Comprehensive Exercises on Systems of Linear Equations

Approach: Use Gaussian elimination to analyse the structure of the solutions.

Steps:

1. The augmented matrix: $\begin{pmatrix} 1 & 1 & | & 2 \\ 2 & 2 & | & 4 \end{pmatrix}$
2. Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & | & 2 \\ 0 & 0 & | & 0 \end{pmatrix}$
3. The rank of the coefficient matrix is 1, the rank of the augmented matrix is 1, and the number of unknowns is 2
4. The rank of the coefficient matrix equals the rank of the augmented matrix, and is less than the number of unknowns

Answer: there are infinitely many solutions

Approach: Follow the steps of Cramer’s rule.

Steps:

1. The coefficient matrix: $A = \begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}$
2. $|A| = 1 \times (-1) - 2 \times 3 = -1 - 6 = -7 \neq 0$
3. $|A_1| = \begin{vmatrix} 5 & 2 \\ 4 & -1 \end{vmatrix} = 5 \times (-1) - 2 \times 4 = -5 - 8 = -13$
4. $|A_2| = \begin{vmatrix} 1 & 5 \\ 3 & 4 \end{vmatrix} = 1 \times 4 - 5 \times 3 = 4 - 15 = -11$
5. $x = \frac{-13}{-7} = \frac{13}{7}$, $y = \frac{-11}{-7} = \frac{11}{7}$

Answer: $x = \frac{13}{7}$, $y = \frac{11}{7}$

Approach: Follow the steps for finding a fundamental system of solutions.

Steps:

1. The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 2 & 2 & 2 & | & 0 \end{pmatrix}$
2. Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
3. Pivot: $x$; free variables: $y, z$
4. Setting $y = 1, z = 0$ gives $\vec{v}_1 = (-1, 1, 0)$
5. Setting $y = 0, z = 1$ gives $\vec{v}_2 = (-1, 0, 1)$

Answer: the fundamental system of solutions is $\{(-1, 1, 0), (-1, 0, 1)\}$

Approach: Follow the steps of Gaussian elimination.

Steps:

1. The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 3 & 1 & | & 13 \\ 1 & -1 & 2 & | & 4 \end{pmatrix}$
2. Forward elimination:
   • $R_2 - 2R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & -1 & | & 1 \\ 1 & -1 & 2 & | & 4 \end{pmatrix}$
   • $R_3 - R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & -1 & | & 1 \\ 0 & -2 & 1 & | & -2 \end{pmatrix}$
   • $R_3 + 2R_2$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & -1 & | & 1 \\ 0 & 0 & -1 & | & 0 \end{pmatrix}$
3. Back substitution:
   • From the third row: $-z = 0$, so $z = 0$
   • From the second row: $y - z = 1$, so $y = 1$
   • From the first row: $x + y + z = 6$, so $x = 5$

Answer: $x = 5$, $y = 1$, $z = 0$

Approach: Use Gaussian elimination to work out which case you are in.

Steps:

1. The augmented matrix: $\begin{pmatrix} 1 & 1 & | & 1 \\ 1 & 1 & | & 2 \end{pmatrix}$
2. Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & | & 1 \\ 0 & 0 & | & 1 \end{pmatrix}$
3. The rank of the coefficient matrix is 1, the rank of the augmented matrix is 2
4. The rank of the coefficient matrix is less than the rank of the augmented matrix

Answer: there is no solution

Approach: Reduce to row echelon form by Gaussian elimination, then express the solution in terms of parameters.

Steps:

1. The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 2 & 2 & 2 & | & 2 \end{pmatrix}$
2. Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
3. Pivot: $x$; free variables: $y, z$
4. Let $y = t$ and $z = s$, then $x = 1 - t - s$

Answer: $x = 1 - t - s$, $y = t$, $z = s$ (where $t, s$ are arbitrary constants)

Approach: Follow the steps of Cramer’s rule.

Steps:

1. The coefficient matrix: $A = \begin{pmatrix} 2 & 1 \\ 1 & -3 \end{pmatrix}$
2. $|A| = 2 \times (-3) - 1 \times 1 = -6 - 1 = -7 \neq 0$
3. $|A_1| = \begin{vmatrix} 7 & 1 \\ -4 & -3 \end{vmatrix} = 7 \times (-3) - 1 \times (-4) = -21 + 4 = -17$
4. $|A_2| = \begin{vmatrix} 2 & 7 \\ 1 & -4 \end{vmatrix} = 2 \times (-4) - 7 \times 1 = -8 - 7 = -15$
5. $x = \frac{-17}{-7} = \frac{17}{7}$, $y = \frac{-15}{-7} = \frac{15}{7}$

Answer: $x = \frac{17}{7}$, $y = \frac{15}{7}$

Approach: Check the determinant of the coefficient matrix.

Steps:

1. The coefficient matrix: $A = \begin{pmatrix} 1 & 1 \\ 2 & 2 \end{pmatrix}$
2. $|A| = 1 \times 2 - 1 \times 2 = 2 - 2 = 0$
3. The determinant is zero, so Cramer’s rule does not apply

Answer: it cannot be solved by Cramer’s rule

Approach: First reduce to row echelon form by Gaussian elimination, then reduce to reduced row echelon form by Gauss-Jordan elimination.

Steps:

1. The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 2 & 3 & 1 & | & 13 \\ 1 & -1 & 2 & | & 4 \end{pmatrix}$
2. Gaussian elimination:
   • $R_2 - 2R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & -1 & | & 1 \\ 1 & -1 & 2 & | & 4 \end{pmatrix}$
   • $R_3 - R_1$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & -1 & | & 1 \\ 0 & -2 & 1 & | & -2 \end{pmatrix}$
   • $R_3 + 2R_2$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & -1 & | & 1 \\ 0 & 0 & -1 & | & 0 \end{pmatrix}$
3. Gauss-Jordan elimination:
   • $R_3 \times (-1)$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & -1 & | & 1 \\ 0 & 0 & 1 & | & 0 \end{pmatrix}$
   • $R_2 + R_3$: $\begin{pmatrix} 1 & 1 & 1 & | & 6 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & 0 \end{pmatrix}$
   • $R_1 - R_2 - R_3$: $\begin{pmatrix} 1 & 0 & 0 & | & 5 \\ 0 & 1 & 0 & | & 1 \\ 0 & 0 & 1 & | & 0 \end{pmatrix}$
4. Read off the solution directly: $x = 5$, $y = 1$, $z = 0$

Answer: $x = 5$, $y = 1$, $z = 0$

Approach: Compute the rank of the coefficient matrix, then use the formula to get the dimension of the solution space.

Steps:

1. The coefficient matrix: $A = \begin{pmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \end{pmatrix}$
2. Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \end{pmatrix}$
3. $r(A) = 2$, $n = 3$
4. The dimension of the solution space: $n - r(A) = 3 - 2 = 1$

Answer: the dimension of the solution space is 1

Approach: Verify the three conditions of a vector space.

Steps:

1. Zero element: $\vec{0}$ is a solution, since $A\vec{0} = \vec{0}$
2. Closure under addition: if $\vec{x}_1, \vec{x}_2$ are solutions, then $A(\vec{x}_1 + \vec{x}_2) = A\vec{x}_1 + A\vec{x}_2 = \vec{0} + \vec{0} = \vec{0}$
3. Closure under scalar multiplication: if $\vec{x}$ is a solution and $k$ is a constant, then $A(k\vec{x}) = k(A\vec{x}) = k\vec{0} = \vec{0}$

Answer: the proof is complete

Approach: First check the determinant, then work out which case you are in.

Steps:

1. The coefficient matrix: $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$
2. $|A| = 1 \times 4 - 2 \times 2 = 4 - 4 = 0$
3. Cramer’s rule does not apply
4. Look at the two equations: the second equation is twice the first
5. The two equations are equivalent, so there are infinitely many solutions

Answer: there are infinitely many solutions

Approach: Reduce to row echelon form by Gaussian elimination, then find the fundamental system of solutions.

Steps:

1. The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 2 & 2 & 2 & | & 2 \\ 3 & 3 & 3 & | & 3 \end{pmatrix}$
2. Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 1 \\ 0 & 0 & 0 & | & 0 \\ 0 & 0 & 0 & | & 0 \end{pmatrix}$
3. Pivot: $x$; free variables: $y, z$
4. Setting $y = 1, z = 0$ gives $\vec{v}_1 = (-1, 1, 0)$
5. Setting $y = 0, z = 1$ gives $\vec{v}_2 = (-1, 0, 1)$

Answer: the fundamental system of solutions is $\{(-1, 1, 0), (-1, 0, 1)\}$

Approach: Use Cramer’s rule together with a proof by contradiction.

Steps:

1. Assume $|A| \neq 0$
2. By Cramer’s rule, the system has a unique solution
3. Since the system is homogeneous, the unique solution is the zero solution
4. This contradicts the assumption that a nonzero solution exists
5. Therefore $|A| = 0$

Answer: the proof is complete

Approach: Reduce to row echelon form by Gaussian elimination, then express the solution in terms of parameters.

Steps:

1. The augmented matrix: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 2 & 3 & 1 & | & 0 \end{pmatrix}$
2. Reduce to row echelon form: $\begin{pmatrix} 1 & 1 & 1 & | & 0 \\ 0 & 1 & -1 & | & 0 \end{pmatrix}$
3. Pivots: $x, y$; free variable: $z$
4. Let $z = t$, then $y = t$ and $x = -2t$

Answer: $x = -2t$, $y = t$, $z = t$ (where $t$ is an arbitrary constant)