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

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

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

Solve the following systems of homogeneous equations.

Question 1.
2x + 3y – z = 0, x – y – 2z = 0, 3x + y + 3z = 0
Solution:
The Coefficient matrix is $$\left[\begin{array}{ccc} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{array}\right]$$
det of $$\left[\begin{array}{ccc} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{array}\right]$$
= 2(-3 + 2) – 3(3 + 6) – 1(1 + 3)
= -2 – 27 – 4
= -33 ≠ 0, ρ(A) = 3
Hence the system has the trivial solution x = y = z = 0 only.

Question 2.
3x + y – 2z = 0, x + y + z = 0, x – 2y + z = 0
Hint: If the determinant of the coefficient matrix ≠ 0 then the system has a trivial solution (i.e.) ρ(A) = 3.
Solution:
The coefficient matrix is $$\left[\begin{array}{ccc} 3 & 1 & -2 \\ 1 & 1 & 1 \\ 1 & -2 & 1 \end{array}\right]$$
$$\left|\begin{array}{ccc} 3 & 1 & -2 \\ 1 & 1 & 1 \\ 1 & -2 & 1 \end{array}\right|$$
= 3(1 + 2) – 1(1 – 1) – 2(-2 – 1)
= 9 + 6
= 15 ≠ 0, ρ(A) = 3
Hence the system has the trivial solutions x = y = z = 0 only.

Question 3.
x + y – 2z = 0, 2x + y – 3z = 0, 5x + 4y – 9z = 0
Solution:
The coefficient matrix is $$\left[\begin{array}{rrr} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right]$$ = A (say)
|A| = $$\left|\begin{array}{rrr} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right|$$
= 1(-9 + 12) – 1(-18 + 15) – 2(8 – 5)
= 3 + 3 – 6
= 0
∴ Rank of A = 2 as the submatrix $$\left[\begin{array}{ll} 1 & 1 \\ 2 & 1 \end{array}\right]$$ is non-singular, ρ(A) < 3
Hence the system has a non-trivial solution.
A = $$\left[\begin{array}{lll} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{array}\right]$$
R2 → R2 – 2R1, R3 → R3 – 3R1
A ~ $$\left[\begin{array}{ccc} 1 & 1 & -2 \\ 0 & -1 & 1 \\ 0 & -1 & -1 \end{array}\right]$$
The system of equation is equivalent to the given system of equations are x + y – 2z = 0, -y + z = 0
Let z = k
⇒ y = k, x = k
∴ x = y = z = k for real number k.

Question 4.
x + y – z = 0, x – 2y + z = 0, 3x + 6y – 5z = 0
Solution:
Coefficient matrix A = $$\left[\begin{array}{ccc} 1 & 1 & -1 \\ 1 & -2 & 1 \\ 3 & 6 & -5 \end{array}\right]$$
R2 → R2 – R1, R3 → R3 – 3R1
A ~ $$\left[\begin{array}{ccc} 1 & 1 & -1 \\ 0 & -3 & 2 \\ 0 & 3 & -2 \end{array}\right]$$
⇒ det A = 0 as R2, R3 are identical.
and rank (A) = 2 as the submatrix $$\left[\begin{array}{cc} 1 & 1 \\ 0 & -3 \end{array}\right]$$ is non-singular.
Hence the system has a non-trivial solution, ∵ ρ(A) < 3
The system of equations equivalent to the given system of equations are
x + y – z = 0
3y – 2z = 0
Let z = k
⇒ y = $$\frac{2 k}{3}$$
x = $$\frac{k}{3}$$
∴ x = $$\frac{k}{3}$$, y = $$\frac{2 k}{3}$$, z = k for any real number of k.