# Matrices - Ex 2.1 | Maharashtra Board 12th Maths Solutions Chapter 2

### Matrices - Ex 2.1 | Maharashtra Board 12th Maths Solutions Chapter 2

1) A = , R1 ↔ R2
Solution:
A = 
By R1 ↔ R2, we get,
A ~ 

Question 2.
B = , R1 → R1 → R2
Solution:
B = ,
R1 → R1 → R2 gives,
B ~ 

Question 3.
A = , C1 ↔ C2; B = , R1 ↔ R2. What do you observe?
Solution:
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.
A = , 2C2
B = , -3R1

Find the addition of the two new matrices.
Solution:

A = 
By 2C2, we get,
A ~ 
B = 
By -3R1, we get,
B ~ 
Now, addition of the two new matrices

Question 5.
A = 123113301, 3R3 and then C3 + 2C2.
Solution:
A = 123113301
By 3R3, we get
A ~ 129119303
By C3 + 2C2, we get,
A ~ 1291193+2(1)0+2(1)3+2(9)
∴ A ~ 1291191221

Question 6.
A = 123113301, C3 + 2C2 and then 3R3. What do you conclude from Ex. 5 and Ex. 6 ?
Solution:

A = 123113301
By C3 + 2C2, we get,
A ~ 1231133+2(1)0+2(1)1+2(3)
∴ A ~ 123113127
By 3R3, we get
A ~ 1291191221
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.
Use suitable transformation on  into an upper triangular matrix.
Solution:
Let A = 
By R2 – 3R1, we get,
A ~ 
This is an upper triangular matrix.

Question 8.
Convert  into an identity matrix by suitable row transformations.
Solution:
Let A = 
By R2 – 2R1, we get,
A ~ 
By (15)R2, we get,
A ~ 
By R1 + R2, we get,
A ~ 
This is an identity matrix.

Question 9.
Transform 123112234 into an upper triangular matrix by suitable row
transformations.
Solution:
Let A = 123112234
By R2 – 2R1 and R3 – 3R1, we get
A ~ 100135212
By R3 – (53)R2, we get,
A ~ 1001302113
This is an upper triangular matrix.