RBSE Solutions Class 12 Maths Chapter 3 Matrix Exercise 3.1

Question 1. If \( A = [a_{ij}]_{2 \times 4} \), then find the number of elements in A.
Answer: For a matrix of order \( m \times n \), the number of elements is \( m \times n \). Here, the matrix A has 2 rows and 4 columns. So, the total number of elements is \( 2 \times 4 = 8 \). This multiplication gives the total count of individual numbers or terms within the matrix.
In simple words: To find how many items are in a matrix, just multiply the number of rows by the number of columns. Here, 2 rows times 4 columns gives 8 elements.

🎯 Exam Tip: Remember that the order of a matrix (rows × columns) directly tells you how to calculate the total number of elements by simple multiplication.

Question 2. Write the identity matrix of order 4 x 4.
Answer: An identity matrix is a square matrix where all the elements on the main diagonal are 1, and all other elements are 0. For a 4x4 matrix, it looks like this:
\[ \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \]
This matrix acts like the number '1' in multiplication, meaning when you multiply any matrix by the identity matrix, the original matrix remains unchanged.
In simple words: An identity matrix has ones along its main diagonal (top-left to bottom-right) and zeros everywhere else. For a 4x4 matrix, this means four ones on the diagonal.

🎯 Exam Tip: Always ensure an identity matrix is square (same number of rows and columns) and has '1's only on the main diagonal, with all other entries as '0'.

Question 3. Find the values of k and a if \( \begin{pmatrix} k+4 & -1 \\ 3 & k-6 \end{pmatrix} = \begin{pmatrix} a & -1 \\ 3 & -4 \end{pmatrix} \).
Answer: When two matrices are equal, their corresponding elements must also be equal. We compare the elements in the same positions in both matrices:
From the first row, first column: \( k+4 = a \) ... (i)
From the second row, second column: \( k-6 = -4 \) ... (ii)
First, we solve equation (ii) for k:
\( k-6 = -4 \)
\( \implies k = -4 + 6 \)
\( \implies k = 2 \)
Now, we substitute the value of k into equation (i) to find a:
\( a = k+4 \)
\( \implies a = 2+4 \)
\( \implies a = 6 \)
So, the values are \( k=2 \) and \( a=6 \). This method ensures that all corresponding elements are equal, satisfying the condition for matrix equality.
In simple words: Because the two matrices are the same, the numbers in the same spots must be equal. We use this to set up small equations and find the values of k and a.

🎯 Exam Tip: When equating matrices, always make sure to compare elements in the exact same positions. Start with equations that have only one unknown variable to solve them easily.

Question 4. What are the possible orders of a matrix having 6 elements?
Answer: The order of a matrix is given by the product of its number of rows (m) and columns (n), i.e., \( m \times n \). If a matrix has 6 elements, we need to find all pairs of positive integers (m, n) whose product is 6. These pairs represent the possible orders of the matrix.
The pairs are:
\( 1 \times 6 \) (1 row, 6 columns)
\( 6 \times 1 \) (6 rows, 1 column)
\( 2 \times 3 \) (2 rows, 3 columns)
\( 3 \times 2 \) (3 rows, 2 columns)
Each of these combinations will result in a matrix with exactly six elements.
In simple words: To find all the ways to arrange 6 items into a matrix, think of all the pairs of numbers that multiply to 6. Each pair gives a possible size (rows by columns) for the matrix.

🎯 Exam Tip: To find possible orders for a matrix with 'x' elements, list all pairs of positive integers that multiply to 'x'. Each pair represents a valid (rows x columns) order.

Question 5. Construct a 2 x 2 matrix \( A = [a_{ij}] \), where elements are given by:
(i) \( a_{ij} = \frac{2i-j}{3i+j} \)
(ii) \( a_{ij} = \frac{(i+2j)^2}{2i} \)
(iii) \( a_{ij} = 2i - 3j \)
Answer: To construct a 2x2 matrix, we need to find the values for \( a_{11}, a_{12}, a_{21}, \) and \( a_{22} \) for each given formula. Here, 'i' represents the row number and 'j' represents the column number.

(i) For \( a_{ij} = \frac{2i-j}{3i+j} \):
\( a_{11} = \frac{2(1)-1}{3(1)+1} = \frac{2-1}{3+1} = \frac{1}{4} \)
\( a_{12} = \frac{2(1)-2}{3(1)+2} = \frac{2-2}{3+2} = \frac{0}{5} = 0 \)
\( a_{21} = \frac{2(2)-1}{3(2)+1} = \frac{4-1}{6+1} = \frac{3}{7} \)
\( a_{22} = \frac{2(2)-2}{3(2)+2} = \frac{4-2}{6+2} = \frac{2}{8} = \frac{1}{4} \)
Thus, the required matrix is: \[ \begin{pmatrix} 1/4 & 0 \\ 3/7 & 1/4 \end{pmatrix} \]

(ii) For \( a_{ij} = \frac{(i+2j)^2}{2i} \):
\( a_{11} = \frac{(1+2(1))^2}{2(1)} = \frac{(1+2)^2}{2} = \frac{3^2}{2} = \frac{9}{2} \)
\( a_{12} = \frac{(1+2(2))^2}{2(1)} = \frac{(1+4)^2}{2} = \frac{5^2}{2} = \frac{25}{2} \)
\( a_{21} = \frac{(2+2(1))^2}{2(2)} = \frac{(2+2)^2}{4} = \frac{4^2}{4} = \frac{16}{4} = 4 \)
\( a_{22} = \frac{(2+2(2))^2}{2(2)} = \frac{(2+4)^2}{4} = \frac{6^2}{4} = \frac{36}{4} = 9 \)
Thus, the required matrix is: \[ \begin{pmatrix} 9/2 & 25/2 \\ 4 & 9 \end{pmatrix} \]

(iii) For \( a_{ij} = 2i - 3j \):
\( a_{11} = 2(1) - 3(1) = 2 - 3 = -1 \)
\( a_{12} = 2(1) - 3(2) = 2 - 6 = -4 \)
\( a_{21} = 2(2) - 3(1) = 4 - 3 = 1 \)
\( a_{22} = 2(2) - 3(2) = 4 - 6 = -2 \)
Thus, the required matrix is: \[ \begin{pmatrix} -1 & -4 \\ 1 & -2 \end{pmatrix} \]
Constructing a matrix involves using the row and column indices in the given formula to find each individual element.
In simple words: For each spot in the 2x2 matrix, use the spot's row number (i) and column number (j) in the given rule to calculate the value for that spot. Then put all these values together to form the matrix.

🎯 Exam Tip: Be very careful with arithmetic when substituting 'i' and 'j' values into the formula. A common mistake is miscalculating squares, fractions, or negative numbers.

Question 6. Construct a matrix \( A = [a_{ij}] \) of order 2 x 3, whose elements are \( a_{ij} = \frac{1}{2} |2i - 3j| \).
Answer: To construct a 2x3 matrix, we need to find the values for \( a_{ij} \) where i can be 1 or 2 (for rows) and j can be 1, 2, or 3 (for columns). The formula for each element is \( a_{ij} = \frac{1}{2} |2i - 3j| \). The absolute value ensures the result is always non-negative.

For \( i=1 \):
\( a_{11} = \frac{1}{2} |2(1) - 3(1)| = \frac{1}{2} |2 - 3| = \frac{1}{2} |-1| = \frac{1}{2}(1) = \frac{1}{2} \)
\( a_{12} = \frac{1}{2} |2(1) - 3(2)| = \frac{1}{2} |2 - 6| = \frac{1}{2} |-4| = \frac{1}{2}(4) = 2 \)
\( a_{13} = \frac{1}{2} |2(1) - 3(3)| = \frac{1}{2} |2 - 9| = \frac{1}{2} |-7| = \frac{1}{2}(7) = \frac{7}{2} \)

For \( i=2 \):
\( a_{21} = \frac{1}{2} |2(2) - 3(1)| = \frac{1}{2} |4 - 3| = \frac{1}{2} |1| = \frac{1}{2}(1) = \frac{1}{2} \)
\( a_{22} = \frac{1}{2} |2(2) - 3(2)| = \frac{1}{2} |4 - 6| = \frac{1}{2} |-2| = \frac{1}{2}(2) = 1 \)
\( a_{23} = \frac{1}{2} |2(2) - 3(3)| = \frac{1}{2} |4 - 9| = \frac{1}{2} |-5| = \frac{1}{2}(5) = \frac{5}{2} \)

So, the required 2x3 matrix is: \[ \begin{pmatrix} 1/2 & 2 & 7/2 \\ 1/2 & 1 & 5/2 \end{pmatrix} \]
This process systematically calculates each element based on its position.
In simple words: We fill in each spot in the 2-row, 3-column matrix by using the given rule \( a_{ij} = \frac{1}{2} |2i - 3j| \). We put the row number for 'i' and the column number for 'j' into the rule to get each value.

🎯 Exam Tip: Pay close attention to the absolute value function \( |...| \); it means you always take the positive value of the number inside, which is crucial for correct calculations.

Question 7. If \( \begin{pmatrix} a+b & 2 \\ 7 & ab \end{pmatrix} = \begin{pmatrix} 6 & 2 \\ 7 & 8 \end{pmatrix} \) then find 'a' and 'b'.
Answer: For two matrices to be equal, their corresponding elements must be equal. By comparing the elements in the first matrix to the second, we get two equations:
1. \( a+b = 6 \) (from the first row, first column)
2. \( ab = 8 \) (from the second row, second column)
We can solve these two simultaneous equations for 'a' and 'b'. From equation (1), we can express \( b \) as \( b = 6-a \).
Substitute this into equation (2):
\( a(6-a) = 8 \)
\( \implies 6a - a^2 = 8 \)
Rearrange the terms to form a quadratic equation:
\( a^2 - 6a + 8 = 0 \)
Factorize the quadratic equation:
\( a^2 - 2a - 4a + 8 = 0 \)
\( a(a-2) - 4(a-2) = 0 \)
\( \implies (a-2)(a-4) = 0 \)
So, the possible values for 'a' are \( a=2 \) or \( a=4 \).

If \( a=2 \), then from \( a+b=6 \), we have \( 2+b=6 \implies b=4 \).
If \( a=4 \), then from \( a+b=6 \), we have \( 4+b=6 \implies b=2 \).
Thus, the solutions are \( (a=2, b=4) \) or \( (a=4, b=2) \). This demonstrates how matrix equality leads to a system of algebraic equations.
In simple words: We make two small equations by matching the numbers in the same spots in both matrices. Then, we solve these equations together to find what 'a' and 'b' are. There are two possible pairs of answers.

🎯 Exam Tip: When solving a system of equations from matrix equality, always check both sets of solutions by plugging them back into the original equations to ensure they are consistent.

Question 8. If \( \begin{pmatrix} 2x & 3x+y \\ -x+z & 3y-2p \end{pmatrix} = \begin{pmatrix} 4 & 5 \\ -4 & -3 \end{pmatrix} \) then find the value of x, y, z and p.
Answer: Since the two matrices are equal, their corresponding elements must be equal. We can set up a system of equations by comparing each element:
1. \( 2x = 4 \)
2. \( 3x+y = 5 \)
3. \( -x+z = -4 \)
4. \( 3y-2p = -3 \)

First, solve for x using equation (1):
\( 2x = 4 \)
\( \implies x = \frac{4}{2} \)
\( \implies x = 2 \)

Next, substitute \( x=2 \) into equation (2) to find y:
\( 3(2)+y = 5 \)
\( 6+y = 5 \)
\( \implies y = 5-6 \)
\( \implies y = -1 \)

Then, substitute \( x=2 \) into equation (3) to find z:
\( -(2)+z = -4 \)
\( -2+z = -4 \)
\( \implies z = -4+2 \)
\( \implies z = -2 \)

Finally, substitute \( y=-1 \) into equation (4) to find p:
\( 3(-1)-2p = -3 \)
\( -3-2p = -3 \)
\( \implies -2p = -3+3 \)
\( \implies -2p = 0 \)
\( \implies p = 0 \)
So, the values are \( x=2, y=-1, z=-2, \) and \( p=0 \). This systematic approach ensures all variables are correctly determined.
In simple words: We match each number in the first matrix with the number in the same spot in the second matrix. This gives us four small math problems to solve, one by one, to find the values for x, y, z, and p.

🎯 Exam Tip: Always solve for variables that appear in single-variable equations first, then substitute those values into equations with more variables to simplify the process.

Question 9. Find the values of a, b and c if matrices A and B are equal, where \( \begin{pmatrix} a-2 & 3 & 2c \\ 12c & b+2 & bc \end{pmatrix} = \begin{pmatrix} b & c & 6 \\ 6b & a & 3b \end{pmatrix} \).
Answer: For the two given matrices to be equal, their corresponding elements must be identical. We compare each position to form equations:
1. \( a-2 = b \)
2. \( 3 = c \)
3. \( 2c = 6 \)
4. \( 12c = 6b \)
5. \( b+2 = a \)
6. \( bc = 3b \)

From equation (2), we immediately get: \( c = 3 \)
Let's check this with equation (3): \( 2c = 6 \implies 2(3) = 6 \implies 6 = 6 \). This is consistent.

Now, use \( c=3 \) in equation (4) to find b:
\( 12c = 6b \)
\( 12(3) = 6b \)
\( 36 = 6b \)
\( \implies b = \frac{36}{6} \)
\( \implies b = 6 \)

Now, use \( b=6 \) in equation (1) or (5) to find a. Using (1):
\( a-2 = b \)
\( a-2 = 6 \)
\( \implies a = 6+2 \)
\( \implies a = 8 \)
Let's verify these values with other equations. From equation (5): \( b+2 = a \implies 6+2 = 8 \), which is consistent. From equation (6): \( bc = 3b \implies (6)(3) = 3(6) \implies 18 = 18 \), also consistent. All equations hold true with these values.
So, the values are \( a=8, b=6, \) and \( c=3 \). This systematic comparison helps solve for multiple unknowns in matrix equations.
In simple words: We set up equations by matching the numbers in the same positions in both matrices. We found 'c' right away, then used that to find 'b', and finally used 'b' to find 'a'. All the values matched up perfectly.

🎯 Exam Tip: When multiple equations for variables exist from matrix equality, always pick the simplest ones (those with fewer variables or direct values) to solve first. Then, use those results in more complex equations.