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

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

Maharashtra Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.1

Question 1.Apply the given elementary transformation on each of the following matrices.

1) A = [1103], R1 ↔ R2
Solution:
A = [1103]
By R1 ↔ R2, we get,
A ~ [1130]

Question 2.
B = [121534], R1 → R1 → R2
Solution:
B = [121534],
R1 → R1 → R2 gives,
B ~ [126514]

Question 3.
A = [5143], C1 ↔ C2; B = [3415], R1 ↔ R2. What do you observe?
Solution:
A = [5143]
By C1 ↔ C2, we get,
A ~ [4351] …(1)
B = [3415]
By R1 ↔ R2, we get,
B ~ [4351] …(2)
From (1) and (2), we observe that the new matrices are equal.

Question 4.
A = [102113], 2C2
B = [120425], -3R1


Find the addition of the two new matrices.
Solution:

A = [102113]
By 2C2, we get,
A ~ [104213]
B = [120425]
By -3R1, we get,
B ~ [320465]
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 [1324] into an upper triangular matrix.
Solution:
Let A = [1324]
By R2 – 3R1, we get,
A ~ [1022]
This is an upper triangular matrix.

Question 8.
Convert [1213] into an identity matrix by suitable row transformations.
Solution:
Let A = [1213]
By R2 – 2R1, we get,
A ~ [1015]
By (15)R2, we get,
A ~ [1011]
By R1 + R2, we get,
A ~ [1001]
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.


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

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