.
Hence, Proved
Solution:
Solution:
= 1 = identity matrix
Solution:
(i) Given:
Adding (1) and (2), we get
(ii) Given:
Now, multiply equation (1) by 2 and equation (2) by 3 we get
Subtracting equation (4) from (3), we get,
Solution:
Solution:
Given:
Equating corresponding entries, we have
2 + y = 5 and 2x + 2 = 8
y = 5 - 2 and 2(x + 1) = 8
y = 3 and x + 1 = 4
Therefore, y = 3 and x = 3
Solution:
Given:
On comparing both sides, we have
2x + 3 = 9 ⇒ 2x = 9 - 3 ⇒ 2x = 6 ⇒ x = 3
2z - 3 = 15 ⇒ 2z = 15 + 3 ⇒ 2z = 18 ⇒ z = 9
2y = 12 ⇒ y = 6
2t + 6 = 18 ⇒ 2t = 18 - 6 ⇒ 2t = 12 ⇒ t = 6
Therefore, x = 3, y = 6, z = 9, t = 6
Solution:
Given:
Equating corresponding entries, we have
2x - y = 10 -(1)
3x + y = 5 -(2)
Adding eq.(1) and (2), we have 5x = 15 ⇒ x = 3
Putting x = 3 in eq.(2)
9 + y = 5 ⇒ y = -4
Therefore, x = 3 and y = -4
Solution:
Given:
Equating corresponding entries, we have
3x = x + 4 ⇒ 2x = 4 ⇒ x = 2
and 3y = 6 + x + y
⇒ 2y = 6 + 2
⇒ 2y = 8
⇒ y = 4
and 3z = -1 + z + w ⇒ 2z - w = - 1 -(1)
and 3w = 2w + 3 ⇒ w = 3
Putting w = 3 in eq(i), 2z - 3 = -1
⇒ 2z = 2 ⇒ z = 1
Therefore, x = 2, y = 4, z = 1, w = 3
Solution:
= F(x + y)
= F(x) F(y) = F(x + y)
Solution:
(i) L.H.S =
R.H.S =
Therefore, from (1) and (2), we get
i.e. L.H.S. ≠ R.H.S
(ii) L.H.S =
Multiply both the matrices
R.H.S.=
Therefore,
L.H.S. ≠ R.H.S.
i.e.
Solution:
Solution:
= 0 (Zero matrix)
= R.H.S.
Hence Proved
Solution:
Given:
Equating corresponding entries, we have
3k - 2 = 1
3k = 3
k = 1
and 4k = 4
k = 1
and -4 = -2k - 2
2k = 2
k = 1
Therefore, k = 1
Solution:
L.H.S. = R.H.S.
Hence, Proved.
Solution:
Let invested in the first bond = Rs x
Then, the sum of money invested in the second bond = ₹(30000 – x)
It is given that the first bond pays 5% interest per year, and the second bond pays 7% interest per year.
Thus, in order to obtain an annual total interest of ₹1800, we get:
⇒ 5x/100 + 7(30000 − x)/100 = 1800
⇒ 5x + 210000 -7x = 180000
⇒ 210000 -2x = 180000
⇒ 2x = 210000 – 180000
⇒ 2x = 30000
⇒ x = 15000
Therefore, in order to obtain an annual total interest of ₹1800, the trust fund should invest ₹15000 in the first bond and the remaining ₹15000 in the second bond.
Hence, the amount invested in each type of the bonds can be represented in matrix form with each column corresponding to a different type of bond as:
X =
Hence, the interest obtained after one year can be expressed in matrix representation as:
⇒ 5x/100 + 7(30000 − x)/100 = 2000
⇒ 5x + 210000 − 7x = 200000
⇒ 210000 − 2x = 200000
⇒ 2x = 210000 – 200000
⇒ 2x = 10000
⇒ x = 5000
Therefore, in order to obtain an annual total interest of ₹2000, the trust fund should invest ₹5000 in the first bond and the remaining ₹(30000 − 5000) = ₹25000 in the second bond.
Solution:
Let the number of books as 1 × 3 matrix = B =
Let the selling prices of each book is a 3 × 1 matrix S =
Therefore, Total amount received by selling all books = BS =
Therefore, Total amount received by selling all the books = Rs 20,160
(A) k = 3, p = n (B) k is arbitrary, p = 2
(C) p is arbitrary, k = 3 (D) k = 2, p = 3
Solution:
Since, Matrices P and Y are of the orders p × k and 3 × k respectively.
Therefore, matrix PY will be defined if k = 3.
Then, PY will be of the order p × k = p × 3.
Matrices W and Y are of the orders n × 3 and 3 × k = 3 × 3 respectively.
As, the number of columns in W is equal to the number of rows in Y, Matrix WY is well-defined and is of the order n × 3.
Matrices PY and WY can be added only when their orders are the same.
Therefore, PY is of the order p × 3 and WY is of the order n × 3.
Thus, we must have p = n.
Therefore, k = 3 and p = n are the restrictions on n, k and p so that PY + WY will be defined.
Therefore, answer is (A)
(A) p × 2 (B) 2 × n
(C) n × 3 (D) p × n
Solution:
Matrix X is of the order 2 × n.
Therefore, matrix 7X is also of the same order.
Matrix Z is of order 2 × p = 2 × n -(∵ p = n)
Then, Matrix 5Z is also of the same order.
Now, both the matrices 7X and 5Z are of the order 2 × n.
Thus, matrix 7X – 5Z is well- defined and is of the order 2 × n.
Therefore, answer is (B)
Also Read,
Matrices are a vital tool for solving real-world problems, from representing data to performing transformations in graphics and solving systems of linear equations. Exercise 3.2 deepens the understanding of fundamental matrix operations, helping students learn how to add, subtract, and multiply matrices. These operations are crucial for many advanced applications, including physics simulations, computer graphics, data analysis, and solving linear equations. By mastering these concepts, students can approach more complex problems involving matrices with confidence.
