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

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

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

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.