Question 1

if product of rows and column of matrix is 27 , then number of possible different ordered matrices are

Accepted Answer

4

Question 2

Horizontally arranged elements in a matrix are called

Accepted Answer

rows

Question 3

who is father of matrices?

Accepted Answer

Arthur Cayley

Question 4

Find area bounded by y=|x|, x=-1, x=2 and x-axis.

Accepted Answer

5/2

Question 5

Find the area bounded by x+y=4 , x-axis, x=1 and x=2

Accepted Answer

5/2

Question 6

Find the area of triangle bounded by 3x - 2y + 1 = 0, 2x + 3y = 21 and x - 5y + 9 = 0

Accepted Answer

13/2

Question 7

Area between the curves y = x 2 and y 2 = x

Accepted Answer

1/3

Question 8

Find the area of the region bounded between the line x = 4 and parabola y 2 = 16x.

Accepted Answer

128/3

Question 9

In the matrix A = \begin{bmatrix} 2 & 5 & 19 & -7 \ 35 & -2 & 12 \ 3 & 1 & -5 & 17 \end{bmatrix}, write:
(i) The order of the matrix,
(ii) The number of elements,
(iii) Write the elements a13, a21, a33, a24, a23.

Accepted Answer

Solution:
(i) The order of the matrix is the number of rows × number of columns. Here, the matrix has 3 rows and 4 columns, so order is 3 × 4.

(ii) The number of elements = number of rows × number of columns = 3 × 4 = 12.

(iii) Elements:
- a13 means element in 1st row and 3rd column = 19
- a21 means element in 2nd row and 1st column = 35
- a33 means element in 3rd row and 3rd column = -5
- a24 means element in 2nd row and 4th column = Since 2nd row has only 3 elements, a24 does not exist.
- a23 means element in 2nd row and 3rd column = 12 — The order is determined by counting rows and columns. Number of elements is product of rows and columns. Elements are identified by their row and column indices.

Question 10

If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Accepted Answer

Solution:
Number of elements = number of rows × number of columns.

For 24 elements, possible orders (m × n) are pairs of positive integers whose product is 24:
1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 8 × 3, 12 × 2, 24 × 1.

For 13 elements, since 13 is prime, possible orders are:
1 × 13 or 13 × 1. — The order of a matrix is determined by factors of the total number of elements.

Question 11

If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Accepted Answer

Solution:
Number of elements = rows × columns.

For 18 elements, possible orders are pairs of positive integers whose product is 18:
1 × 18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1.

For 5 elements (5 is prime), possible orders are:
1 × 5 or 5 × 1. — The order is determined by factor pairs of the total number of elements.

Question 12

Construct a 2 × 2 matrix, A = [aij], whose elements are given by:
(i) aij = (i + j)^2
(ii) aij = i
(iii) aij = (i + 2j)^2

Accepted Answer

Solution:
For a 2 × 2 matrix, i = 1, 2 and j = 1, 2.

(i) aij = (i + j)^2
- a11 = (1+1)^2 = 2^2 = 4
- a12 = (1+2)^2 = 3^2 = 9
- a21 = (2+1)^2 = 3^2 = 9
- a22 = (2+2)^2 = 4^2 = 16
Matrix A = \begin{bmatrix} 4 & 9 \ 9 & 16 \end{bmatrix}

(ii) aij = i
- a11 = 1
- a12 = 1
- a21 = 2
- a22 = 2
Matrix A = \begin{bmatrix} 1 & 1 \ 2 & 2 \end{bmatrix}

(iii) aij = (i + 2j)^2
- a11 = (1 + 2*1)^2 = (1+2)^2 = 3^2 = 9
- a12 = (1 + 2*2)^2 = (1+4)^2 = 5^2 = 25
- a21 = (2 + 2*1)^2 = (2+2)^2 = 4^2 = 16
- a22 = (2 + 2*2)^2 = (2+4)^2 = 6^2 = 36
Matrix A = \begin{bmatrix} 9 & 25 \ 16 & 36 \end{bmatrix} — Substitute values of i and j in the given formula for each element to construct the matrix.

Question 13

Construct a 3 × 4 matrix, whose elements are given by:
(i) aij = |−3i + j|
(ii) aij = 2i − j

Accepted Answer

Solution:
For a 3 × 4 matrix, i = 1, 2, 3 and j = 1, 2, 3, 4.

(i) aij = |−3i + j|
Calculate each element:
- i=1: a11=|−3*1+1|=|−3+1|=2, a12=|−3+2|=1, a13=|−3+3|=0, a14=|−3+4|=1
- i=2: a21=|−6+1|=5, a22=|−6+2|=4, a23=|−6+3|=3, a24=|−6+4|=2
- i=3: a31=|−9+1|=8, a32=|−9+2|=7, a33=|−9+3|=6, a34=|−9+4|=5
Matrix:
\begin{bmatrix} 2 & 1 & 0 & 1 \ 5 & 4 & 3 & 2 \ 8 & 7 & 6 & 5 \end{bmatrix}

(ii) aij = 2i − j
Calculate each element:
- i=1: a11=2*1−1=1, a12=2*1−2=0, a13=2*1−3=−1, a14=2*1−4=−2
- i=2: a21=4−1=3, a22=4−2=2, a23=4−3=1, a24=4−4=0
- i=3: a31=6−1=5, a32=6−2=4, a33=6−3=3, a34=6−4=2
Matrix:
\begin{bmatrix} 1 & 0 & -1 & -2 \ 3 & 2 & 1 & 0 \ 5 & 4 & 3 & 2 \end{bmatrix} — Substitute values of i and j in the given formulas and compute each element.

Question 14

Find the values of x, y and z from the following equations:
(i) \begin{bmatrix} x + y + z \ x \end{bmatrix} = \begin{bmatrix} 9 \ 4 \end{bmatrix}
(ii) \begin{bmatrix} y & z \ 1 & 5 \end{bmatrix} = \begin{bmatrix} x + y & 2 \ 6 & 2 \end{bmatrix}
(iii) \begin{bmatrix} x + z \ 5 + z \ xy \ y + z \end{bmatrix} = \begin{bmatrix} 5 \ 8 \ 5 \ 7 \end{bmatrix}

Accepted Answer

Solution:
(i) \begin{bmatrix} x + y + z \ x \end{bmatrix} = \begin{bmatrix} 9 \ 4 \end{bmatrix}
From the second row: x = 4
From the first row: x + y + z = 9
Substitute x=4: 4 + y + z = 9 => y + z = 5

(ii) \begin{bmatrix} y & z \ 1 & 5 \end{bmatrix} = \begin{bmatrix} x + y & 2 \ 6 & 2 \end{bmatrix}
Equate corresponding elements:
First row, first column: y = x + y => x = 0
First row, second column: z = 2
Second row, first column: 1 = 6 (contradiction)
Second row, second column: 5 = 2 (contradiction)
Since the matrices are equal, the given matrices cannot be equal as per the second row elements. Possibly a typo.

(iii) \begin{bmatrix} x + z \ 5 + z \ xy \ y + z \end{bmatrix} = \begin{bmatrix} 5 \ 8 \ 5 \ 7 \end{bmatrix}
From first element: x + z = 5
Second element: 5 + z = 8 => z = 3
Third element: x y = 5
Fourth element: y + z = 7
Substitute z=3 in first: x + 3 = 5 => x = 2
Substitute z=3 in fourth: y + 3 = 7 => y = 4
Check third: x y = 2 × 4 = 8, but given 5, contradiction.
Hence, no consistent solution unless there is a typo. — Equate corresponding elements of matrices and solve the resulting equations stepwise.

Question 15

Find the value of a, b, c and d from the equation:
\begin{bmatrix} a - b & 2a + c \ 2a - b & 3c + d \end{bmatrix} = \begin{bmatrix} -1 & 5 \ 0 & 13 \end{bmatrix}

Accepted Answer

Solution:
Equate corresponding elements:
1) a - b = -1
2) 2a + c = 5
3) 2a - b = 0
4) 3c + d = 13

From (1) and (3):
(1) a - b = -1 => b = a + 1
(3) 2a - b = 0 => b = 2a
From above, a + 1 = 2a => 1 = a => a = 1
Then b = 2a = 2 × 1 = 2

From (2): 2a + c = 5 => 2 × 1 + c = 5 => c = 3
From (4): 3c + d = 13 => 3 × 3 + d = 13 => 9 + d = 13 => d = 4

Therefore, a=1, b=2, c=3, d=4. — Equate corresponding elements and solve the system of linear equations stepwise.

Matrices

Matrices — Study Notes

Introduction

Types of Matrices

Operations on Matrices

Practice Questions — Matrices

All 6 Chapters in Mathematics Part-I