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

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Matrices Solutions Exercise 3(f) will help students to clear their doubts quickly.

## Intermediate 1st Year Maths 1A Matrices Solutions Exercise 3(f)

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

Question 1.
$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right|$$ = 0 – 0 = 0
and |1| = 1 ≠ 0
∴ ρ(A) = 1

Question 2.
$$\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right|$$ = 1 – 0 = 1 ≠ 0
∴ ρ(A) = 2

Question 3.
$$\left[\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ll} 1 & 1 \\ 0 & 0 \end{array}\right|$$ = 0 – 0 = 0
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 4.
$$\left[\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ll} 1 & 1 \\ 1 & 0 \end{array}\right|$$ = 0 – 1 = -1 ≠ 0
∴ ρ(A) = 2

Question 5.
$$\left[\begin{array}{ccc} 1 & 0 & -4 \\ 2 & -1 & 3 \end{array}\right]$$
Solution:
$$\left|\begin{array}{cc} 1 & -4 \\ 2 & 3 \end{array}\right|$$ = 3 + 8 = -11 ≠ 0
∴ ρ(A) = 2

Question 6.
$$\left[\begin{array}{lll} 1 & 2 & 6 \\ 2 & 4 & 3 \end{array}\right]$$
Solution:
$$\left|\begin{array}{ll} 2 & 6 \\ 4 & 3 \end{array}\right|$$ = 6 – 24 = -18 ≠ 0
∴ ρ(A) = 2

II.

Question 1.
$$\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right|$$
= 1(1 – 0) – 0(0 – 0) + 0(0 – 0)
= 1 – 0 + 0
= 1 ≠ 0
∴ ρ(A) = 3

Question 2.
$$\left[\begin{array}{ccc} 1 & 4 & -1 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{ccc} 1 & 4 & -1 \\ 2 & 3 & 0 \\ 0 & 1 & 2 \end{array}\right|$$
= 1(6 – 0) – 2(8 + 1) + 0(0 + 3)
= 6 – 18
= -12 ≠ 0
∴ ρ(A) = 3

Question 3.
$$\left[\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]$$
Solution:
Det A = $$\left|\begin{array}{lll} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right|$$
= 1(6 – 4) – 2(4 – 3) + 0(8 – 9)
= 2 – 2 + 0
= 0
∴ ρ(A) ≠ 3, ρ(A) < 3
Take $$\left|\begin{array}{ll} 1 & 2 \\ 2 & 3 \end{array}\right|$$ = 3 – 4 = -1 ≠ 0
∴ ρ(A) = 2

Question 4.
$$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$
Solution:
Let A = $$\left[\begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right]$$, det A = 0, ρ(A) ≠ 3
All 2 × 2 sub-matrix det. is zero
∴ ρ(A) ≠ 2
|1| = 1 ≠ 0
∴ ρ(A) = 1

Question 5.
$$\left[\begin{array}{cccc} 1 & 2 & 0 & -1 \\ 3 & 4 & 1 & 2 \\ -2 & 3 & 2 & 5 \end{array}\right]$$
Solution:
Take sub-matrix B = $$\left|\begin{array}{ccc} 1 & 2 & 0 \\ 3 & 4 & 1 \\ -2 & 3 & 2 \end{array}\right|$$
= 1(8 – 3) – 2(6 + 2)
= 5 – 16
= -11 ≠ 0
Rank of the given matrix is 3.

Question 6.
$$\left[\begin{array}{cccc} 0 & 1 & 1 & -2 \\ 4 & 0 & 2 & 5 \\ 2 & 1 & 3 & 1 \end{array}\right]$$
Solution:
Take sub matrix A = $$\left[\begin{array}{lll} 0 & 1 & 1 \\ 4 & 0 & 2 \\ 2 & 1 & 3 \end{array}\right]$$
= -1(12 – 4) + 1(4 – 0)
= -8 + 4
= -4 ≠ 0
∴ ρ(A) = 3