Matrices Class 12 Maths 1 Exercise 2.1 Solutions Maharashtra Board

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.1 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.1 Questions And Answers Maharashtra Board

Question 1.
Apply the given elementary transformation on each of the following matrices.
A = $\left[\begin{array}{cc} 1 & 0 \\ -1 & 3 \end{array}\right]$, R₁ ↔ R₂
Solution:
A = $\left[\begin{array}{cc} 1 & 0 \\ -1 & 3 \end{array}\right]$
By R₁ ↔ R₂, we get,
A ~ $\left[\begin{array}{rr} -1 & 3 \\ 1 & 0 \end{array}\right]$

Question 2.
B = $\left[\begin{array}{ccc} 1 & -1 & 3 \\ 2 & 5 & 4 \end{array}\right]$, R₁ → R₁ → R₂
Solution:
B = $\left[\begin{array}{ccc} 1 & -1 & 3 \\ 2 & 5 & 4 \end{array}\right]$,
R₁ → R₁ → R₂ gives,
B ~ $\left[\begin{array}{rrr} -1 & -6 & -1 \\ 2 & 5 & 4 \end{array}\right]$

Question 3.
A = $\left[\begin{array}{ll} 5 & 4 \\ 1 & 3 \end{array}\right]$, C₁ ↔ C₂; B = $\left[\begin{array}{ll} 3 & 1 \\ 4 & 5 \end{array}\right]$, R₁ ↔ R₂. What do you observe?
Solution:
A = $\left[\begin{array}{ll} 5 & 4 \\ 1 & 3 \end{array}\right]$
By C₁ ↔ C₂, we get,
A ~ $\left[\begin{array}{ll} 4 & 5 \\ 3 & 1 \end{array}\right]$ …(1)
B = $\left[\begin{array}{ll} 3 & 1 \\ 4 & 5 \end{array}\right]$
By R₁ ↔ R₂, we get,
B ~ $\left[\begin{array}{ll} 4 & 5 \\ 3 & 1 \end{array}\right]$ …(2)
From (1) and (2), we observe that the new matrices are equal.

Question 4.
A = $\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & 3 \end{array}\right]$, 2C₂
B = $\left[\begin{array}{lll} 1 & 0 & 2 \\ 2 & 4 & 5 \end{array}\right]$, -3R₁
Find the addition of the two new matrices.
Solution:
A = $\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 1 & 3 \end{array}\right]$
By 2C₂, we get,
A ~ $\left[\begin{array}{rrr} 1 & 4 & -1 \\ 0 & 2 & 3 \end{array}\right]$
B = $\left[\begin{array}{lll} 1 & 0 & 2 \\ 2 & 4 & 5 \end{array}\right]$
By -3R₁, we get,
B ~ $\left[\begin{array}{rrr} -3 & 0 & -6 \\ 2 & 4 & 5 \end{array}\right]$
Now, addition of the two new matrices

Question 5.
A = $\left[\begin{array}{ccc} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right]$, 3R₃ and then C₃ + 2C₂.
Solution:
A = $\left[\begin{array}{ccc} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right]$
By 3R₃, we get
A ~ $\left[\begin{array}{rrr} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 9 & 9 & 3 \end{array}\right]$
By C₃ + 2C₂, we get,
A ~ $\left(\begin{array}{rrr} 1 & -1 & 3+2(-1) \\ 2 & 1 & 0+2(1) \\ 9 & 9 & 3+2(9) \end{array}\right)$
∴ A ~ $\left(\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 9 & 9 & 21 \end{array}\right)$

Question 6.
A = $\left(\begin{array}{rrr} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right)$, C₃ + 2C₂ and then 3R₃. What do you conclude from Ex. 5 and Ex. 6 ?
Solution:
A = $\left(\begin{array}{rrr} 1 & -1 & 3 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{array}\right)$
By C₃ + 2C₂, we get,
A ~ $\left(\begin{array}{rrr} 1 & -1 & 3+2(-1) \\ 2 & 1 & 0+2(1) \\ 3 & 3 & 1+2(3) \end{array}\right)$
∴ A ~ $\left(\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 3 & 3 & 7 \end{array}\right)$
By 3R₃, we get
A ~ $\left(\begin{array}{rrr} 1 & -1 & 1 \\ 2 & 1 & 2 \\ 9 & 9 & 21 \end{array}\right)$
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 $\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]$ into an upper triangular matrix.
Solution:
Let A = $\left[\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right]$
By R₂ – 3R₁, we get,
A ~ $\left[\begin{array}{rr} 1 & 2 \\ 0 & -2 \end{array}\right]$
This is an upper triangular matrix.

Question 8.
Convert $\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]$ into an identity matrix by suitable row transformations.
Solution:
Let A = $\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]$
By R₂ – 2R₁, we get,
A ~ $\left[\begin{array}{rr} 1 & -1 \\ 0 & 5 \end{array}\right]$
By $\left(\frac{1}{5}\right)$R₂, we get,
A ~ $\left[\begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right]$
By R₁ + R₂, we get,
A ~ $\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$
This is an identity matrix.

Question 9.
Transform $\left[\begin{array}{rrr} 1 & -1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{array}\right]$ into an upper triangular matrix by suitable row transformations.
Solution:
Let A = $\left[\begin{array}{rrr} 1 & -1 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{array}\right]$
By R₂ – 2R₁ and R₃ – 3R₁, we get
A ~ $\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 3 & -1 \\ 0 & 5 & -2 \end{array}\right]$
By R₃ – $\left(\frac{5}{3}\right)$R₂, we get,
A ~ $\left(\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 3 & -1 \\ 0 & 0 & -\frac{1}{3} \end{array}\right)$
This is an upper triangular matrix.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 12 Maths 1
• Exercise 2.2 Class 12 Maths 1
• Exercise 2.3 Class 12 Maths 1
• Miscellaneous Exercise 2A Class 12 Maths 1
• Miscellaneous Exercise 2B Class 12 Maths 1