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]\)

R_{2} → R_{2} – 2R_{1}, R_{3} → R_{3} – 3R_{1}

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]\)

R_{2} → R_{2} – R_{1}, R_{3} → R_{3} – 3R_{1}

A ~ \(\left[\begin{array}{ccc}

1 & 1 & -1 \\

0 & -3 & 2 \\

0 & 3 & -2

\end{array}\right]\)

⇒ det A = 0 as R_{2}, R_{3} 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.