CBSE Class 12 Mathematics Matrices And Determinants Worksheet Set 03

Question. To promote the making of toilets for women, an organisation tried to generate awareness through (i) house calls (ii) letters and (iii) announcements. The cost for each mode per attempt is given below : (i) ₹ 50 (ii) ₹ 20 (iii) ₹ 40 The number of attempts made in three villages X, Y and Z are given below: village (i) (ii) (iii) X 400 300 100 Y 300 250 75 Z 500 400 150 Find the total cost incurred by the organisation for the three villages separately, using matrices. Write one value generated by the organisation in the society.
Answer: Let ₹ \(A\), ₹ \(B\) and ₹ \(C\) be the cost incurred by the organisation for villages X, Y and Z respectively. Then we get the matrix equation as \[ \begin{bmatrix} 400 & 300 & 100 \\ 300 & 250 & 75 \\ 500 & 400 & 150 \end{bmatrix} \begin{bmatrix} 50 \\ 20 \\ 40 \end{bmatrix} = \begin{bmatrix} A \\ B \\ C \end{bmatrix} \] \[ \Rightarrow \begin{bmatrix} A \\ B \\ C \end{bmatrix} = \begin{bmatrix} 400 \times 50 + 300 \times 20 + 100 \times 40 \\ 300 \times 50 + 250 \times 20 + 75 \times 40 \\ 500 \times 50 + 400 \times 20 + 150 \times 40 \end{bmatrix} \] \[ = \begin{bmatrix} 20,000 + 6,000 + 4,000 \\ 15,000 + 5,000 + 3,000 \\ 25,000 + 8,000 + 6,000 \end{bmatrix} = \begin{bmatrix} 30,000 \\ 23,000 \\ 39,000 \end{bmatrix} \] \( \therefore A = \text{₹ } 30,000 \); \( B = \text{₹ } 23,000 \) and \( C = \text{₹ } 39,000 \). These are respectively the costs incurred by the organisation on villages X, Y and Z respectively. The value generated by the organisation in the society is cleanliness.

Question. If \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \) and \( (A+B)^2 = A^2 + B^2 \), then find the values of \( a \) and \( b \).
Answer: We have, \( A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \) and \( B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \). Consider, \( (A + B) = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} + \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} = \begin{bmatrix} 1+a & 0 \\ 2+b & -2 \end{bmatrix} \). Now, \( (A + B)^2 = \begin{bmatrix} 1+a & 0 \\ 2+b & -2 \end{bmatrix} \begin{bmatrix} 1+a & 0 \\ 2+b & -2 \end{bmatrix} \) \[ = \begin{bmatrix} (1+a)^2 & 0 \\ (2+b)(1+a) - 2(2+b) & 4 \end{bmatrix} = \begin{bmatrix} (1+a)^2 & 0 \\ (2+b)(a-1) & 4 \end{bmatrix} \] Now, consider \( A^2 = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 1-2 & -1+1 \\ 2-2 & -2+1 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \) and \( B^2 = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix} = \begin{bmatrix} a^2+b & a-1 \\ ab-b & b+1 \end{bmatrix} \) \( \therefore A^2 + B^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} + \begin{bmatrix} a^2+b & a-1 \\ ab-b & b+1 \end{bmatrix} = \begin{bmatrix} a^2+b-1 & a-1 \\ ab-b & b \end{bmatrix} \) It is given that \( (A + B)^2 = A^2 + B^2 \) \( \therefore \begin{bmatrix} (1+a)^2 & 0 \\ (2+b)(a-1) & 4 \end{bmatrix} = \begin{bmatrix} a^2+b-1 & a-1 \\ ab-b & b \end{bmatrix} \) By equality of matrices, comparing the corresponding elements, we get: \( a - 1 = 0 \Rightarrow a = 1 \) and \( b = 4 \). Also, \( (1+a)^2 = a^2+b-1 \) and \( (2+b)(a-1) = ab-b \) are satisfied by \( a=1 \) and \( b=4 \). Therefore, \( a = 1 \) and \( b = 4 \).

Question. In a parliament election, a political party hired a public relations firm to promote its candidates in three ways-telephone, house calls and letters. The cost per contact (in paise) is given in matrix A as \[ A = \begin{bmatrix} 140 \\ 200 \\ 150 \end{bmatrix} \begin{matrix} \text{Telephone} \\ \text{House call} \\ \text{Letters} \end{matrix} \] The number of contacts of each type made in two cities X and Y is given in matrix B as \[ B = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{matrix} \text{City X} \\ \text{City Y} \end{matrix} \] Find the total amount spent by the party in the two cities. What should one consider before casting his/her vote-party's promotional activity or their social activities?
Answer: The total amount spent by the party in two cities X and Y is represented in the matrix equation by matrix C as, \( C = BA \) \[ \Rightarrow \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} 1000 & 500 & 5000 \\ 3000 & 1000 & 10000 \end{bmatrix} \begin{bmatrix} 140 \\ 200 \\ 150 \end{bmatrix} \] \[ \Rightarrow \begin{bmatrix} X \\ Y \end{bmatrix} = \begin{bmatrix} 1000 \times 140 + 500 \times 200 + 5000 \times 150 \\ 3000 \times 140 + 1000 \times 200 + 10000 \times 150 \end{bmatrix} \] \[ = \begin{bmatrix} 990000 \\ 2120000 \end{bmatrix} \] \( \therefore X = \text{₹ } 990000 \) and \( Y = \text{₹ } 2120000 \) (Note: Expressed in paise, this is \( \text{₹ } 9900 \) and \( \text{₹ } 21200 \) respectively). One should consider the social activity of the party before casting his/her vote.

Question. If \( \begin{bmatrix} 2x & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ 3 \end{bmatrix} = O \), find \( x \).
Answer: Here, \( \begin{bmatrix} 2x & 3 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -3 & 0 \end{bmatrix} \begin{bmatrix} x \\ 3 \end{bmatrix} = O \) \[ \Rightarrow \begin{bmatrix} 2x & 3 \end{bmatrix} \begin{bmatrix} x+6 \\ -3x \end{bmatrix} = O \] \[ \Rightarrow 2x(x + 6) + 3(-3x) = 0 \] \[ \Rightarrow 2x^2 + 12x - 9x = 0 \] \[ \Rightarrow 2x^2 + 3x = 0 \Rightarrow x(2x + 3) = 0 \] \[ \Rightarrow x = 0, -\frac{3}{2} \]

Question. A trust fund, ₹ 35,000 is to be invested in two different types of bonds. The first bond pays 8% interest per annum which will be given to orphanage and second bond pays 10% interest per annum which will be given to an N.G.O. (Cancer Aid Society). Using matrix multiplication, determine how to divide ₹ 35,000 among two types of bonds if the trust fund obtains an annual total interest of ₹ 3,200. What are the values reflected in this question?
Answer: Trust fund = ₹ 35,000. Let ₹ \( x \) be invested in the first bond and then ₹ \( (35,000 - x) \) will be invested in the second bond. Interest paid on the first bond = \( 8\% = 0.08 \) Interest paid on the second bond = \( 10\% = 0.10 \) Total annual interest = ₹ 3,200. \( \therefore \) In matrices, \( \begin{bmatrix} x & 35,000 - x \end{bmatrix} \begin{bmatrix} 0.08 \\ 0.10 \end{bmatrix} = [3,200] \) \[ \Rightarrow x \times 0.08 + (35,000 - x) \times 0.10 = 3,200 \] \[ \Rightarrow x \times \frac{8}{100} + (35,000 - x) \times \frac{10}{100} = 3,200 \] \[ \Rightarrow 8x + 3,50,000 - 10x = 3,20,000 \] \[ \Rightarrow 2x = 30,000 \] \[ \Rightarrow x = 15,000 \] \( \therefore \) ₹ 15,000 should be invested in the first bond and ₹ 35,000 - ₹ 15,000 = ₹ 20,000 be invested in the second bond. The values reflected in this question are: (i) Spirit of investment. (ii) Giving charity to cancer patients. (iii) Helping the orphans living in the society.

Question. If \( A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \), then find the value of \( A^2 - 3A + 2I \).
Answer: Refer to answer 43.

Question. If \( A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \), then verify that \( A^2 - 4A - 5I = O \).
Answer: Given, \( A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \) Now, \( A^2 = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} \) \( \therefore A^2 - 4A - 5I \) \[ = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - 4 \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix} - 5 \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \] \[ = \begin{bmatrix} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 8 & 8 \\ 8 & 4 & 8 \\ 8 & 8 & 4 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \end{bmatrix} \] \[ = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} = O \]

Question. If \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \), then show that \( A^2 - 5A + 7I = O \).
Answer: Given, \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \) \( \Rightarrow A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 9-1 & 3+2 \\ -3-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} \) and \( 5A = 5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} \) Now, \( A^2 - 5A + 7I \) \[ = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \] \[ = \begin{bmatrix} -7 & 0 \\ 0 & -7 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \]

Transpose of a Matrix

Question. If \( A^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} \), then find \( A^T - B^T \).
Answer: Given, \( A^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{bmatrix} \) \( \Rightarrow B^T = \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \) \( \therefore A^T - B^T = \begin{bmatrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \end{bmatrix} \)

Question. If \( \begin{bmatrix} a+b & 2 \\ 5 & b \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 2 & 2 \end{bmatrix}^T \), then find \( a \).
Answer: Given, \( \begin{bmatrix} a+b & 2 \\ 5 & b \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 2 & 2 \end{bmatrix}^T \) \( \Rightarrow \begin{bmatrix} a+b & 2 \\ 5 & b \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 2 \end{bmatrix} \) On comparing corresponding elements of the matrices, we get \( a + b = 6 \) and \( b = 2 \) \( \Rightarrow a = 4 \).

Question. If \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), find \( A + A' \).
Answer: We have, \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \Rightarrow A' = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \) Now, \( A + A' = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 5 & 8 \end{bmatrix} \).

Question. If \( \begin{bmatrix} 2x+y & 3y \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 6 & 4 \end{bmatrix}^T \), then find \( x \).
Answer: We have \( \begin{bmatrix} 2x+y & 3y \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 6 & 4 \end{bmatrix}^T \) \( \Rightarrow \begin{bmatrix} 2x+y & 3y \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 6 \\ 0 & 4 \end{bmatrix} \) By equality of two matrices, we have \( 2x + y = 6 \) and \( 3y = 6 \Rightarrow y = 2 \). Putting the value of \( y \), we get \( 2x + 2 = 6 \Rightarrow 2x = 4 \Rightarrow x = 2 \).

Question. If matrix \( A = [1\ 2\ 3] \), then find \( AA' \) where \( A' \) is the transpose of matrix A.
Answer: \( A = [1\ 2\ 3] \Rightarrow A' = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \) \( AA' = [1\ 2\ 3] \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \Rightarrow AA' = [1+4+9] = [14] \).

Question. If \( A = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \), then find \( A + A' \) where \( A' \) is the transpose of matrix A.
Answer: \( A = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} \), \( A' = \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix} \), \( \therefore A + A' = \begin{bmatrix} 3 & 4 \\ 2 & 3 \end{bmatrix} + \begin{bmatrix} 3 & 2 \\ 4 & 3 \end{bmatrix} = \begin{bmatrix} 6 & 6 \\ 6 & 6 \end{bmatrix} \).

Question. For the following matrices A and B, verify that \( (AB)' = B'A' \). \( A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \), \( B = [-1\ 2\ 1] \)
Answer: Given \( A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} \), \( B = [-1\ 2\ 1] \) \( AB = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix} [-1\ 2\ 1] = \begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \end{bmatrix} \) \( \dots (AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \) Now, \( A' = [1\ -4\ 3] \) and \( B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \) \( B'A' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} [1\ -4\ 3] = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix} \) \( \therefore (AB)' = B'A' \). Hence proved.

Symmetric and Skew Symmetric Matrices

Question. Matrix \( A = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} \) is given to be symmetric, find values of \( a \) and \( b \).
Answer: Given, \( A = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} \). \( \because A \) is symmetric. \( \therefore A' = A \) \[ \Rightarrow \begin{bmatrix} 0 & 3 & 3a \\ 2b & 1 & 3 \\ -2 & 3 & -1 \end{bmatrix} = \begin{bmatrix} 0 & 2b & -2 \\ 3 & 1 & 3 \\ 3a & 3 & -1 \end{bmatrix} \] On comparing the corresponding elements of the matrices, we get: \( 3a = -2 \Rightarrow a = -\frac{2}{3} \) and \( 2b = 3 \Rightarrow b = \frac{3}{2} \).

Question. If \( A = \begin{bmatrix} 3 & 5 \\ 7 & 9 \end{bmatrix} \) is written as \( A = P + Q \), where \( P \) is a symmetric matrix and \( Q \) is a skew symmetric matrix, then write the matrix \( P \).
Answer: Given, \( A = \begin{bmatrix} 3 & 5 \\ 7 & 9 \end{bmatrix} \Rightarrow A' = \begin{bmatrix} 3 & 7 \\ 5 & 9 \end{bmatrix} \) \( \because P \) is symmetric matrix. So, \( P = \frac{1}{2}(A + A') \) \( \therefore P = \frac{1}{2}\left( \begin{bmatrix} 3 & 5 \\ 7 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 7 \\ 5 & 9 \end{bmatrix} \right) \) \[ = \frac{1}{2} \begin{bmatrix} 3+3 & 5+7 \\ 7+5 & 9+9 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 6 & 12 \\ 12 & 18 \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 6 & 9 \end{bmatrix} \] Hence, the matrix \( P = \begin{bmatrix} 3 & 6 \\ 6 & 9 \end{bmatrix} \).

Question. Express the matrix \( A = \begin{bmatrix} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{bmatrix} \) as the sum of a symmetric and a skew symmetric matrix.
Answer: We know that a square matrix \( A \) can be written as \( A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') \) Out of which \( \frac{1}{2}(A + A') \) is symmetric and \( \frac{1}{2}(A - A') \) is skew symmetric. \( \therefore \) For the given matrix: \( A = \begin{bmatrix} 2 & 4 & -6 \\ 7 & 3 & 5 \\ 1 & -2 & 4 \end{bmatrix} \), \( A' = \begin{bmatrix} 2 & 7 & 1 \\ 4 & 3 & -2 \\ -6 & 5 & 4 \end{bmatrix} \) \( \therefore A + A' = \begin{bmatrix} 4 & 11 & -5 \\ 11 & 6 & 3 \\ -5 & 3 & 8 \end{bmatrix} \) and \( A - A' = \begin{bmatrix} 0 & -3 & -7 \\ 3 & 0 & 7 \\ 7 & -7 & 0 \end{bmatrix} \) Hence, \( A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') \) \[ = \begin{bmatrix} 2 & 11/2 & -5/2 \\ 11/2 & 3 & 3/2 \\ -5/2 & 3/2 & 4 \end{bmatrix} + \begin{bmatrix} 0 & -3/2 & -7/2 \\ 3/2 & 0 & 7/2 \\ 7/2 & -7/2 & 0 \end{bmatrix} \] First of which is symmetric and the second is skew symmetric matrix.

Question. Write a 2 × 2 matrix which is both symmetric and skew symmetric.
Answer: \( \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) is a 2 × 2 symmetric as well as skew symmetric matrix.

Question. For what value of \( x \), is the matrix \( A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix} \) a skew-symmetric matrix?
Answer: The matrix \( A = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 3 \\ x & -3 & 0 \end{bmatrix} \) is skew-symmetric. \( \therefore A' = -A \Rightarrow \begin{bmatrix} 0 & -1 & x \\ 1 & 0 & -3 \\ -2 & 3 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 2 \\ 1 & 0 & -3 \\ -x & 3 & 0 \end{bmatrix} \) \( \Rightarrow x = 2 \).

Question. Express the following matrix as the sum of a symmetric and skew symmetric matrix and verify your result. \( A = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \)
Answer: We know that \( A = \frac{1}{2}(A + A') + \frac{1}{2}(A - A') \) Here, \( \frac{1}{2}(A + A') \) is symmetric matrix and \( \frac{1}{2}(A - A') \) is skew symmetric matrix. Now, \( A = \begin{bmatrix} 3 & -2 & -4 \\ 3 & -2 & -5 \\ -1 & 1 & 2 \end{bmatrix} \Rightarrow A' = \begin{bmatrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{bmatrix} \) \( \therefore \frac{1}{2}(A + A') = \frac{1}{2}\begin{bmatrix} 3+3 & -2+3 & -4-1 \\ 3-2 & -2-2 & -5+1 \\ -1-4 & 1-5 & 2+2 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} \), which is symmetric. and \( \frac{1}{2}(A - A') = \frac{1}{2}\begin{bmatrix} 3-3 & -2-3 & -4+1 \\ 3+2 & -2+2 & -5-1 \\ -1+4 & 1+5 & 2-2 \end{bmatrix} = \frac{1}{2}\begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} \), which is skew symmetric. \( \therefore A = \frac{1}{2}\begin{bmatrix} 6 & 1 & -5 \\ 1 & -4 & -4 \\ -5 & -4 & 4 \end{bmatrix} + \frac{1}{2}\begin{bmatrix} 0 & -5 & -3 \\ 5 & 0 & -6 \\ 3 & 6 & 0 \end{bmatrix} \) \[ \Rightarrow A = \begin{bmatrix} 3 & 1/2 & -5/2 \\ 1/2 & -2 & -2 \\ -5/2 & -2 & 2 \end{bmatrix} + \begin{bmatrix} 0 & -5/2 & -3/2 \\ 5/2 & 0 & -3 \\ 3/2 & 3 & 0 \end{bmatrix} \]

Question. Let \( A = \begin{bmatrix} 3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7 \end{bmatrix} \), express A as a sum of two matrices such that one is symmetric and the other is skew symmetric.
Answer: Refer to answer 65.

Elementary Operation (Transformation) of a Matrix

Question. Use elementary column operation \( C_2 \to C_2 + 2C_1 \) in the following matrix equation: \( \begin{bmatrix} 2 & 1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \)
Answer: We have, \( \begin{bmatrix} 2 & 1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \) On applying \( C_2 \to C_2 + 2C_1 \), we get \( \begin{bmatrix} 2 & 5 \\ 2 & 4 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \).

Question. Use elementary column operations \( C_2 \to C_2 - 2C_1 \) in the matrix equation \( \begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \)
Answer: We have, \( \begin{bmatrix} 4 & 2 \\ 3 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} \) Applying \( C_2 \to C_2 - 2C_1 \), we get \( \begin{bmatrix} 4 & -6 \\ 3 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} \begin{bmatrix} 2 & -4 \\ 1 & -1 \end{bmatrix} \).

Invertible Matrices

Question. Using elementary row transformations, find the inverse of the following matrix. \( A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \)
Answer: We have, \( A = \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} \) We know that, \( A = IA \) \[ \Rightarrow \begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} A \] Applying \( R_1 \to R_1 - R_2 \), we get \[ \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to R_2 - R_1 \), we get \[ \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} A \] Applying \( R_1 \to R_1 - 2 R_2 \), we get \[ \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} A \] Hence, \( A^{-1} = \begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix} \).

Question. Using elementary operations, find the inverse of the following matrix: \( \begin{bmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} \)
Answer: We have \( A = IA \) \[ \Rightarrow \begin{bmatrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to R_2 + R_1, R_3 \to R_3 + 3R_1 \), we get \[ \begin{bmatrix} -1 & 1 & 2 \\ 0 & 3 & 5 \\ 0 & 4 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 3 & 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to R_2 - R_3 \), we get \[ \begin{bmatrix} -1 & 1 & 2 \\ 0 & -1 & -2 \\ 0 & 4 & 7 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & -1 \\ 3 & 0 & 1 \end{bmatrix} A \] Applying \( R_1 \to R_1 + R_2 \), we get \[ \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & -2 \\ 0 & 4 & 7 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ -2 & 1 & -1 \\ 3 & 0 & 1 \end{bmatrix} A \] Applying \( R_3 \to R_3 + 4R_2 \), we get \[ \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & -2 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ -2 & 1 & -1 \\ -5 & 4 & -3 \end{bmatrix} A \] Applying \( R_2 \to R_2 - 2R_3 \), we get \[ \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -7 & 5 \\ -5 & 4 & -3 \end{bmatrix} A \] Applying \( R_1 \to (-1)R_1, R_2 \to (-1)R_2, R_3 \to (-1)R_3 \) \[ \Rightarrow \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{bmatrix} A \] Hence, \( A^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{bmatrix} \).

Question. Using elementary transformations, find the inverse of the matrix. \( \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \)
Answer: Consider \( A = \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} \) We have \( A = IA \) \[ \begin{bmatrix} 1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to R_2 + 3R_1 \) and \( R_3 \to R_3 - 2R_1 \), we get \[ \begin{bmatrix} 1 & 3 & -2 \\ 0 & 9 & -7 \\ 0 & -5 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to \frac{R_2}{9} \), we get \[ \begin{bmatrix} 1 & 3 & -2 \\ 0 & 1 & -7/9 \\ 0 & -5 & 4 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1/3 & 1/9 & 0 \\ -2 & 0 & 1 \end{bmatrix} A \] Applying \( R_3 \to R_3 + 5R_2 \), we get \[ \begin{bmatrix} 1 & 3 & -2 \\ 0 & 1 & -7/9 \\ 0 & 0 & 1/9 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1/3 & 1/9 & 0 \\ -1/3 & 5/9 & 1 \end{bmatrix} A \] Applying \( R_3 \to 9R_3 \), we get \[ \begin{bmatrix} 1 & 3 & -2 \\ 0 & 1 & -7/9 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1/3 & 1/9 & 0 \\ -3 & 5 & 9 \end{bmatrix} A \] Applying \( R_1 \to R_1 - 3R_2 \), we get \[ \begin{bmatrix} 1 & 0 & 1/3 \\ 0 & 1 & -7/9 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -1/3 & 0 \\ 1/3 & 1/9 & 0 \\ -3 & 5 & 9 \end{bmatrix} A \] Applying \( R_1 \to R_1 - \frac{1}{3}R_3, R_2 \to R_2 + \frac{7}{9}R_3 \), we get \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{bmatrix} A \] \( \therefore A^{-1} = \begin{bmatrix} 1 & -2 & -3 \\ -2 & 4 & 7 \\ -3 & 5 & 9 \end{bmatrix} \).

Question. Find the inverse of the following matrix using elementary operations: \( A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} \)
Answer: Since, \( A = IA \) \( \therefore \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \) Applying \( R_2 \to R_2 + R_1 \), we get \[ \begin{bmatrix} 1 & 2 & -2 \\ 0 & 5 & -2 \\ 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to \frac{1}{5}R_2 \), we get \[ \begin{bmatrix} 1 & 2 & -2 \\ 0 & 1 & -2/5 \\ 0 & -2 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 1/5 & 1/5 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_1 \to R_1 - 2R_2, R_3 \to R_3 + 2R_2 \), we get \[ \begin{bmatrix} 1 & 0 & -6/5 \\ 0 & 1 & -2/5 \\ 0 & 0 & 1/5 \end{bmatrix} = \begin{bmatrix} 3/5 & -2/5 & 0 \\ 1/5 & 1/5 & 0 \\ 2/5 & 2/5 & 1 \end{bmatrix} A \] Applying \( R_3 \to 5R_3 \), we get \[ \begin{bmatrix} 1 & 0 & -6/5 \\ 0 & 1 & -2/5 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3/5 & -2/5 & 0 \\ 1/5 & 1/5 & 0 \\ 2 & 2 & 5 \end{bmatrix} A \] Applying \( R_1 \to R_1 + \frac{6}{5}R_3, R_2 \to R_2 + \frac{2}{5}R_3 \), we get \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} A \] Hence, \( A^{-1} = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \).

Question. Obtain the inverse of the following matrix using elementary operations: \( A = \begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix} \)
Answer: Since, \( A = IA \) \[ \Rightarrow \begin{bmatrix} 3 & 0 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_1 \to R_1 - R_2 \), we get \[ \begin{bmatrix} 1 & -3 & -1 \\ 2 & 3 & 0 \\ 0 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to R_2 - 2R_1 \), we get \[ \begin{bmatrix} 1 & -3 & -1 \\ 0 & 9 & 2 \\ 0 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2 & 3 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to \frac{1}{9}R_2 \), we get \[ \begin{bmatrix} 1 & -3 & -1 \\ 0 & 1 & 2/9 \\ 0 & 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 & 0 \\ -2/9 & 1/3 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_1 \to R_1 + 3R_2, R_3 \to R_3 - 4R_2 \), we get \[ \begin{bmatrix} 1 & 0 & -1/3 \\ 0 & 1 & 2/9 \\ 0 & 0 & 1/9 \end{bmatrix} = \begin{bmatrix} 1/3 & 0 & 0 \\ -2/9 & 1/3 & 0 \\ 8/9 & -4/3 & 1 \end{bmatrix} A \] Applying \( R_3 \to 9R_3 \), we get \[ \begin{bmatrix} 1 & 0 & -1/3 \\ 0 & 1 & 2/9 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1/3 & 0 & 0 \\ -2/9 & 1/3 & 0 \\ 8 & -12 & 9 \end{bmatrix} A \] Applying \( R_1 \to R_1 + \frac{1}{3}R_3, R_2 \to R_2 - \frac{2}{9}R_3 \), we get \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix} A \] Hence, \( A^{-1} = \begin{bmatrix} 3 & -4 & 3 \\ -2 & 3 & -2 \\ 8 & -12 & 9 \end{bmatrix} \).

Question. Using elementary row transformations, find the inverse of the following matrix: \( A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix} \)
Answer: Since, \( A = IA \) \[ \Rightarrow \begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} A \] Applying \( R_2 \to R_2 - 2R_1, R_3 \to R_3 + 2R_1 \), we get \[ \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix} A \] Applying \( R_1 \to R_1 - 2R_2 \), we get \[ \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & -2 & 0 \\ -2 & 1 & 0 \\ 2 & 0 & 1 \end{bmatrix} A \] Applying \( R_1 \to R_1 - R_3, R_2 \to R_2 - R_3 \), we get \[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{bmatrix} A \] \( \therefore A^{-1} = \begin{bmatrix} 3 & -2 & -1 \\ -4 & 1 & -1 \\ 2 & 0 & 1 \end{bmatrix} \).

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