Maharashtra Board Class 12 Maths Part 1 Chapter 2 Matrices 2.1 Solutions

Question 1. Construct a matrix \( A = [a_{ij}]_{3 \times 2} \) whose elements \( a_{ij} \) is given by
(i) \( a_{ij} = \frac{(i-j)^2}{5-i} \)
Answer:
A matrix \( A \) of order \( 3 \times 2 \) has 3 rows and 2 columns, and is represented as:
\( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} \)
The formula for the elements is given by \( a_{ij} = \frac{(i-j)^2}{5-i} \).
For \( i = 1, j = 1 \):
\( a_{11} = \frac{(1-1)^2}{5-1} = \frac{0}{4} = 0 \)
For \( i = 1, j = 2 \):
\( a_{12} = \frac{(1-2)^2}{5-1} = \frac{(-1)^2}{4} = \frac{1}{4} \)
For \( i = 2, j = 1 \):
\( a_{21} = \frac{(2-1)^2}{5-2} = \frac{1^2}{3} = \frac{1}{3} \)
For \( i = 2, j = 2 \):
\( a_{22} = \frac{(2-2)^2}{5-2} = \frac{0}{3} = 0 \)
For \( i = 3, j = 1 \):
\( a_{31} = \frac{(3-1)^2}{5-3} = \frac{2^2}{2} = \frac{4}{2} = 2 \)
For \( i = 3, j = 2 \):
\( a_{32} = \frac{(3-2)^2}{5-3} = \frac{1^2}{2} = \frac{1}{2} \)
Therefore, the required matrix is:
\( A = \begin{bmatrix} 0 & \frac{1}{4} \\ \frac{1}{3} & 0 \\ 2 & \frac{1}{2} \end{bmatrix} \)
In simple words: To build this matrix, we find the value of each position by putting its row number \( i \) and column number \( j \) into the given formula.

🎯 Exam Tip: Always write down the general form of the matrix with its dimensions first to ensure you calculate the correct number of rows and columns without any confusion.

Question 1. Construct a matrix \( A = [a_{ij}]_{3 \times 2} \) whose elements \( a_{ij} \) are given by:
(i) \( a_{ij} = \frac{(i-j)^2}{5-i} \)
(ii) \( a_{ij} = i - 3j \)
Answer:
In general, a \( 3 \times 2 \) matrix has 3 rows and 2 columns, and is represented as:
\( A = [a_{ij}]_{3 \times 2} = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} \)

(i) \( a_{ij} = \frac{(i-j)^2}{5-i} \)
For \( i = 1, 2, 3 \) and \( j = 1, 2 \):
\( a_{11} = \frac{(1-1)^2}{5-1} = \frac{0}{4} = 0 \)
\( a_{12} = \frac{(1-2)^2}{5-1} = \frac{(-1)^2}{4} = \frac{1}{4} \)
\( a_{21} = \frac{(2-1)^2}{5-2} = \frac{1^2}{3} = \frac{1}{3} \)
\( a_{22} = \frac{(2-2)^2}{5-2} = \frac{0}{3} = 0 \)
\( a_{31} = \frac{(3-1)^2}{5-3} = \frac{2^2}{2} = \frac{4}{2} = 2 \)
\( a_{32} = \frac{(3-2)^2}{5-3} = \frac{1^2}{2} = \frac{1}{2} \)
Substituting these values into the matrix format gives us the desired array of numbers.
\( \therefore A = \begin{bmatrix} 0 & \frac{1}{4} \\ \frac{1}{3} & 0 \\ 2 & \frac{1}{2} \end{bmatrix} \)

(ii) \( a_{ij} = i - 3j \)
For \( i = 1, 2, 3 \) and \( j = 1, 2 \):
\( a_{11} = 1 - 3(1) = 1 - 3 = -2 \)
\( a_{12} = 1 - 3(2) = 1 - 6 = -5 \)
\( a_{21} = 2 - 3(1) = 2 - 3 = -1 \)
\( a_{22} = 2 - 3(2) = 2 - 6 = -4 \)
\( a_{31} = 3 - 3(1) = 3 - 3 = 0 \)
\( a_{32} = 3 - 3(2) = 3 - 6 = -3 \)
\( \therefore A = \begin{bmatrix} -2 & -5 \\ -1 & -4 \\ 0 & -3 \end{bmatrix} \)
In simple words: To build these matrices, we look at the row number \( i \) and column number \( j \) for each position. We plug these numbers into the given formula to calculate the value for each spot in our 3-row, 2-column grid.

🎯 Exam Tip: Double-check the order of the matrix (rows \( \times \) columns) before calculating elements to avoid finding values for non-existent positions like \( a_{23} \).

Question 1. (iii) Construct a matrix \( A = [a_{ij}]_{3 \times 2} \) whose elements are given by \( a_{ij} = \frac{(i+j)^3}{5} \)
Answer: Given: \( a_{ij} = \frac{(i+j)^3}{5} \) for a \( 3 \times 2 \) matrix.
\( \therefore a_{11} = \frac{(1+1)^3}{5} = \frac{2^3}{5} = \frac{8}{5} \)
\( a_{12} = \frac{(1+2)^3}{5} = \frac{3^3}{5} = \frac{27}{5} \)
\( a_{21} = \frac{(2+1)^3}{5} = \frac{3^3}{5} = \frac{27}{5} \)
\( a_{22} = \frac{(2+2)^3}{5} = \frac{4^3}{5} = \frac{64}{5} \)
\( a_{31} = \frac{(3+1)^3}{5} = \frac{4^3}{5} = \frac{64}{5} \)
\( a_{32} = \frac{(3+2)^3}{5} = \frac{5^3}{5} = \frac{125}{5} = 25 \)
\( \therefore A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix} = \begin{bmatrix} \frac{8}{5} & \frac{27}{5} \\ \frac{27}{5} & \frac{64}{5} \\ \frac{64}{5} & 25 \end{bmatrix} = \frac{1}{5} \begin{bmatrix} 8 & 27 \\ 27 & 64 \\ 64 & 125 \end{bmatrix} \)
In simple words: To find each element of the matrix, we substitute the row number for i and the column number for j into the given formula, and then arrange these values into a grid with 3 rows and 2 columns.

🎯 Exam Tip: Double-check your calculations for each element by carefully substituting the row index \( i \) and column index \( j \) into the formula to avoid simple arithmetic errors.

Question 2. Classify each of the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular matrix:
(i) \( \begin{bmatrix} 3 & -2 & 4 \\ 0 & 0 & -5 \\ 0 & 0 & 0 \end{bmatrix} \)
(ii) \( \begin{bmatrix} 5 \\ 4 \\ -3 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 9 & \sqrt{2} & -3 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \)
(v) \( \begin{bmatrix} 2 & 0 & 0 \\ 3 & -1 & 0 \\ -7 & 3 & 1 \end{bmatrix} \)
(vi) \( \begin{bmatrix} 3 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & \frac{1}{3} \end{bmatrix} \)
(vii) \( \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Answer:
(i) Since all the elements below the diagonal are zero, it is an upper triangular matrix.
(ii) This matrix has only one column, it is a column matrix.
(iii) This matrix has only one row, it is a row matrix.
(iv) Since diagonal elements are equal and non-diagonal elements are zero, it is a scalar matrix.
(v) Since all the elements above the diagonal are zero, it is a lower triangular matrix.
(vi) Since all non-diagonal elements are zero, it is a diagonal matrix.
(vii) Since diagonal elements are 1 and non-diagonal elements are zero, it is a unit (or identity) matrix.
In simple words: We classify matrices based on their shape and the arrangement of their numbers: a single row is a row matrix, a single column is a column matrix, and special patterns of zeros around the diagonal define triangular, diagonal, scalar, or unit matrices.

🎯 Exam Tip: Memorize the definitions of special matrices like scalar and identity matrices, as identifying them correctly is crucial for solving matrix multiplication and determinant problems.

Question 3. Which of the following matrices are singular or non-singular:
(i) \( \begin{bmatrix} a & b & c \\ p & q & r \\ 2a - p & 2b - q & 2c - r \end{bmatrix} \)
(ii) \( \begin{bmatrix} 5 & 0 & 5 \\ 1 & 99 & 100 \\ 6 & 99 & 105 \end{bmatrix} \)
Answer:
To determine if a matrix is singular, we must calculate its determinant and check if it equals zero.
(i) Let \( A = \begin{bmatrix} a & b & c \\ p & q & r \\ 2a - p & 2b - q & 2c - r \end{bmatrix} \)
\( \therefore |A| = \begin{vmatrix} a & b & c \\ p & q & r \\ 2a - p & 2b - q & 2c - r \end{vmatrix} \)
By performing the row operation \( R_3 + R_2 \), we get:
\( |A| = \begin{vmatrix} a & b & c \\ p & q & r \\ 2a & 2b & 2c \end{vmatrix} \)
Taking out the common factor 2 from the third row, we have:
\( = 2 \begin{vmatrix} a & b & c \\ p & q & r \\ a & b & c \end{vmatrix} \)
Since rows \( R_1 \) and \( R_3 \) are identical, the value of the determinant becomes zero.
\( = 2 \times 0 = 0 \)
\( \therefore \) A is a singular matrix.

(ii) Let \( B = \begin{bmatrix} 5 & 0 & 5 \\ 1 & 99 & 100 \\ 6 & 99 & 105 \end{bmatrix} \)
\( \therefore |B| = \begin{vmatrix} 5 & 0 & 5 \\ 1 & 99 & 100 \\ 6 & 99 & 105 \end{vmatrix} \)
By performing the row operation \( R_3 - R_2 \), we get:
\( |B| = \begin{vmatrix} 5 & 0 & 5 \\ 1 & 99 & 100 \\ 5 & 0 & 5 \end{vmatrix} \)
Since rows \( R_1 \) and \( R_3 \) are identical, the determinant of this matrix is also zero.
\( = 0 \)
\( \therefore \) The given matrix is a singular matrix.
In simple words: A matrix is called singular if its determinant is zero, and non-singular if its determinant is any other number. By using simple row operations, we found that the determinant for both matrices is zero, which means both of them are singular matrices.

🎯 Exam Tip: Always look for row or column operations that can make two rows or columns identical, as this immediately proves the determinant is zero without tedious calculations.

Question 1. Determine whether the following matrices are singular or non-singular:
(ii) \( \begin{bmatrix} 5 & 0 & 5 \\ 1 & 99 & 100 \\ 6 & 99 & 105 \end{bmatrix} \)
(iii) \( \begin{bmatrix} 3 & 5 & 7 \\ -2 & 1 & 4 \\ 3 & 2 & 5 \end{bmatrix} \)
(iv) \( \begin{bmatrix} 7 & 5 \\ -4 & 7 \end{bmatrix} \)
Answer:
(ii) Let \( B = \begin{bmatrix} 5 & 0 & 5 \\ 1 & 99 & 100 \\ 6 & 99 & 105 \end{bmatrix} \)
\( \therefore |B| = \begin{vmatrix} 5 & 0 & 5 \\ 1 & 99 & 100 \\ 6 & 99 & 105 \end{vmatrix} \)
By \( R_3 - R_2 \), we get
\( |B| = \begin{vmatrix} 5 & 0 & 5 \\ 1 & 99 & 100 \\ 5 & 0 & 5 \end{vmatrix} \)
\( = 0 \) [\( \because R_1 \equiv R_3 \)]
\( \therefore B \) is a singular matrix.

(iii) Let \( C = \begin{bmatrix} 3 & 5 & 7 \\ -2 & 1 & 4 \\ 3 & 2 & 5 \end{bmatrix} \)
\( \therefore |C| = \begin{vmatrix} 3 & 5 & 7 \\ -2 & 1 & 4 \\ 3 & 2 & 5 \end{vmatrix} \)
\( = 3(5 - 8) - 5(-10 - 12) + 7(-4 - 3) \)
\( = -9 + 110 - 49 \)
\( = 52 \neq 0 \)
\( \therefore C \) is a non-singular matrix.

(iv) Let \( D = \begin{bmatrix} 7 & 5 \\ -4 & 7 \end{bmatrix} \)
\( \dots |D| = \begin{vmatrix} 7 & 5 \\ -4 & 7 \end{vmatrix} \)
\( = 49 - (-20) \)
\( = 69 \neq 0 \)
\( \therefore D \) is a non-singular matrix.
In simple words: To find if a matrix is singular or non-singular, we calculate its determinant. If the determinant is equal to 0, the matrix is singular; if the determinant is any other number, it is non-singular.

🎯 Exam Tip: Always state the condition for singularity (\( |A| = 0 \)) clearly before solving, and use row or column operations to simplify large determinants quickly.

Question 4. Find k, if the following matrices are singular:
(i) \( \begin{bmatrix} 7 & 3 \\ -2 & k \end{bmatrix} \)
(ii) \( \begin{bmatrix} 4 & 3 & 1 \\ 7 & k & 1 \\ 10 & 9 & 1 \end{bmatrix} \)
Answer:
(i) Let \( A = \begin{bmatrix} 7 & 3 \\ -2 & k \end{bmatrix} \)
Since \( A \) is a singular matrix, its determinant must be equal to zero, which means \( |A| = 0 \). This fundamental property helps us set up a simple algebraic equation to solve for the unknown variable.
\( \implies \begin{vmatrix} 7 & 3 \\ -2 & k \end{vmatrix} = 0 \)
\( \implies 7k - 3(-2) = 0 \)
\( \implies 7k - (-6) = 0 \)
\( \implies 7k = -6 \)
\( \implies k = -\frac{6}{7} \)

(ii) Let \( B = \begin{bmatrix} 4 & 3 & 1 \\ 7 & k & 1 \\ 10 & 9 & 1 \end{bmatrix} \)
Since \( B \) is a singular matrix, its determinant is also zero, meaning \( |B| = 0 \).
\( \implies \begin{vmatrix} 4 & 3 & 1 \\ 7 & k & 1 \\ 10 & 9 & 1 \end{vmatrix} = 0 \)
\( \implies 4(k(1) - 1(9)) - 3(7(1) - 1(10)) + 1(7(9) - k(10)) = 0 \)
\( \implies 4(k - 9) - 3(7 - 10) + 1(63 - 10k) = 0 \)
\( \implies 4k - 36 - 3(-3) + 63 - 10k = 0 \)
\( \implies 4k - 36 + 9 + 63 - 10k = 0 \)
\( \implies -6k + 36 = 0 \)
\( \implies 6k = 36 \)
\( \implies k = 6 \)
In simple words: A matrix is called singular when its determinant is equal to zero. To find the missing value of \( k \), we calculate the determinant of the matrix, set it to zero, and solve the resulting equation.

🎯 Exam Tip: Always remember that a matrix is singular if and only if its determinant is zero. Be extremely careful with negative signs when expanding \( 3 \times 3 \) determinants to avoid calculation errors.

Question. Find the value of \( k \) if the following matrix is singular:
(iii) \( \begin{bmatrix} k - 1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{bmatrix} \)
Answer:
Let \( C = \begin{bmatrix} k - 1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{bmatrix} \)
Since \( C \) is a singular matrix, its determinant must be equal to zero, which means \( |C| = 0 \).
\( \therefore \begin{vmatrix} k - 1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{vmatrix} = 0 \)
\( \therefore (k - 1)(4 + 4) - 2(12 - 2) + 3(-6 - 1) = 0 \)
\( \therefore 8k - 8 - 20 - 21 = 0 \)
\( \therefore 8k = 49 \)
\( \therefore k = \frac{49}{8} \)
In simple words: A singular matrix is a special square matrix whose determinant is equal to zero. By setting the determinant of this matrix to zero and solving the algebraic equation, we find the value of \( k \).

🎯 Exam Tip: Always write down the condition \( |C| = 0 \) clearly before expanding the determinant to ensure you get full marks for the method. Be extra careful with signs when expanding \( 3(-6-1) \).