Question

Find the condition(s) on $\alpha$ such that the simultaneous equations $$\begin{array}{rl}{x}_1+\alpha x_2& =1\\ x_1-x_2+3x_3& =-1\\ 2x_1-2x_2+\alpha x_3& =-2\end{array}$$ have (a) exactly one solution, (b) no solutions, or (c) an infinite number of solutions; give all solutions where they exist.

Short answer

(a) $\alpha eq\frac{-4}{3}$.(b) No solutions for \alpha = 0\).(c) Infinite solutions for $\alpha =1$}}}}

Step-by-step solution

- Write the system in matrix form

Express the given system of linear equations as a matrix equation. $$\left(\begin{array}{ccccccc}1& \alpha & 01& -1& 32& -2& \alpha \end{array}\right)\left(\begin{array}{c}{x}_1{x}_2{x}_3\end{array}\right)=\left(\begin{array}{c}1-1-2\end{array}\right)$$

- Form the augmented matrix

Combine the coefficient matrix with the constants on the right-hand side to form the augmented matrix.$$\left(\begin{array}{ccccccccccccc}1& \alpha & 0& |& 11& -1& 3& |& -12& -2& \alpha & |& -2\end{array}\right)$$

- Apply Gaussian elimination

Use row operations to simplify the augmented matrix into row echelon form.1. Subtract the first row from the second row:$$\left(\begin{array}{ccccccccccccc}1& \alpha & 0& |& 10& -1-\alpha & 3& |& -22& -2& \alpha & |& -2\end{array}\right)$$2. Subtract 2 times the first row from the third row:$$\left(\begin{array}{ccccccccccccc}1& \alpha & 0& |& 10& -1-\alpha & 3& |& -20& -2-2\alpha & \alpha & |& -4\end{array}\right)$$

- Continue Gaussian elimination

Subtract $2+2(\alpha )$ times the second row from the third row to eliminate the leading coefficient in the third row.$$\left(\begin{array}{ccccccccccccc}1& \alpha & 0& |& 10& -(1+\alpha )& 3& |& -20& 0& \alpha -(-2-2\ast \alpha )3& |& -2+4\ast \alpha \end{array}\right)$$

- Find conditions for \alpha

(a) For exactly one solution, the system must have non-zero pivots in each row, so $\alpha eq1$.(b) For no solutions, the system must be inconsistent, so $\alpha =(3\alpha -2)$.(c) For infinitely many solutions, the equation must be consistent and have at least one row with all zeros, so $\alpha =1$.

- Solve for cases with solutions

(a) For $\alpha eq\frac{-4}{3}$, apply back substitution to find the exact solutions.(b) For \alpha = \frac{-4}{3}\ ,apply back substitution to confirm no solutions exists.(c) For \alpha = 1, solve for an infinite number of solutions to confirm that system is consistent.

Key concepts

Matrix form

To tackle simultaneous equations, a powerful method is to express them in matrix form. This allows us to use more sophisticated tools, like matrix operations, to find solutions. For instance, take the set of equations given. It can be represented as: This can be written concisely as where 'A' is the matrix of coefficients, 'x' is the vector of variables, and 'b' is the constants vector. This notation saves time and provides a neat way to apply methods like Gaussian elimination.

Gaussian elimination

Gaussian elimination is key to solving matrix equations. The aim is to transform the matrix into a simpler form, known as Row Echelon Form. This process involves a series of row operations that simplify our work without changing the solution set. The main steps include:

• Swap rows to position pivot elements.
• Multiply rows to scale the pivot to 1.
• Subtract multiples of rows to eliminate entries below the pivot.

These operations lead us toward an upper triangular matrix, which is much easier to handle.

Row echelon form

Row Echelon Form (REF) is a simpler form of a matrix that makes solving linear equations straightforward. In REF, each non-zero row starts with '1', and every leading coefficient (pivot) in a row is to the right of the pivot in the row above. Here's an example structure: i.e. each leading '1' ensures that everything below it is zero, resulting in a step-like pattern. Converting the matrix to this form is crucial because it lets us use back-substitution to easily solve for the variables.

Unique solution

For a system of equations to have a unique solution, every variable must have exactly one distinct value. This means all rows in the matrix must have non-zero pivots after applying Gaussian elimination. Mathematically, this means our resulting Row Echelon Form should have all pivot columns from left to right. For our specific problem, this condition means avoiding specific values of the parameter

No solution

A system of linear equations has no solution if it is inconsistent. This usually happens when, during Gaussian elimination, we end up with a row that effectively translates to a false statement (e.g., '0 = 1'). In terms of our matrix, if the augmented part of the matrix introduces a contradiction, then no combination of variables satisfies all equations. For our problem, this occurs when

Infinite solutions

Sometimes, a system of equations can have infinitely many solutions. This happens when there are fewer independent equations than variables. In matrix terms, it implies that not every row contains a pivot after Gaussian elimination. For infinite solutions to occur, our matrix in Row Echelon Form must have at least one row with all zeros, showing that not all variables are fully constrained by the equations. For the given problem, this scenario occurs when