Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 2 Matrices Ex 2.1 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 2 Matrices Ex 2.1
Question 1
Maharashtra Board Solution
Apply the given elementary transformation on each of the following matrices. A = , R1 ↔ R2
Solution & Step-by-Step Answer:
A = By R1 ↔ R2, we get, A ~
Question 2
Maharashtra Board Solution
B = , R1 → R1 → R2
Solution & Step-by-Step Answer:
B = , R1 → R1 → R2 gives, B ~
Question 3
Maharashtra Board Solution
A = , C1 ↔ C2; B = , R1 ↔ R2. What do you observe?
Solution & Step-by-Step Answer:
A = By C1 ↔ C2, we get, A ~ …(1) B = By R1 ↔ R2, we get, B ~ …(2) From (1) and (2), we observe that the new matrices are equal.
Question 4
Maharashtra Board Solution
A = , 2C2 B = , -3R1 Find the addition of the two new matrices.
Solution & Step-by-Step Answer:
A = By 2C2, we get, A ~ B = By -3R1, we get, B ~ Now, addition of the two new matrices

Question 5
Maharashtra Board Solution
A = , 3R3 and then C3 + 2C2.
Solution & Step-by-Step Answer:
A = By 3R3, we get A ~ By C3 + 2C2, we get, A ~ ∴ A ~
Question 6
Maharashtra Board Solution
A = , C3 + 2C2 and then 3R3. What do you conclude from Ex. 5 and Ex. 6 ?
Solution & Step-by-Step Answer:
A = By C3 + 2C2, we get, A ~ ∴ A ~ By 3R3, we get A ~ We conclude from Ex. 5 and Ex. 6 that the matrix remains same by interchanging the order of the elementary transformations. Hence, the transformations are commutative.
Question 7
Maharashtra Board Solution
Use suitable transformation on into an upper triangular matrix.
Solution & Step-by-Step Answer:
Let A = By R2 – 3R1, we get, A ~ This is an upper triangular matrix.
Question 8
Maharashtra Board Solution
Convert into an identity matrix by suitable row transformations.
Solution & Step-by-Step Answer:
Let A = By R2 – 2R1, we get, A ~ By R2, we get, A ~ By R1 + R2, we get, A ~ This is an identity matrix.
Question 9
Maharashtra Board Solution
Transform into an upper triangular matrix by suitable row transformations.
Solution & Step-by-Step Answer:
Let A = By R2 – 2R1 and R3 – 3R1, we get A ~ By R3 – R2, we get, A ~ This is an upper triangular matrix.