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.
[latex]\left[\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right][/latex]
Solution:
Det A = [latex]\left|\begin{array}{ll}
1 & 0 \\
0 & 0
\end{array}\right|[/latex] = 0 – 0 = 0
and |1| = 1 ≠ 0
∴ ρ(A) = 1
Question 2.
[latex]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right][/latex]
Solution:
Det A = [latex]\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|[/latex] = 1 – 0 = 1 ≠ 0
∴ ρ(A) = 2
Question 3.
[latex]\left[\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right][/latex]
Solution:
Det A = [latex]\left|\begin{array}{ll}
1 & 1 \\
0 & 0
\end{array}\right|[/latex] = 0 – 0 = 0
|1| = 1 ≠ 0
∴ ρ(A) = 1
Question 4.
[latex]\left[\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right][/latex]
Solution:
Det A = [latex]\left|\begin{array}{ll}
1 & 1 \\
1 & 0
\end{array}\right|[/latex] = 0 – 1 = -1 ≠ 0
∴ ρ(A) = 2
Question 5.
[latex]\left[\begin{array}{ccc}
1 & 0 & -4 \\
2 & -1 & 3
\end{array}\right][/latex]
Solution:
[latex]\left|\begin{array}{cc}
1 & -4 \\
2 & 3
\end{array}\right|[/latex] = 3 + 8 = -11 ≠ 0
∴ ρ(A) = 2
Question 6.
[latex]\left[\begin{array}{lll}
1 & 2 & 6 \\
2 & 4 & 3
\end{array}\right][/latex]
Solution:
[latex]\left|\begin{array}{ll}
2 & 6 \\
4 & 3
\end{array}\right|[/latex] = 6 – 24 = -18 ≠ 0
∴ ρ(A) = 2
II.
Question 1.
[latex]\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right][/latex]
Solution:
Det A = [latex]\left|\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right|[/latex]
= 1(1 – 0) – 0(0 – 0) + 0(0 – 0)
= 1 – 0 + 0
= 1 ≠ 0
∴ ρ(A) = 3
Question 2.
[latex]\left[\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right][/latex]
Solution:
Det A = [latex]\left|\begin{array}{ccc}
1 & 4 & -1 \\
2 & 3 & 0 \\
0 & 1 & 2
\end{array}\right|[/latex]
= 1(6 – 0) – 2(8 + 1) + 0(0 + 3)
= 6 – 18
= -12 ≠ 0
∴ ρ(A) = 3
Question 3.
[latex]\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right][/latex]
Solution:
Det A = [latex]\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right|[/latex]
= 1(6 – 4) – 2(4 – 3) + 0(8 – 9)
= 2 – 2 + 0
= 0
∴ ρ(A) ≠ 3, ρ(A) < 3
Take [latex]\left|\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right|[/latex] = 3 – 4 = -1 ≠ 0
∴ ρ(A) = 2
Question 4.
[latex]\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right][/latex]
Solution:
Let A = [latex]\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{array}\right][/latex], 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.
[latex]\left[\begin{array}{cccc}
1 & 2 & 0 & -1 \\
3 & 4 & 1 & 2 \\
-2 & 3 & 2 & 5
\end{array}\right][/latex]
Solution:
Take sub-matrix B = [latex]\left|\begin{array}{ccc}
1 & 2 & 0 \\
3 & 4 & 1 \\
-2 & 3 & 2
\end{array}\right|[/latex]
= 1(8 – 3) – 2(6 + 2)
= 5 – 16
= -11 ≠ 0
Rank of the given matrix is 3.
Question 6.
[latex]\left[\begin{array}{cccc}
0 & 1 & 1 & -2 \\
4 & 0 & 2 & 5 \\
2 & 1 & 3 & 1
\end{array}\right][/latex]
Solution:
Take sub matrix A = [latex]\left[\begin{array}{lll}
0 & 1 & 1 \\
4 & 0 & 2 \\
2 & 1 & 3
\end{array}\right][/latex]
= -1(12 – 4) + 1(4 – 0)
= -8 + 4
= -4 ≠ 0
∴ ρ(A) = 3