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

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

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