OP Malhotra Class 12 Maths Solutions Chapter 6 Matrices Exercise 6 (A)

S Chand Class 12 ICSE Maths Solutions Chapter 6 Matrices Ex 6(a)

Question 1. If a matrix has 8 elements, what are the possible orders it can have? What if it has 5 elements?
Answer: We know that a matrix of order \( m \times n \) has \( mn \) elements. To find all possible orders for a matrix with 8 elements, we need to find all ordered pairs of positive integers whose product is 8. These ordered pairs are (1, 8), (8, 1), (2, 4), and (4, 2). So, the possible orders for a matrix with 8 elements are \( 1 \times 8, 8 \times 1, 2 \times 4 \), and \( 4 \times 2 \). Now, if a matrix has 5 elements, we find ordered pairs whose product is 5. Since 5 is a prime number, the only ordered pairs are (1, 5) and (5, 1). Thus, the possible orders for a matrix with 5 elements are \( 1 \times 5 \) and \( 5 \times 1 \). A matrix's order tells us its size and shape, defining how many rows and columns it has.In simple words: To find the possible sizes (orders) of a matrix, find all the pairs of numbers that multiply to give the total number of elements. For 8 elements, pairs like (1,8), (8,1), (2,4), (4,2) work. For 5 elements, only (1,5) and (5,1) work.

🎯 Exam Tip: Remember that the order of a matrix \( m \times n \) means \( m \) rows and \( n \) columns. When finding possible orders, always list both \( m \times n \) and \( n \times m \).

Question 2. How many entries are there in (i) a 3 x 3 matrix, (ii) a 3 x 4 matrix, (iii) an m x n matrix, (iv) a square matrix of order n?
Answer: We know that a matrix of order \( m \times n \) contains \( mn \) elements. 1. A \( 3 \times 3 \) matrix contains \( 3 \times 3 = 9 \) entries. 2. A \( 3 \times 4 \) matrix contains \( 3 \times 4 = 12 \) entries. 3. An \( m \times n \) matrix contains \( mn \) entries. 4. A square matrix of order \( n \) means it has \( n \) rows and \( n \) columns. So, it contains \( n \times n = n^2 \) entries. The number of entries in a matrix is simply the product of its number of rows and columns.In simple words: To find how many items are in a matrix, just multiply the number of rows by the number of columns. For a 3x3 matrix, there are 9 items. For a square matrix of order 'n', there are \( n^2 \) items.

🎯 Exam Tip: Always remember that "order n" for a square matrix means it is an \( n \times n \) matrix. This is a common shortcut in matrix notation.

Question 3. Write out the matrix \[ \left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right] \] given that \( a_{ij} = 4i - 3j \)
Answer: We are given the formula \( a_{ij} = 4i - 3j \) for each element in the matrix. We need to find the values for a \( 3 \times 3 \) matrix. For \( i=1, j=1 \): \( a_{11} = 4(1) - 3(1) = 4 - 3 = 1 \) For \( i=1, j=2 \): \( a_{12} = 4(1) - 3(2) = 4 - 6 = -2 \) For \( i=1, j=3 \): \( a_{13} = 4(1) - 3(3) = 4 - 9 = -5 \) For \( i=2, j=1 \): \( a_{21} = 4(2) - 3(1) = 8 - 3 = 5 \) For \( i=2, j=2 \): \( a_{22} = 4(2) - 3(2) = 8 - 6 = 2 \) For \( i=2, j=3 \): \( a_{23} = 4(2) - 3(3) = 8 - 9 = -1 \) For \( i=3, j=1 \): \( a_{31} = 4(3) - 3(1) = 12 - 3 = 9 \) For \( i=3, j=2 \): \( a_{32} = 4(3) - 3(2) = 12 - 6 = 6 \) For \( i=3, j=3 \): \( a_{33} = 4(3) - 3(3) = 12 - 9 = 3 \) Now, we can write the required matrix: \[ A = \left[\begin{array}{ccc} 1 & -2 & -5 \\ 5 & 2 & -1 \\ 9 & 6 & 3 \end{array}\right] \] Each element's position \( (i, j) \) determines its value based on the given rule.In simple words: To build this matrix, use the rule \( a_{ij} = 4i - 3j \). Put in the row number for 'i' and the column number for 'j' to find each value. For example, for the first spot in the first row (\( a_{11} \)), use \( i=1 \) and \( j=1 \).

🎯 Exam Tip: Be careful with the signs (positive or negative) when substituting values, especially with subtraction. A small error can change the entire matrix element.

Question 4. Construct a 2 x 2 matrix B – \( [b_{ij}] \) whose elements are given by
(i) \( b_{ij} = \frac{(i-2 j)^2}{2} \)
(ii) \( b_{ij} = \frac{1}{2}|-3 i+j| \)
Answer: Given that B is a \( 2 \times 2 \) matrix, so \( 1 \le i, j \le 2 \). (i) For \( b_{ij} = \frac{(i-2 j)^2}{2} \): For \( i=1, j=1 \): \( b_{11} = \frac{(1-2(1))^2}{2} = \frac{(1-2)^2}{2} = \frac{(-1)^2}{2} = \frac{1}{2} \) For \( i=1, j=2 \): \( b_{12} = \frac{(1-2(2))^2}{2} = \frac{(1-4)^2}{2} = \frac{(-3)^2}{2} = \frac{9}{2} \) For \( i=2, j=1 \): \( b_{21} = \frac{(2-2(1))^2}{2} = \frac{(2-2)^2}{2} = \frac{0^2}{2} = 0 \) For \( i=2, j=2 \): \( b_{22} = \frac{(2-2(2))^2}{2} = \frac{(2-4)^2}{2} = \frac{(-2)^2}{2} = \frac{4}{2} = 2 \) So the matrix B is: \[ B = \left[\begin{array}{cc} 1/2 & 9/2 \\ 0 & 2 \end{array}\right] \] (ii) For \( b_{ij} = \frac{1}{2}|-3 i+j| \): For \( i=1, j=1 \): \( b_{11} = \frac{1}{2}|-3(1)+1| = \frac{1}{2}|-3+1| = \frac{1}{2}|-2| = \frac{1}{2}(2) = 1 \) For \( i=1, j=2 \): \( b_{12} = \frac{1}{2}|-3(1)+2| = \frac{1}{2}|-3+2| = \frac{1}{2}|-1| = \frac{1}{2}(1) = \frac{1}{2} \) For \( i=2, j=1 \): \( b_{21} = \frac{1}{2}|-3(2)+1| = \frac{1}{2}|-6+1| = \frac{1}{2}|-5| = \frac{1}{2}(5) = \frac{5}{2} \) For \( i=2, j=2 \): \( b_{22} = \frac{1}{2}|-3(2)+2| = \frac{1}{2}|-6+2| = \frac{1}{2}|-4| = \frac{1}{2}(4) = 2 \) So the matrix B is: \[ B = \left[\begin{array}{cc} 1 & 1/2 \\ 5/2 & 2 \end{array}\right] \] The absolute value function \( |x| \) always gives a non-negative result, which is important for calculations like these.In simple words: For each part, use the given formula for \( b_{ij} \) to find each number in the \( 2 \times 2 \) matrix. Remember that 'i' is the row number and 'j' is the column number. For the second part, the absolute value symbol \( | \; | \) means you always take the positive version of the number inside.

🎯 Exam Tip: When using the absolute value function, ensure you first calculate the expression inside the bars and then take its positive value before performing other operations.

Question 5. Construct a 3 x 4 matrix whose elements are:
(i) \( a_{ij} = i - j \)
(ii) \( a_{ij} = ij \)
(iii) \( a_{ij} = \frac{i}{j} \)
Answer: Let A be a \( 3 \times 4 \) matrix, so \( 1 \le i \le 3 \) and \( 1 \le j \le 4 \). (i) For \( a_{ij} = i - j \): When \( i=1 \): \( a_{11} = 1 - 1 = 0 \) \( a_{12} = 1 - 2 = -1 \) \( a_{13} = 1 - 3 = -2 \) \( a_{14} = 1 - 4 = -3 \) When \( i=2 \): \( a_{21} = 2 - 1 = 1 \) \( a_{22} = 2 - 2 = 0 \) \( a_{23} = 2 - 3 = -1 \) \( a_{24} = 2 - 4 = -2 \) When \( i=3 \): \( a_{31} = 3 - 1 = 2 \) \( a_{32} = 3 - 2 = 1 \) \( a_{33} = 3 - 3 = 0 \) \( a_{34} = 3 - 4 = -1 \) So the matrix A is: \[ A = \left[\begin{array}{cccc} 0 & -1 & -2 & -3 \\ 1 & 0 & -1 & -2 \\ 2 & 1 & 0 & -1 \end{array}\right] \] (ii) For \( a_{ij} = ij \): When \( i=1 \): \( a_{11} = 1 \times 1 = 1 \) \( a_{12} = 1 \times 2 = 2 \) \( a_{13} = 1 \times 3 = 3 \) \( a_{14} = 1 \times 4 = 4 \) When \( i=2 \): \( a_{21} = 2 \times 1 = 2 \) \( a_{22} = 2 \times 2 = 4 \) \( a_{23} = 2 \times 3 = 6 \) \( a_{24} = 2 \times 4 = 8 \) When \( i=3 \): \( a_{31} = 3 \times 1 = 3 \) \( a_{32} = 3 \times 2 = 6 \) \( a_{33} = 3 \times 3 = 9 \) \( a_{34} = 3 \times 4 = 12 \) So the matrix A is: \[ A = \left[\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & 8 \\ 3 & 6 & 9 & 12 \end{array}\right] \] (iii) For \( a_{ij} = \frac{i}{j} \): When \( i=1 \): \( a_{11} = \frac{1}{1} = 1 \) \( a_{12} = \frac{1}{2} \) \( a_{13} = \frac{1}{3} \) \( a_{14} = \frac{1}{4} \) When \( i=2 \): \( a_{21} = \frac{2}{1} = 2 \) \( a_{22} = \frac{2}{2} = 1 \) \( a_{23} = \frac{2}{3} \) \( a_{24} = \frac{2}{4} = \frac{1}{2} \) When \( i=3 \): \( a_{31} = \frac{3}{1} = 3 \) \( a_{32} = \frac{3}{2} \) \( a_{33} = \frac{3}{3} = 1 \) \( a_{34} = \frac{3}{4} \) So the matrix A is: \[ A = \left[\begin{array}{cccc} 1 & 1/2 & 1/3 & 1/4 \\ 2 & 1 & 2/3 & 1/2 \\ 3 & 3/2 & 1 & 3/4 \end{array}\right]_{3 \times 4} \] The construction of a matrix from an element-generating rule is a fundamental skill.In simple words: For each part, use the given rule to calculate every spot in the \( 3 \times 4 \) matrix. 'i' is the row number (from 1 to 3) and 'j' is the column number (from 1 to 4). Fill in the matrix with these calculated values.

🎯 Exam Tip: When constructing matrices, always double-check the dimensions (number of rows and columns) and carefully substitute the values of \( i \) and \( j \) into the given formula for each element.

Question 6. Construct a 2 x 3 matrix whose elements are given by
(i) \( a_{ij} = \frac{3 i-j}{2} \)
(ii) \( a_{ij} = \frac{i+3 j}{2} \)
Answer: Let A be a \( 2 \times 3 \) matrix, so \( 1 \le i \le 2 \) and \( 1 \le j \le 3 \). (i) For \( a_{ij} = \frac{3 i-j}{2} \): When \( i=1 \): \( a_{11} = \frac{3(1)-1}{2} = \frac{2}{2} = 1 \) \( a_{12} = \frac{3(1)-2}{2} = \frac{1}{2} \) \( a_{13} = \frac{3(1)-3}{2} = \frac{0}{2} = 0 \) When \( i=2 \): \( a_{21} = \frac{3(2)-1}{2} = \frac{5}{2} \) \( a_{22} = \frac{3(2)-2}{2} = \frac{4}{2} = 2 \) \( a_{23} = \frac{3(2)-3}{2} = \frac{3}{2} \) So the matrix A is: \[ A = \left[\begin{array}{ccc} 1 & 1/2 & 0 \\ 5/2 & 2 & 3/2 \end{array}\right] \] (ii) For \( a_{ij} = \frac{i+3 j}{2} \): When \( i=1 \): \( a_{11} = \frac{1+3(1)}{2} = \frac{4}{2} = 2 \) \( a_{12} = \frac{1+3(2)}{2} = \frac{7}{2} \) \( a_{13} = \frac{1+3(3)}{2} = \frac{10}{2} = 5 \) When \( i=2 \): \( a_{21} = \frac{2+3(1)}{2} = \frac{5}{2} \) \( a_{22} = \frac{2+3(2)}{2} = \frac{8}{2} = 4 \) \( a_{23} = \frac{2+3(3)}{2} = \frac{11}{2} \) So the matrix A is: \[ A = \left[\begin{array}{ccc} 2 & 7/2 & 5 \\ 5/2 & 4 & 11/2 \end{array}\right] \] Understanding how to correctly apply the formulas for \( a_{ij} \) is essential for matrix construction.In simple words: For each formula given, calculate all the numbers that go into the \( 2 \times 3 \) matrix. Remember, 'i' is the row (1 or 2) and 'j' is the column (1, 2, or 3). Write down the matrix once all numbers are found.

🎯 Exam Tip: Pay close attention to the order of operations when calculating each element, especially with fractions and multiple arithmetic steps. Be precise with fractions and do not convert to decimals unless specified.

Question 6 (b). Construct a 3 x 2 matrix whose elements are given by
(i) \( a_{ij} = \frac{i+3 j}{2} \)
(ii) \( a_{ij} = \frac{(i+2 j)^2}{2} \)
Answer: Let A be a \( 3 \times 2 \) matrix, so \( 1 \le i \le 3 \) and \( 1 \le j \le 2 \). (i) For \( a_{ij} = \frac{i+3 j}{2} \): When \( i=1 \): \( a_{11} = \frac{1+3(1)}{2} = \frac{4}{2} = 2 \) \( a_{12} = \frac{1+3(2)}{2} = \frac{7}{2} \) When \( i=2 \): \( a_{21} = \frac{2+3(1)}{2} = \frac{5}{2} \) \( a_{22} = \frac{2+3(2)}{2} = \frac{8}{2} = 4 \) When \( i=3 \): \( a_{31} = \frac{3+3(1)}{2} = \frac{6}{2} = 3 \) \( a_{32} = \frac{3+3(2)}{2} = \frac{9}{2} \) So the matrix A is: \[ A = \left[\begin{array}{cc} 2 & 7/2 \\ 5/2 & 4 \\ 3 & 9/2 \end{array}\right]_{3 \times 2} \] (ii) For \( a_{ij} = \frac{(i+2 j)^2}{2} \): When \( i=1 \): \( a_{11} = \frac{(1+2(1))^2}{2} = \frac{(1+2)^2}{2} = \frac{3^2}{2} = \frac{9}{2} \) \( a_{12} = \frac{(1+2(2))^2}{2} = \frac{(1+4)^2}{2} = \frac{5^2}{2} = \frac{25}{2} \) When \( i=2 \): \( a_{21} = \frac{(2+2(1))^2}{2} = \frac{(2+2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8 \) \( a_{22} = \frac{(2+2(2))^2}{2} = \frac{(2+4)^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18 \) When \( i=3 \): \( a_{31} = \frac{(3+2(1))^2}{2} = \frac{(3+2)^2}{2} = \frac{5^2}{2} = \frac{25}{2} \) \( a_{32} = \frac{(3+2(2))^2}{2} = \frac{(3+4)^2}{2} = \frac{7^2}{2} = \frac{49}{2} \) So the matrix A is: \[ A = \left[\begin{array}{cc} 9/2 & 25/2 \\ 8 & 18 \\ 25/2 & 49/2 \end{array}\right] \] Understanding matrix dimensions is key: a \( 3 \times 2 \) matrix means 3 rows and 2 columns.In simple words: For each rule, calculate the numbers for a \( 3 \times 2 \) matrix. 'i' will be the row number (1, 2, or 3), and 'j' will be the column number (1 or 2). Write down each matrix after finding all its numbers.

🎯 Exam Tip: When the formula involves squaring, remember to perform the addition inside the parenthesis first, then square the result, and finally divide. Be mindful of potential calculation errors.

Question 7. If A = \[ \left[\begin{array}{rrrrr} 5 & -2 & 1 & 0 & 3 \\ 7 & 6 & 4 & 2 & -1 \\ 0 & 8 & 3 & 5 & 6 \end{array}\right], \] then State (i) the order of A ; (ii) The number of elements (iii) Write down the entries of the second row of A; (iv) Write down the entries of the third column of A ; (v) State the entries \( a_{12}, a_{22}, a_{34}, a_{14} \) of the above matrix ; (vi) If \( a_{ij} = 4 \), find \( i, j \).
Answer: Given the matrix: \[ A = \left[\begin{array}{rrrrr} 5 & -2 & 1 & 0 & 3 \\ 7 & 6 & 4 & 2 & -1 \\ 0 & 8 & 3 & 5 & 6 \end{array}\right] \] (i) The matrix has 3 rows and 5 columns. So, the order of A is \( 3 \times 5 \). (ii) The number of elements in a matrix of order \( m \times n \) is \( mn \). Therefore, matrix A contains \( 3 \times 5 = 15 \) elements. (iii) The entries of the second row of A are 7, 6, 4, 2, and -1. (iv) The entries of the third column of A are 1, 4, and 3. (v) The entries are: \( a_{12} \) (first row, second column) = -2 \( a_{22} \) (second row, second column) = 6 \( a_{34} \) (third row, fourth column) = 5 \( a_{14} \) (first row, fourth column) = 0 (v) If \( a_{ij} = 4 \), we look for the element '4' in the matrix. We find '4' at the position of the second row and third column. So, \( a_{ij} = 4 \implies a_{23} = 4 \).
\( \implies \) Therefore, \( i=2 \) and \( j=3 \). A matrix organizes numbers in rows and columns, and each element has a unique address.In simple words: Look at the matrix and answer each question. The "order" is rows by columns. The "number of elements" is rows multiplied by columns. For specific entries like \( a_{12} \), find the number in row 1, column 2. To find \( i, j \) when \( a_{ij} = 4 \), simply locate the number 4 in the matrix and note its row and column.

🎯 Exam Tip: When identifying elements \( a_{ij} \), always remember that \( i \) represents the row number and \( j \) represents the column number. A common mistake is to reverse these. Also, count rows from top to bottom and columns from left to right, starting from 1.

Question 8. Find x and y such that
(i) \( \left[\begin{array}{rr} x & y \\ -1 & 5 \end{array}\right]=\left[\begin{array}{ll} -2 & 0 \\ -1 & 5 \end{array}\right] \)
(ii) \( \left[\begin{array}{ll} x & 3 \end{array}\right]=\left[\begin{array}{ll} -1 & y \end{array}\right] \)
(iii) \( \left[\begin{array}{r} x+1 \\ -3+y \end{array}\right]=\left[\begin{array}{r} -2 \\ 0 \end{array}\right] \)
Answer: When two matrices are equal, their corresponding elements must be equal. (i) Given \( \left[\begin{array}{rr} x & y \\ -1 & 5 \end{array}\right]=\left[\begin{array}{ll} -2 & 0 \\ -1 & 5 \end{array}\right] \) By comparing corresponding elements: \( x = -2 \) \( y = 0 \) (ii) Given \( \left[\begin{array}{ll} x & 3 \end{array}\right]=\left[\begin{array}{ll} -1 & y \end{array}\right] \) By comparing corresponding elements: \( x = -1 \) \( y = 3 \) (iii) Given \( \left[\begin{array}{r} x+1 \\ -3+y \end{array}\right]=\left[\begin{array}{r} -2 \\ 0 \end{array}\right] \) By comparing corresponding elements: \( x+1 = -2 \)
\( \implies x = -2 - 1 \)
\( \implies x = -3 \) And \( -3+y = 0 \)
\( \implies y = 3 \) When matrices are equal, every element in the first matrix must match the element in the same position in the second matrix.In simple words: If two matrices are exactly the same, it means each number in one matrix is equal to the number in the same spot in the other matrix. Just match up the numbers in the same positions to find 'x' and 'y'.

🎯 Exam Tip: Remember the fundamental rule of matrix equality: corresponding elements must be equal. This allows you to set up simple algebraic equations to solve for unknown variables like x and y.

Question 9. If \( \left[\begin{array}{rr} x+y & y-z \\ z-2 x & y-x \end{array}\right]=\left[\begin{array} {rr} 3 & -1 \\ 1 & 1 \end{array}\right], \) find x, y, z.
Answer: Given that the matrices are equal, their corresponding elements are equal. We can set up a system of equations: 1. \( x+y = 3 \) 2. \( y-z = -1 \) 3. \( z-2x = 1 \) 4. \( y-x = 1 \) Let's solve the system: Add equation (1) and equation (4): \( (x+y) + (y-x) = 3 + 1 \)
\( \implies 2y = 4 \)
\( \implies y = 2 \) Now substitute \( y=2 \) into equation (1): \( x+2 = 3 \)
\( \implies x = 3 - 2 \)
\( \implies x = 1 \) Substitute \( y=2 \) into equation (2): \( 2-z = -1 \)
\( \implies -z = -1 - 2 \)
\( \implies -z = -3 \)
\( \implies z = 3 \) We can check our values with equation (3): \( z-2x = 3 - 2(1) = 3 - 2 = 1 \). This matches. So, the values are \( x=1, y=2 \), and \( z=3 \). Solving systems of linear equations is a common application of matrix equality.In simple words: Because the two matrices are equal, we can set up four small math problems by matching the numbers in the same spots. Then, we solve these problems one by one to find the values of 'x', 'y', and 'z'.

🎯 Exam Tip: When solving a system of equations from matrix equality, look for equations that can be easily solved by addition or substitution to simplify the process. Always check your final answers by plugging them back into all original equations.

Question 10.
(i) If matrix \( \left[\begin{array}{rr} x-y & 2 x+z \\ 2 x-y & 3 z+w \end{array}\right]=\left[\begin{array} {rr} -1 & 5 \\ 0 & 13 \end{array}\right], \) find x, y, z, w.
(ii) If matrix \( \left[\begin{array}{cr} a+b & 2 \\ 5 & a b \end{array}\right]=\left[\begin{array} {rr} 6 & 2 \\ 5 & 8 \end{array}\right], \) find the values of a and b.
Answer:(i) Given: \( \left[\begin{array}{rr} x-y & 2 x+z \\ 2 x-y & 3 z+w \end{array}\right]=\left[\begin{array} {rr} -1 & 5 \\ 0 & 13 \end{array}\right] \) By comparing corresponding elements, we get the following equations: 1. \( x-y = -1 \) 2. \( 2x+z = 5 \) 3. \( 2x-y = 0 \) 4. \( 3z+w = 13 \) From equation (3), we can write \( y = 2x \). Substitute \( y = 2x \) into equation (1): \( x - (2x) = -1 \)
\( \implies -x = -1 \)
\( \implies x = 1 \) Now substitute \( x=1 \) into \( y=2x \): \( y = 2(1) \)
\( \implies y = 2 \) Substitute \( x=1 \) into equation (2): \( 2(1) + z = 5 \)
\( \implies 2 + z = 5 \)
\( \implies z = 5 - 2 \)
\( \implies z = 3 \) Substitute \( z=3 \) into equation (4): \( 3(3) + w = 13 \)
\( \implies 9 + w = 13 \)
\( \implies w = 13 - 9 \)
\( \implies w = 4 \) So, the values are \( x=1, y=2, z=3, w=4 \). (ii) Given: \( \left[\begin{array}{cr} a+b & 2 \\ 5 & a b \end{array}\right]=\left[\begin{array} {rr} 6 & 2 \\ 5 & 8 \end{array}\right] \) By comparing corresponding elements, we get: 1. \( a+b = 6 \) 2. \( ab = 8 \) We have a system of two equations with two variables. We can use the identity \( (a-b)^2 = (a+b)^2 - 4ab \). Substitute the known values: \( (a-b)^2 = (6)^2 - 4(8) \)
\( \implies (a-b)^2 = 36 - 32 \)
\( \implies (a-b)^2 = 4 \)
\( \implies a-b = \pm \sqrt{4} \)
\( \implies a-b = \pm 2 \) Case I: \( a-b = 2 \) We have two linear equations: \( a+b = 6 \) (Equation from above) \( a-b = 2 \) (From Case I) Adding these two equations: \( (a+b) + (a-b) = 6 + 2 \)
\( \implies 2a = 8 \)
\( \implies a = 4 \) Substitute \( a=4 \) into \( a+b=6 \): \( 4+b = 6 \)
\( \implies b = 6 - 4 \)
\( \implies b = 2 \) So, one possible solution is \( a=4, b=2 \). Case II: \( a-b = -2 \) We have two linear equations: \( a+b = 6 \) (Equation from above) \( a-b = -2 \) (From Case II) Adding these two equations: \( (a+b) + (a-b) = 6 + (-2) \)
\( \implies 2a = 4 \)
\( \implies a = 2 \) Substitute \( a=2 \) into \( a+b=6 \): \( 2+b = 6 \)
\( \implies b = 6 - 2 \)
\( \implies b = 4 \) So, another possible solution is \( a=2, b=4 \). Both sets of solutions are valid, showing that algebraic identities can simplify finding values.In simple words: For both parts, match the numbers in the same positions in the equal matrices to create small math problems. Solve these problems to find the values of 'x', 'y', 'z', 'w', 'a', and 'b'. For the second part, use a special algebra trick involving \( (a+b)^2 \) to help find 'a' and 'b'.

🎯 Exam Tip: For problems involving products and sums of variables (like \( a+b \) and \( ab \)), remember to use algebraic identities such as \( (a-b)^2 = (a+b)^2 - 4ab \) or \( (a+b)^2 = (a^2+b^2) + 2ab \) to simplify finding the unknown variables.

Question 11. Find the values of x, y, and z from the following equations:
\[ \left[\begin{array}{r} x+y+z \\ x+z \\ y+z \end{array}\right]=\left[\begin{array}{l} 9 \\ 5 \\ 7 \end{array}\right] \]
Answer: Given that the two matrices are equal, their corresponding elements must be equal. This gives us a system of three linear equations: 1. \( x+y+z = 9 \) 2. \( x+z = 5 \) 3. \( y+z = 7 \) Substitute equation (2) into equation (1): \( (x+z) + y = 9 \)
\( \implies 5 + y = 9 \)
\( \implies y = 9 - 5 \)
\( \implies y = 4 \) Substitute equation (3) into equation (1): \( x + (y+z) = 9 \)
\( \implies x + 7 = 9 \)
\( \implies x = 9 - 7 \)
\( \implies x = 2 \) Now substitute the values of \( x=2 \) into equation (2): \( 2 + z = 5 \)
\( \implies z = 5 - 2 \)
\( \implies z = 3 \) So, the values are \( x=2, y=4 \), and \( z=3 \). This problem demonstrates how to solve a system of equations by carefully substituting known values.In simple words: Since the matrices are equal, each number in the left matrix is the same as the number in the right matrix, in the same position. Use these matches to make three small math problems. Solve them by replacing parts of one problem with numbers from another problem until you find 'x', 'y', and 'z'.

🎯 Exam Tip: For systems of equations like this, look for terms that appear in multiple equations (e.g., \( x+z \) or \( y+z \)). Substituting these simpler expressions into more complex ones can quickly reduce the number of variables and simplify the solution.

Question 12. Find the values of a, b, c and d, if
\[ \left[\begin{array}{rr} a-b & 2 a+c \\ 2 a-b & 3 c+d \end{array}\right]=\left[\begin{array}{rr} -1 & 5 \\ 0 & 13 \end{array}\right] \]
Answer: Given that the two matrices are equal, their corresponding elements must be equal. This gives us a system of four linear equations: 1. \( a-b = -1 \) 2. \( 2a+c = 5 \) 3. \( 2a-b = 0 \) 4. \( 3c+d = 13 \) From equation (3), we can express \( b \) in terms of \( a \): \( b = 2a \) Substitute \( b = 2a \) into equation (1): \( a - (2a) = -1 \)
\( \implies -a = -1 \)
\( \implies a = 1 \) Now substitute \( a=1 \) into \( b=2a \): \( b = 2(1) \)
\( \implies b = 2 \) Substitute \( a=1 \) into equation (2): \( 2(1) + c = 5 \)
\( \implies 2 + c = 5 \)
\( \implies c = 5 - 2 \)
\( \implies c = 3 \) Finally, substitute \( c=3 \) into equation (4): \( 3(3) + d = 13 \)
\( \implies 9 + d = 13 \)
\( \implies d = 13 - 9 \)
\( \implies d = 4 \) So, the values are \( a=1, b=2, c=3 \), and \( d=4 \). This illustrates a step-by-step approach to solving simultaneous equations derived from matrix equality.In simple words: Make four separate math problems by matching the numbers in the same spots in both matrices. Solve these problems in order. First, find 'a' and 'b' using the first and third problems. Then, use 'a' to find 'c' from the second problem. Finally, use 'c' to find 'd' from the fourth problem.

🎯 Exam Tip: Always solve for the variables in an organized manner. Start with equations that contain fewer variables or allow for direct substitution, and then use those solutions to find other unknowns. This reduces complexity and helps avoid errors.