TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(f)

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Students must practice these TS Intermediate Maths 1A Solutions Chapter 3 Matrices Ex 3(f) to find a better approach to solving the problems.

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(f)

I.
Find the rank of each of the following matrices.

Question 1.
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right]\) and det A = 1
∴ ρ(A) = Rank of the matrix A = 1.

Question 2.
\(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) and det A = 1 ≠ 0.
∴ ρ(A) = 2

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(f)

Question 3.
\(\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right]\) and det A = 0
∴ ρ(A) = 1

Question 4.
\(\left[\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right]\) and det A = -1 ≠ 0.

Question 5.
\(\left[\begin{array}{rrr}
1 & 0 & -4 \\
2 & -1 & 3
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{rrr}
1 & 0 & -4 \\
2 & -1 & 3
\end{array}\right]\)
The determinant of a submatrix of order 2 × 2 of A = \(\left|\begin{array}{ll}
1 & -4 \\
2 & 3
\end{array}\right|\) = 3 + 8 = 11 ≠ 0
∴ ρ(A) = 2

Question 6.
\(\left[\begin{array}{lll}
1 & 3 & 6 \\
2 & 4 & 3
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 3 & 6 \\
2 & 4 & 3
\end{array}\right]\)
The determinant of a submatrix order 2 × 2 of A is = \(\left|\begin{array}{ll}
1 & 3 \\
2 & 4
\end{array}\right|\) = -2 ≠ 0

II.
Question 1.
\(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) and det A = 1(1 – 0) = 1 ≠ 0
∴ ρ(A) = 3

Question 2.
\(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right]\)
and det A = 1(6) – 4(4) – 1(2)
= 6 – 16 – 12 = -12 ≠ 0
∴ ρ(A) = 3

Question 3.
\(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\) (March 2015 T.S)
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\)
and det A = 1(6 – 4) – 2(4) + 3(2)
= 2 – 8 + 6 = 0
The determinant of submatrix of order 2 × 2 of A = \(\left|\begin{array}{ll}
2 & 3 \\
3 & 4
\end{array}\right|\) = 8 – 9 = – 1 ≠ 0
Hence ρ(A) = 2

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(f)

Question 4.
\(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right]\) and
det A = 1(0) – 1(0) + 1(0) = 0
The determinant of submatrix of order 2 × 2 of A is \(\left|\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right|\) = 0
Hence ρ(A) = 1

Question 5.
\(\left[\begin{array}{rrrr}
1 & 2 & 0 & -1 \\
3 & 4 & 1 & 2 \\
-2 & 3 & 2 & 5
\end{array}\right]\)
Answer:
Consider 3 × 3 submatrix of above matrix
|A| = \(\left|\begin{array}{ccc}
1 & 2 & 0 \\
3 & 4 & 1 \\
-2 & 3 & 2
\end{array}\right|\)
= 1(8 – 3) – 2(9 + 8)
= 5 – 34 = -29 ≠ 0
∴ ρ(A) = 3

Question 6.
\(\left[\begin{array}{rrrr}
0 & 1 & 1 & -2 \\
4 & 0 & 2 & 5 \\
2 & 1 & 3 & 1
\end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{rrrr}
0 & 1 & 1 & -2 \\
4 & 0 & 2 & 5 \\
2 & 1 & 3 & 1
\end{array}\right]\) and
Consider a submatrix B of order 3 × 3 of above matrix ‘A’.
Then |B| = \(\left|\begin{array}{lll}
0 & 1 & 1 \\
4 & 0 & 2 \\
2 & 1 & 3
\end{array}\right|\)
= -1(12 – 4) + 1(4)
= -8 + 4 = -4
Hence ρ(A) = 3

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