How to Find Position of an Element in a Matrix? | Examples on Elements Position in a Matrix
A matrix is the rectangular array of m x n numbers in the form of rows and columns. Those numbers are enclosed by [] or (). The order of the matrix m and n is written in the form m x n which means m no of rows, n no of columns. The numbers in the matrix are called the elements of the matrix. We can learn about the position of an element in a matrix on this page along with solved questions.
What are Elements in Matrix?
Entries in a matrix are called the elements of the matrix. The address or position of one element in the matrices is given by listing the rows number and column number. While learning the element position in a matrix, you also have to learn the order of matrices which defines the number of rows, elements in a matrix.
The order of a two-dimensional matrix is the number of rows followed by the number of columns. If a matrix has ‘m’ number of rows and ‘n’ number of columns, then its size or order is m x n and it is read as m by n.
Consider a matrix \( A =\left[
\begin{matrix}
10 & 4 & 6\cr
5 & 1 & 3 \cr
2 & 7 & 8 \cr
\end{matrix}
\right]
\)
The order of the above matrix A is 3 x 3 as it has 3 rows and 3 columns.
The elements of A are 10, 4, 6, 5, 1, 3, 2, 7, and 3.
Step by Step Process to Find Position of Elements in a Matrix
The following are the steps to get the position of an element in a matrix. Go through these simple steps and find the address easily.
- Take any matrix of any order to get the position of elements.
- Know the column number and row number of a particular element by counting it.
- The position of an element is the row number of the element followed by the column number.
- The way to represent the position of an element in matrices are (row number, column number) or Matrix_Nameij.
Example on Position of an Element in a Matrix
Let us consider a matrix \( Z =\left[
\begin{matrix}
1 & 3 & 7\cr
2 & 8 & 15 \cr
11 & 21 & 26 \cr
\end{matrix}
\right]
\)
The elements in the first row are 1, 3, 7, the second row is 2, 8, 15 and the third row are 11, 21, 26.
The position of 15 is this element falls on 2nd row and 3rd column.
So, 15 is (2, 3)th element of Z.
Problems on Address of Elements in Matrices
Problem 1:
Find the position of element 12 in \( X =\left[
\begin{matrix}
8 & 17 \cr
4 & 12 \cr
5 & 13 \cr
\end{matrix}
\right]
\)
Solution:
Given matrix is \( X =\left[
\begin{matrix}
8 & 17 \cr
4 & 12 \cr
5 & 13 \cr
\end{matrix}
\right]
\)
The element 12 falls on 2nd row, 2nd column.
So, the 12 is (2,2)th element of X.
Problem 2:
Find the position of elements 2, 1, 18 in \( Y =\left[
\begin{matrix}
16 & 18 \cr
14 & 7 \cr
4 & 2 \cr
1 & 3 \cr
\end{matrix}
\right]
\)
Solution:
Given matrix is \( Y =\left[
\begin{matrix}
16 & 18 \cr
14 & 7 \cr
4 & 2 \cr
1 & 3 \cr
\end{matrix}
\right]
\)
Element 2 falls on the 3rd row, 2nd column. So, 2 is (2, 3)th element of Y.
Element 1 falls on 4th row, 1st column. So, 1 is (4, 1)th element
Element 18 falls on 1st row, 2nd column. So, 18 is (1, 2)th element.
Problem 3:
Find the position of all elements in the matrix \( A =\left[
\begin{matrix}
1 & 10 & 20 \cr
3 & 13 & 17 \cr
7 & 21 & 28 \cr
\end{matrix}
\right]
\)
Solution:
Given matrix is \( A =\left[
\begin{matrix}
1 & 10 & 20 \cr
3 & 13 & 17 \cr
7 & 21 & 28 \cr
\end{matrix}
\right]
\)
Here, element 1 falls on row number 1 and column 1.
We say, 1 is the (1, 1)th element. Similarly,
(1, 2)th element = 10
(1, 3)th element = 20
(2, 1)th element = 3
(2, 2)th element = 13
(2, 3)th element = 17
(3, 1)th element = 7
(3, 2)th element = 21
(3, 3)th element = 28
Frequently Asked Question’s
1. How to determine the position o an element in a matrix?
The position of an element in the matrix is determined by checking the row number, column number where it falls. The row number, column number is the exact address of the element.
2. What is the order of the matrix?
The order of a matrix is the size of the matrix. It is the number of rows x number of columns in it.
3. What are elements in the matrix?
Every number in a matrix are called the element of a matrix. The number of elements in the matrix is found by multiplying the number of rows by the number of columns.
4. What is element A23 in the matrix?
A23 is nothing the element or number or entry which is present at 2nd row and 3rd column of the matrix A.
Probability and Playing Cards – Examples | Solved Questions on Playing Cards Probability
In maths, the probability is nothing but the chance of occurrence of an event successfully. Examples of probability include tossing a coin, rolling a die, playing cards and so on. This probability and playing cards page contains the example questions and how to calculate the probability of playing cards.
Basics About Playing Cards Probability
In a pack of playing cards, we can see 52 cards which are divided into 4 suits having each 13 cards. The shape names of the suits are spades, diamonds, hearts and clubs. Again these suits are available in two different colours like red and black. The color of diamonds and hearts in red and spades, clubs color is black.
Each suit have Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3, 2. Based on the definition, the formula to calculate the probability is here:
Probability = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \) (or)
P(E) = \(\frac { N(E) }{ N(S) } \)
For playing cards, N(S) is always 52.
Also, Check
How to Calculate Probability and Playing Cards?
Below mentioned are the simple steps to find the probability of playing cards easily.
- Find the number of favorable events.
- We already know that a total number of possible outcomes for playing cards is 52.
- So, divide the number of favorable events by 52 to get the answer.
Worked out Problems on Probability and Playing Cards
Problem 1:
A card is drawn from a pack of 52 cards. What is the probability that the drawn card is king?
Solution:
Let E be the event of drawing a king card.
There are 4 king cards in playing cards.
Number of favorable outcomes N(E) = 4
Total number of outcomes N(S) = 52
Probability of drawing a king from pack of cards = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 4 }{ 52 } \) = \(\frac { 1 }{ 13 } \)
Problem 2:
A card is drawn at random from a well-shuffled pack of cards numbered 1 to 20. Find the probability of
(i) getting a number less than 5
(ii) getting a number divisible by 3
Solution:
Total number of possible outcomes N(S) = 20
(i) Number of favorable outcomes = Number of cards showing less than 5 = 4 i.e {1, 2, 3, 4}
Probability of getting a number less than 5 P(E) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 4 }{ 20 } \) = \(\frac { 1 }{ 5 } \)
(ii) Number of favorable outcomes = Number of cards showing a number divisible by 3 = 6 i.e {3, 6, 9, 12, 15, 18}
Probability of getting a number divisible by 3 P(E) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 6 }{ 20 } \) = \(\frac { 3 }{ 10 } \)
Problem 3:
All kings, jacks, diamonds have been removed from a pack of 52 playing cards and the remaining cards are well shuffled. A card is drawn from the remaining pack. Find the probability that the card drawn is:
(i) a red queen
(ii) a face card
(iii) a black card
(iv) a heart
Solution:
Number of kings in a pack of 52 cards = 4
Number of jacks in a pack of 52 cards = 4
Number of diamonds in a pack of 52 cards = 13
Total number of removed cards = (4 + 4 + 11) = 19 cards
[Excluding the diamond king and jack there are 11 diamonds]
Remaining total cards after removing kings, jacks and diamonds = 52 – 19 = 33
(i)
Queen of heart and queen of diamond are two red queens
The queen of diamond is already removed.
So, there is 1 red queen out of 33 cards
Therefore, the probability of getting ‘a red queen’ P(A) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 1 }{ 33 } \)
(ii)
Number of face cards after removing all kings, jacks, diamonds = 3
Therefore, the probability of getting ‘a face card’ P(B) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 3 }{ 33 } \) = \(\frac { 1 }{ 11 } \)
(iii)
Cards of spades and clubs are black cards.
Number of spades = 13 – 2 = 11, since king and jack are removed
Number of clubs = 13 – 2 = 11, since king and jack are removed
Therefore, in this case, total number of black cards = 11 + 11 = 22
The probability of getting ‘a black card’ P(C) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 22 }{ 33 } \) = \(\frac { 2 }{ 3 } \)
(iv)
Number of hearts = 13
Therefore, in this case, total number of hearts = 13 – 2 = 11, since king and jack are removed
Therefore, the probability of getting ‘a heart card’ P(D) = \(\frac { Number of favorable outcomes }{ Total number of outcomes } \)
= \(\frac { 11 }{ 33 } \) = \(\frac { 1 }{ 3 } \).
Frequently Asked Question’s
1. How are 52 cards divided?
A standard pack of playing cards have 52 cards. All those cards are divided into two colors red & black. Deck of cards contains 4 suits “spades”, “hearts”, “clubs”, “diamonds”. Each suit has 13 cards Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King. Hearts & diamonds are in red color and spades, clubs are in black color.
2. What is the probability of getting a number card from a deck of 52 cards?
Total number of cards = 52
Number of number cards in pack = 36
Probability of getting a number card = \(\frac { 36 }{ 52 } \) = \(\frac { 9 }{ 13 } \)
3. What are the 4 types of cards?
The 4 different types of cards are clubs, diamonds, hearts and spades.
4. How many kings are in a pack of cards?
A total of 4 kings are there in a pack of cards.
Eureka Math Geometry Module 2 Lesson 26 Answer Key
Engage NY Eureka Math Geometry Module 2 Lesson 26 Answer Key
Eureka Math Geometry Module 2 Lesson 26 Exercise Answer Key
Exercise 1.
Identify the \(\frac{\text { opp }}{\text { hyp }}\) ratios for ∠A and ∠B.
Answer:
For ∠A: \(\frac{12}{13}\)
For ∠B: \(\frac{5}{13}\)
Exercise 2.
Identify the \(\frac{\text { adj }}{\text { hyp }}\) ratios for ∠A and ∠B.
Answer:
For ∠A: \(\frac{5}{13}\)
For ∠B: \(\frac{12}{13}\)
Exercise 3.
Describe the relationship between the ratios for ∠A and ∠B.
Answer:
The \(\frac{\text { opp }}{\text { hyp }}\) ratio for ∠A is equal to the \(\frac{\text { adj }}{\text { hyp }}\) ratio for ∠B.
The \(\frac{\text { opp }}{\text { hyp }}\) ratio for ∠B is equal to the \(\frac{\text { adj }}{\text { hyp }}\) ratio for ∠A.
Exercise 4.
In ∆ PQR, m∠P = 53.2° and m∠Q = 36.8°. Complete the following table.
Answer:
Exercise 5.
In the triangle below, m∠A = 33.7° and m∠B = 56.3°. Complete the following table.
Answer:
Exercise 6.
In the triangle below, let e be the measure of ∠E and d be the measure of ∠D. Complete the following table.
Answer:
Exercise 7.
In the triangle below, let x be the measure of ∠X and y be the measure of ∠Y. Complete the following table.
Answer:
Exercise 8.
Tamer did not finish completing the table below for a diagram similar to the previous problems that the teacher had on the board where p was the measure of ∠P and q was the measure of ∠Q. Use any patterns you notice from Exercises 1-4 to complete the table for Tamer.
Answer:
Exercise 9.
Explain how you were able to determine the sine, cosine, and tangent of ∠Q in Exercise 8.
Answer:
I was able to complete the table for Tamer by observing the patterns of previous problems. For example, I noticed that the sine of one angle was always equal to the ratio that represented the cosine of the other angle. Since I was given sin p, I knew the ratio \(\frac{11}{\sqrt{157}}\) would be the cos q. Similarly, cos p = sin q = \(\frac{6}{\sqrt{157}}\) Finally, I noticed that the tangents of the angles were always reciprocals of each other. Since i was given the tan p = \(\frac{11}{6}\) knew that the tan q must be equal to \(\frac{6}{11}\).
Eureka Math Geometry Module 2 Lesson 26 Problem Set Answer Key
Question 1.
Given the triangle in the diagram, complete the following table.
Answer:
Question 2.
Given the table of values below (not in simplest radical form), label the sides and angles in the right triangle.
Answer:
Question 3.
Given sin α and sin β, complete the missing values in the table. You may draw a diagram to help you.
Answer:
Question 4.
Given the triangle shown to the right, fill in the missing values in the table.
Answer:
Using the Pythagorean theorem:
hyp2 = 22 + 62
hyp2 = 4 + 36
hyp2 = 40
hyp = √40
hyp = 2√10
Question 5.
Jules thinks that if α and β are two different acute angle measures, then sin α ≠ sin β. Do you agree or disagree? Explain.
Answer:
I agree. If α and β are different acute angle measures, then either α > β or β > α. A right triangle with acute angle a cannot be similar to a right triangle with acute angle f? (unless α + β = 90) because the triangles fail the AA criterion. If the triangles are not similar, then their corresponding sides are not in proportion, meaning their within-figure ratios are not in proportion; therefore, sin α ≠ sin β. In the case where α + β = 90, the given right triangles are similar; however, a and fi must be alternate acute angles, meaning sin α = cos β, and sin β = cos α, but sin α ≠ sin β.
Question 6.
Given the triangle in the diagram, complete the following table.
Answer:
Rewrite the values from the table in simplest terms.
Answer:
Draw and label the sides and angles of a right triangle using the values of the ratios sin and cos. How is the new triangle related to the original triangle?
Answer:
The triangles are similar by SSS criterion because the new triangle has sides that are \(\frac{1}{3}\) of the length of their corresponding sides in the original triangle.
Question 7.
Given tan α and cos β, in simplest terms, find the missing side lengths of the right triangle if one leg of the triangle has a length of 4. Draw and label the sides and angles of the right triangle.
Answer:
The problem does not specify which leg is 4, so there are two possible solutions to this problem. The values given in the table do not represent the actual lengths of the sides of the triangles; however, they do represent the lengths of the sides of a similar triangle, which is a 30 – 60 – 90 right triangle with side lengths 1, 2, and √3.
Case 1: The short leg of the right triangle is 4:
Case 2: The long leg of the right triangle is 4:
Question 8.
Eric wants to hang a rope bridge over a small ravine so that it is easier to cross. To hang the bridge, he needs to know how much rope Is needed to span the distance between two trees that are directly across from each other on either side of the ravine. Help Eric devise a plan using sine, cosine, and tangent to determine the approximate distance from tree A to tree B without having to cross the ravine.
Answer:
Student solutions will vary. Possible solution:
If Eric walks a path parallel to the ravine to a point P at a convenient distance from A, he could measure the angle formed by his line of sight to both trees. Using the measured angle and distance, he could use the value of the tangent ratio of the angle to determine the length of the opposite leg of the triangle. The length of the opposite leg of the triangle represents the distance between the two trees.
Question 9.
A fisherman is at point F on the open sea and has three favorite fishing locations. The locations are indicated by points A, B, and C. The fisherman plans to sail from F to A, then toB, then to C, and then back to F. If the fisherman is 14 miles from \(\overline{A C}\), find the total distance that he will sail.
Answer:
FP = 14 and can be considered the adjacent side to the 35° angle shown in triangle APF.
Using cosine:
cos 35 = \(\frac{14}{A F}\)
AF = \(\frac{14}{\cos 35}\)
AF ≈ 17.09
Using tangent:
tan 35 = \(\frac{A P}{14}\)
AP = 14 tan 35
AP ≈ 9. 8029
\(\overline{P C}\) is the leg opposite angle PFC in triangle PFC and has a degree measure of 42.5.
Using tangent:
tan 42.5 = \(\frac{P C}{14}\)
PC = 14 tan 42.5
PC ≈ 12.8286
Using cosine:
cos 42.5 = \(\frac{14}{F C}\)
FC = \(\frac{14}{\cos 42.5}\)
FC ≈ 18. 9888
The total distance that the fisherman will sail:
distance = AF + AP + PC + FC
distance = \(\frac{14}{\cos 35}\) +14 tan 35 + 14 tan 42.5 + \(\frac{14}{\cos 42.5}\)
distance ≈ 58.7
The total distance that the fisherman will sail is approximately 58.7 miles.
Eureka Math Geometry Module 2 Lesson 26 Exit Ticket Answer Key
Question 1.
Given the diagram of the triangle, complete the following table.
Answer:
a. Which values are equal?
Answer:
sin s = cos t and cos s = sin t
b. How are tan s and tan t related?
Answer:
They are reciprocals: \(\frac{5}{6} \cdot \frac{6}{5}\) = 1.
Question 2.
If u and y are the measures of complementary angles such that sin u = \(\frac{2}{5}\) and tan v = \(\frac{\sqrt{21}}{2}\) label the sides and angles of the right triangle In the diagram below with possible side lengths:
Answer:
A possible solution is shown below; however, any similar triangle having a shorter leg with length of 2x, longer leg with length of x√21, and hypotenuse with length of 5x, for some positive number x, is also correct.
Skip Counting by 100S
Skip Counting by 25S
