Question

Suppose, the system of linear equations

-2x + y + z = l

x - 2y + z = m

x + y - 2z = n

is such that l + m + n = 0, then the system has:

1. A non-zero unique solution.
2. Trivial solution.
3. Infinitely many solutions.
4. No solution.

Answer 1

Correct Answer - Option 3 : Infinitely many solutions.

Concept:

 

• A system of equations is consistent if it has at least one solution. An inconsistent system has no solutions.
• Rouché–Capelli theorem: A system of linear equations with n variables has a solution if and only if the rank of its coefficient matrix [A] is equal to the rank of its augmented matrix [A|B].
  • If n = rank(A) = rank(A|B) and det(A) ≠ 0: consistent; there is a unique solution.
  • If n > rank(A) = rank(A|B) and det(A) = 0: consistent and dependent; there are infinitely many solutions.
  • If rank(A) ≠ rank(A|B): inconsistent; there are no solutions.
• Rank of a matrix A corresponds to the maximal number of linearly independent rows/columns of A. In other words, it is the number of non-zero rows when the matrix is expressed in row echelon form (all rows consisting of only zeroes are at the bottom).

 

Calculation:

The coefficient matrix for the given system of equations is:

[A] = \(\rm \begin{bmatrix} -2 &\ \ \ 1 &\ \ \ 1\\ \ \ \ 1 & -2 & \ \ \ 1\\ \ \ \ 1 & \ \ \ 1 & -2 \end{bmatrix}\)

The augmented matrix will be:

[A|B] = \(\rm \begin{bmatrix} -2 &\ \ \ 1 &\ \ \ 1&|&\rm l\\ \ \ \ 1 & -2 & \ \ \ 1&|&\rm m\\ \ \ \ 1 & \ \ \ 1 & -2&|&\rm n \end{bmatrix}\)

In order to find the rank of both the matrices, lets convert into row echelon form using the following elementary row operations:

R₃ → R₁ + R₂ + R₃

[A|B] = \(\rm \begin{bmatrix} -2 &\ \ \ 1 &1&|&\rm l\\ \ \ \ 1 & -2 &1&|&\rm m\\ \ \ \ 0 & \ \ \ 0 & 0&|&\rm l+m+n \end{bmatrix}\)

Since l + m + n = 0 (given), we have rank(A) = rank(A|B) = 2 which is less than the number of variables 3.

Therefore, there are infinitely many solutions.

 

On further reduction, we get:

[A|B] = \(\rm \begin{bmatrix} -2 &\ \ \ 1 &1&|&\rm l\\ \ \ \ 1 & -2 &1&|&\rm m\\ \ \ \ 0 & \ \ \ 0 & 0&|&\rm l+m+n \end{bmatrix}\)

R2 → R1 + 2R2

[A|B] = \(\rm \begin{bmatrix} -2 &\ \ \ 1 &1&|&\rm l\\ \ \ \ 0 & -3 &3&|&\rm l+2m\\ \ \ \ 0 & \ \ \ 0 & 0&|&0 \end{bmatrix}\)

R1 → 3R1 + R2

[A|B] = \(\rm \begin{bmatrix} -6 &\ \ \ 0 &3&|&\rm 4l+2m\\ \ \ \ 0 & -3 &3&|&\rm l+2m\\ \ \ \ 0 & \ \ \ 0 & 0&|&0\end{bmatrix}\)

The solution is \(\rm y = -\frac{2 m}{3} - \frac{n}{3} + x\) and \(\rm z = -\frac{m}{3} - \frac{2n}{3} + x\).