Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.16

Question 1. In the matrix \( A = \begin{pmatrix} 8 & 9 & 4 & 3 \\ -1 & \sqrt{7} & \frac{\sqrt{3}}{2} & 5 \\ 1 & 4 & 3 & 0 \\ 6 & 8 & -11 & 1 \end{pmatrix} \), write (i) The number of elements (ii) The order of the matrix (iii) The elements \( a_{22}, a_{23}, a_{24}, a_{34}, a_{43}, a_{44} \).
Answer:
(i) The total number of elements in the matrix is 16. This is found by multiplying the number of rows by the number of columns.
(ii) The matrix has 4 rows and 4 columns, so its order is \( 4 \times 4 \).
(iii) The specific elements corresponding to their positions are:
\( a_{22} = \sqrt{7} \) (element in the 2nd row, 2nd column)
\( a_{23} = \frac{\sqrt{3}}{2} \) (element in the 2nd row, 3rd column)
\( a_{24} = 5 \) (element in the 2nd row, 4th column)
\( a_{34} = 0 \) (element in the 3rd row, 4th column)
\( a_{43} = -11 \) (element in the 4th row, 3rd column)
\( a_{44} = 1 \) (element in the 4th row, 4th column)
In simple words: This matrix has 16 items inside it, arranged in 4 rows and 4 columns. We also listed some specific items by their row and column numbers, matching each number to its exact place in the matrix.

🎯 Exam Tip: To find the number of elements, multiply the number of rows by the number of columns. To find a specific element \( a_{ij} \), look for the item located at the intersection of the i-th row and j-th column.

Question 2. If a matrix has 18 elements, what are the possible orders it can have? What if it has 6 elements?
Answer:
If a matrix has 18 elements, its possible orders are all pairs of numbers that multiply to 18. These are:
\( 1 \times 18 \) (1 row, 18 columns)
\( 2 \times 9 \) (2 rows, 9 columns)
\( 3 \times 6 \) (3 rows, 6 columns)
\( 6 \times 3 \) (6 rows, 3 columns)
\( 9 \times 2 \) (9 rows, 2 columns)
\( 18 \times 1 \) (18 rows, 1 column)

If a matrix has 6 elements, its possible orders are:
\( 1 \times 6 \) (1 row, 6 columns)
\( 2 \times 3 \) (2 rows, 3 columns)
\( 3 \times 2 \) (3 rows, 2 columns)
\( 6 \times 1 \) (6 rows, 1 column)
In simple words: A matrix's order tells us its size by stating how many rows and columns it has. To find all possible orders for a given total number of elements, you need to list all the pairs of whole numbers that multiply to that total.

🎯 Exam Tip: Remember that an order of \( m \times n \) (m rows, n columns) is generally considered different from \( n \times m \) (n rows, m columns) for matrix description, as they represent different shapes.

Question 3. Construct a \( 3 \times 3 \) matrix whose elements are given by
(i) \( a_{ij} = |i - 2j| \)
(ii) \( a_{ij} = \frac{(i+j)^{3}}{3} \)
Answer:
(i) To construct a \( 3 \times 3 \) matrix where each element \( a_{ij} \) is given by the formula \( |i - 2j| \), we calculate each element:
\( a_{11} = |1 - 2(1)| = |1 - 2| = |-1| = 1 \)
\( a_{12} = |1 - 2(2)| = |1 - 4| = |-3| = 3 \)
\( a_{13} = |1 - 2(3)| = |1 - 6| = |-5| = 5 \)
\( a_{21} = |2 - 2(1)| = |2 - 2| = |0| = 0 \)
\( a_{22} = |2 - 2(2)| = |2 - 4| = |-2| = 2 \)
\( a_{23} = |2 - 2(3)| = |2 - 6| = |-4| = 4 \)
\( a_{31} = |3 - 2(1)| = |3 - 2| = |1| = 1 \)
\( a_{32} = |3 - 2(2)| = |3 - 4| = |-1| = 1 \)
\( a_{33} = |3 - 2(3)| = |3 - 6| = |-3| = 3 \)
The required matrix is:
\[ A = \begin{pmatrix} 1 & 3 & 5 \\ 0 & 2 & 4 \\ 1 & 1 & 3 \end{pmatrix} \] Each element is calculated based on its position using the given formula, ensuring all values are positive due to the absolute value function.

(ii) To construct a \( 3 \times 3 \) matrix where each element \( a_{ij} \) is given by the formula \( \frac{(i+j)^{3}}{3} \), we calculate each element:
\( a_{11} = \frac{(1+1)^{3}}{3} = \frac{2^{3}}{3} = \frac{8}{3} \)
\( a_{12} = \frac{(1+2)^{3}}{3} = \frac{3^{3}}{3} = \frac{27}{3} = 9 \)
\( a_{13} = \frac{(1+3)^{3}}{3} = \frac{4^{3}}{3} = \frac{64}{3} \)
\( a_{21} = \frac{(2+1)^{3}}{3} = \frac{3^{3}}{3} = \frac{27}{3} = 9 \)
\( a_{22} = \frac{(2+2)^{3}}{3} = \frac{4^{3}}{3} = \frac{64}{3} \)
\( a_{23} = \frac{(2+3)^{3}}{3} = \frac{5^{3}}{3} = \frac{125}{3} \)
\( a_{31} = \frac{(3+1)^{3}}{3} = \frac{4^{3}}{3} = \frac{64}{3} \)
\( a_{32} = \frac{(3+2)^{3}}{3} = \frac{5^{3}}{3} = \frac{125}{3} \)
\( a_{33} = \frac{(3+3)^{3}}{3} = \frac{6^{3}}{3} = \frac{216}{3} = 72 \)
The required matrix is:
\[ A = \begin{pmatrix} \frac{8}{3} & 9 & \frac{64}{3} \\ 9 & \frac{64}{3} & \frac{125}{3} \\ \frac{64}{3} & \frac{125}{3} & 72 \end{pmatrix} \] Each element in the matrix is found by adding its row and column numbers, cubing the sum, and then dividing by 3.
In simple words: For the first part, we make a 3 by 3 box of numbers where each number is found by taking its row number, subtracting two times its column number, and then making the result positive. For the second part, each number is found by adding its row and column numbers, multiplying the sum by itself three times, and then dividing by three.

🎯 Exam Tip: When constructing a matrix based on a formula, carefully substitute the row (i) and column (j) indices for each position to avoid calculation errors. Pay special attention to absolute values and exponents.

Question 4. If \( A = \begin{pmatrix} 5 & 4 & 3 \\ 1 & -7 & 9 \\ 3 & 8 & 2 \end{pmatrix} \) then find the transpose of A.
Answer:
Given the matrix A:
\[ A = \begin{pmatrix} 5 & 4 & 3 \\ 1 & -7 & 9 \\ 3 & 8 & 2 \end{pmatrix} \] To find the transpose of A, denoted as \( A^T \), we simply swap its rows with its columns. The first row becomes the first column, the second row becomes the second column, and so on. Transposing a matrix essentially "flips" it over its main diagonal, which runs from the top-left to the bottom-right.
The transpose of A is:
\[ A^T = \begin{pmatrix} 5 & 1 & 3 \\ 4 & -7 & 8 \\ 3 & 9 & 2 \end{pmatrix} \]In simple words: To transpose a matrix, you just turn all its rows into columns. The first row becomes the first column, the second row becomes the second column, and so on.

🎯 Exam Tip: A quick way to check if your transpose is correct is to verify that the element \( a_{ij} \) in the original matrix is the same as \( a_{ji} \) in the transposed matrix.

Question 5. If \( A = \begin{pmatrix} \sqrt{7} & -3 \\ -\sqrt{5} & 2 \\ \sqrt{3} & -5 \end{pmatrix} \) then find the transpose of \( -A \).
Answer:
Given the matrix A:
\[ A = \begin{pmatrix} \sqrt{7} & -3 \\ -\sqrt{5} & 2 \\ \sqrt{3} & -5 \end{pmatrix} \] First, we find \( -A \) by multiplying every element in matrix A by -1:
\[ -A = \begin{pmatrix} -\sqrt{7} & -(-3) \\ -(-\sqrt{5}) & -(2) \\ -(\sqrt{3}) & -(-5) \end{pmatrix} = \begin{pmatrix} -\sqrt{7} & 3 \\ \sqrt{5} & -2 \\ -\sqrt{3} & 5 \end{pmatrix} \] Next, we find the transpose of \( -A \), denoted as \( (-A)^T \), by swapping its rows and columns. Another way to solve this is to find \( A^T \) first, and then multiply by -1. Both methods give the same result.
\[ (-A)^T = \begin{pmatrix} -\sqrt{7} & \sqrt{5} & -\sqrt{3} \\ 3 & -2 & 5 \end{pmatrix} \]In simple words: First, change the sign of every number in matrix A to get -A. Then, flip this new matrix over so its rows become columns, which gives you the transpose of -A.

🎯 Exam Tip: The transpose of \( -A \) is the same as \( -(A^T) \). You can calculate either way to double-check your answer and ensure all signs are correct.

Question 6. If \( A = \begin{pmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{pmatrix} \) then verify \( (A^T)^T = A \).
Answer:
Given the matrix A:
\[ A = \begin{pmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{pmatrix} \] First, we find the transpose of A, \( A^T \), by swapping its rows and columns:
\[ A^T = \begin{pmatrix} 5 & -\sqrt{17} & 8 \\ 2 & 0.7 & 3 \\ 2 & \frac{5}{2} & 1 \end{pmatrix} \] Next, we find the transpose of \( A^T \), which is \( (A^T)^T \), by swapping the rows and columns of \( A^T \). This property means that transposing a matrix twice brings it back to its original form.
\[ (A^T)^T = \begin{pmatrix} 5 & 2 & 2 \\ -\sqrt{17} & 0.7 & \frac{5}{2} \\ 8 & 3 & 1 \end{pmatrix} \] By comparing \( (A^T)^T \) with the original matrix A, we see that they are identical.
Therefore, it is verified that \( (A^T)^T = A \).
In simple words: When you flip a matrix (turn rows into columns) and then flip it again, you get the same matrix you started with. We showed this by doing the flips one by one and checking the final matrix.

🎯 Exam Tip: This property, \( (A^T)^T = A \), is fundamental in matrix algebra and is often used in proofs and simplifying expressions because it means the operation is its own inverse.

Question 7. Find the values of x, y and z from the following equations
(i) If \( \begin{pmatrix} 12 & 3 \\ 3 & 5 \end{pmatrix} = \begin{pmatrix} y & z \\ x & 5 \end{pmatrix} \)
Answer:
Given the matrix equation where two matrices are equal:
\[ \begin{pmatrix} 12 & 3 \\ 3 & 5 \end{pmatrix} = \begin{pmatrix} y & z \\ x & 5 \end{pmatrix} \] When two matrices are equal, their corresponding elements must be equal. By comparing the elements in the same positions from both matrices, we can find the values of x, y, and z.
From the first row, first column: \( y = 12 \)
From the first row, second column: \( z = 3 \)
From the second row, first column: \( x = 3 \)
From the second row, second column: \( 5 = 5 \) (This confirms the equality.)
Thus, the values are \( x = 3, y = 12, \) and \( z = 3 \). Each element must match up exactly.
In simple words: If two matrices are exactly the same, then the number in the top-left corner of the first matrix must be the same as the number in the top-left corner of the second matrix, and so on for all positions. We use this idea to find the missing letters.

🎯 Exam Tip: Always ensure the matrices are of the same order before equating them. Carefully match corresponding elements to solve for variables, paying close attention to their exact positions.

Question 7. (ii) \( \begin{pmatrix} x+y & 2 \\ 5+z & xy \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 5 & 8 \end{pmatrix} \)
Answer:
Given the matrix equality:
\[ \begin{pmatrix} x+y & 2 \\ 5+z & xy \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 5 & 8 \end{pmatrix} \] By equating the corresponding elements of the matrices, we get the following system of equations:
1. \( x + y = 6 \)
2. \( 5 + z = 5 \)
3. \( xy = 8 \)

From equation (2):
\( 5 + z = 5 \)
\( \implies z = 5 - 5 \)
\( \implies z = 0 \)

Now we have a system of two equations for x and y:
\( x + y = 6 \)
\( xy = 8 \)

From the first equation, we can write \( y = 6 - x \).
Substitute this into the second equation:
\( x(6 - x) = 8 \)
\( \implies 6x - x^2 = 8 \)
\( \implies x^2 - 6x + 8 = 0 \)

This is a quadratic equation. We can solve it by factoring:
\( (x - 4)(x - 2) = 0 \)
This means either \( x - 4 = 0 \) or \( x - 2 = 0 \).
So, \( x = 4 \) or \( x = 2 \).

Now find the corresponding y values:
If \( x = 4 \), then \( y = 6 - 4 = 2 \).
If \( x = 2 \), then \( y = 6 - 2 = 4 \).

So, the possible sets of values for (x, y, z) are:
\( x = 4, y = 2, z = 0 \) or \( x = 2, y = 4, z = 0 \). The solution shows both possibilities clearly.
In simple words: We have two boxes of numbers that are equal, so numbers in the same spots are equal. We make math problems from these equal spots. First, we find 'z' easily. Then, we solve for 'x' and 'y' using two equations, which gives us two possible pairs of answers for x and y.

🎯 Exam Tip: When solving for multiple variables from matrix equality, first identify independent equations (like for z), then use substitution or other algebraic methods for systems of equations (like for x and y). Always check for multiple possible solutions when dealing with quadratic equations.

Question 7. (iii) \( \begin{pmatrix} x+y+z \\ x+z \\ y+z \end{pmatrix} = \begin{pmatrix} 9 \\ 5 \\ 7 \end{pmatrix} \)
Answer:
Given the matrix equality:
\[ \begin{pmatrix} x+y+z \\ x+z \\ y+z \end{pmatrix} = \begin{pmatrix} 9 \\ 5 \\ 7 \end{pmatrix} \] By equating the corresponding elements, we get the following system of linear equations:
1. \( x + y + z = 9 \)
2. \( x + z = 5 \)
3. \( y + z = 7 \)

Substitute equation (2) into equation (1):
\( (x + z) + y = 9 \)
Since \( x + z = 5 \), we have:
\( 5 + y = 9 \)
\( \implies y = 9 - 5 \)
\( \implies y = 4 \)

Now substitute the value of \( y = 4 \) into equation (3):
\( y + z = 7 \)
\( 4 + z = 7 \)
\( \implies z = 7 - 4 \)
\( \implies z = 3 \)

Finally, substitute the value of \( z = 3 \) into equation (2):
\( x + z = 5 \)
\( x + 3 = 5 \)
\( \implies x = 5 - 3 \)
\( \implies x = 2 \)

Thus, the values are \( x = 2, y = 4, \) and \( z = 3 \). These values satisfy all three original equations.
In simple words: We have a list of math problems that are equal. We match up each line to make three simple equations. We solve these step by step. First, we use two equations to find 'y'. Then we use 'y' to find 'z'. Lastly, we use 'z' to find 'x'.

🎯 Exam Tip: For systems of linear equations derived from matrix equality, substitution is often an efficient method. Clearly label your equations and steps, and double-check your final values by plugging them back into the original equations.