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11.6 Solving equation

  1. Let \(\mathbb{F}\) be the field of complex numbers \(\mathbb{C}\). Decide if each of the following pairs of systems of linear equations are equivalent and, if so, express the equations in each system as a linear combination of the equations in the other system.

    (a) \(\begin{cases} x_1 - x_2 = 0 \\ 2x_1 + x_2 = 0 \end{cases}\) and \(\begin{cases} 3x_1 + x_2 = 0 \\ x_1 + x_2 = 0 \end{cases}\)

    Yes,\(\begin{pmatrix}1 & -1\\ 2 & 1\end{pmatrix}\Rightarrow\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix};\begin{pmatrix}3 & 1\\ 1 & 1\end{pmatrix}\Rightarrow\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\)

    (a) It is enough to write each equation as a linear combination of the equations in the other system.

    • \(x_1 - x_2 = 1 \cdot (3x_1 + x_2) + (-2) \cdot (x_1 + x_2)\)
    • \(3x_1 + x_2 = \frac{1}{3}(x_1 - x_2) + \frac{4}{3}(2x_1 + x_2)\)
    • \(2x_1 + x_2 = \frac{1}{2}(3x_1 + x_2) + \frac{1}{2}(x_1 + x_2)\)
    • \(x_1 + x_2 = -\frac{1}{3}(x_1 - x_2) + \frac{2}{3}(2x_1 + x_2)\)

    Therefore, the systems are equivalent.

    (b) \(\begin{cases} -x_1 + x_2 + 4x_3 = 0 \\ x_1 + 3x_2 + 8x_3 = 0 \\ \frac{1}{2}x_1 + x_2 + \frac{5}{2}x_3 = 0 \end{cases}\) and \(\begin{cases} x_1 - x_3 = 0 \\ x_2 + 3x_3 = 0 \end{cases}\)

    Yes,\(\begin{pmatrix}-1 & 1 & 4\\ 1 & 3 & 8\\ \frac12 & 1 & \frac52\end{pmatrix}\Rightarrow\begin{pmatrix}0 & 0 & 0\\ 0 & 1 & 3\\ 1 & 0 & -1\end{pmatrix};\begin{pmatrix}1 & 0 & -1\\ 0 & 1 & 3\\ 0 & 0 & 0\end{pmatrix}\)

    (c) \(\begin{cases} 2x_1 + (-1 + i)x_2 + x_4 = 0 \\ 2x_2 - 2ix_3 + 5x_4 = 0 \end{cases}\) and \(\begin{cases} \left(1 + \frac{i}{2}\right)x_1 + 8x_2 - ix_3 - x_4 = 0 \\ \frac{3}{2}x_1 - \frac{1}{2}x_2 + x_3 + 7x_4 = 0 \end{cases}\)

    No,\(\begin{pmatrix}2 & -1+i & 0 & 4\\ 0 & 2 & -2i & 5\end{pmatrix}\Rightarrow\begin{pmatrix}1 & \frac12+\frac{i}{2} & -i & \frac92\\ 0 & 1 & -i & \frac52\end{pmatrix}\)

    The systems are NOT equivalent.

  2. Find the space of solutions of the following systems of linear equations by applying elementary row operations.

    (a) \(\begin{cases} x - 3y + 5z = 0 \\ 2x - 3y + z = 0 \\ -y + 3z = 0 \end{cases}\)

    Apply row operations:

    1. \(E_2 - 2E_1\): \(\begin{cases} x - 3y + 5z = 0 \\ 3y - 9z = 0 \\ -y + 3z = 0 \end{cases}\)​
    2. \(E_3 + \frac{1}{3}E_2\): \(\begin{cases} x - 3y + 5z = 0 \\ 3y - 9z = 0 \\ 0 = 0 \end{cases}\)​
    3. \(\frac{1}{3}E_2\): \(\begin{cases} x - 3y + 5z = 0 \\ y - 3z = 0 \end{cases}\)​

    From the reduced system:

    • We get \(y = 3z\).
    • Substitute \(y = 3z\) into the first equation to get \(x - 3(3z) + 5z = x - 4z = 0\), so \(x = 4z\).

    Thus, the solutions are \((4z, 3z, z)\) for \(z \in \mathbb{R}\).

    (b) \(\begin{cases} x + y - z = 0 \\ 4x - y + 5z = 0 \\ 6x + y + 3z = 0 \end{cases}\)

    \(\begin{cases}x+y-z=0\\ -5y+9z=0\end{cases}\Rightarrow\begin{cases}x=-\frac45z\\ y=\frac95z\end{cases}\)

    Thus the solutions are \((-\frac45z,\frac95z,z)\)​

  3. For each of the following systems of linear equations, describe the set of vectors \((b_1, b_2)\) or \((b_1, b_2, b_3)\) for which the system has solutions.

    (a) \(\begin{cases} x + y = b_1 \\ 2x - 2y = b_2 \end{cases}\)

    Apply row operations:

    1. \(E_2 - 2E_1\): \(\begin{cases} x + y = b_1 \\ -4y = b_2 - 2b_1 \end{cases}\)​
    2. \(\begin{cases}y=\frac{b_2-2b_1}{-4}=\frac{2b_1-b_2}{4}\\ x=b_1+\frac{b_2}{4}-\frac{2b_1}{4}=\frac{b_1}{2}+\frac{b_2}{4}\end{cases}\)

    So the solutions are of the form \((x, y) = \left( \frac{b_1}{2} + \frac{b_2}{4}, \frac{2b_1 - b_2}{4} \right)\).

    Since this makes sense for all \((b_1, b_2)\), we can take any \((b_1, b_2) \in \mathbb{R}^2\).

    (b) \(\begin{cases} x + y = b_1 \\ 2x + 2y = b_2 \end{cases}\)

    \((x,y)=(b_1,2b_2)\)​

    (c) \(\begin{cases} x - y + 2z + w = b_1 \\ 2x + 2y + z + w = b_2 \\ 3x + y + 3z = b_3 \end{cases}\)

    1. \(E_2 - 2E_1\), \(E_3 - 3E_1\)​: \(\begin{cases}x-y+2z+w=b_1\\ 4y-3z-3w=b_2-2b_1\\ 4y-3z-3w=b_3-3b_1\end{cases}\)
    2. \(E_3-E_2\): \(\begin{cases} x - y + 2z + w = b_1 \\ 4y - 3z - 3w = b_2 - 2b_1 \\ 0 = b_3 - b_1 - b_2 \end{cases}\)

    We should have \(b_3 = b_1 + b_2\) to have a solution.

    Thus, \(\left(x,y,z)=(b_1,b_2,b_1+b_2\right)\).

  4. Find the row-reduced echelon form of the following matrices.

    (a) \(A=\begin{pmatrix}0 & 1 \\ 0 & 0 \\ 5 & 9 \end{pmatrix}\)

    1. \(R_1\leftrightarrow\)​ \(R_2\)
      ​\(\begin{pmatrix}0 & 1\\ 5 & 9\\ 0 & 0\end{pmatrix}\)
    2. \(R_2-9R_1\):
      ​\(\begin{pmatrix}0 & 1\\ 5 & 0\\ 0 & 0\end{pmatrix}\)
    3. \(\frac15R_2\)
      ​\(\begin{pmatrix}0 & 1\\ 1 & 0\\ 0 & 0\end{pmatrix}\)
    4. \(R_1\leftrightarrow R_2\)​
      ​\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}\)

    (b) \(B=\begin{pmatrix}1 & -1 & -2 & 4\\ 2 & -1 & -1 & 2\\ 2 & 1 & 4 & 16\end{pmatrix}\)​

    \(\begin{pmatrix}1 & 0 & 0 & 24\\ 0 & 1 & 0 & 72\\ 0 & 0 & 1 & -26\end{pmatrix}\)

  5. Find a row-reduced echelon matrix which is row-equivalent to \(A = \begin{pmatrix} 1 & -i \\ 2 & 2 \\ i & 1 + i \end{pmatrix}\).

    What are the solutions of \(Ax = 0\)?

    1. \(R_2 - 2R_1\) and \(R_3 - iR_1\): \(\begin{pmatrix} 1 & -i \\ 0 & 2 + 2i \\ 0 & i \end{pmatrix}\)
    2. \(\frac{R_2}{2 + 2i}\) and \(\frac{R_3}{i}\): \(\begin{pmatrix} 1 & -i \\ 0 & 1 \\ 0 & 1 \end{pmatrix}\)
    3. \(R_3 - R_2\): \(\begin{pmatrix} 1 & -i \\ 0 & 1 \\ 0 & 0 \end{pmatrix}\)
    4. \(R_1 + iR_2\): \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}\)

    If we look at \(Bx = 0 \Rightarrow x = (0, 0, 0)\).

  6. Give the following system of linear equations in its matrix form and find the space of solutions by row-reducing said associated matrix.
    ​\(\begin{cases} \frac{1}{3} x_1 + 2x_2 - 6x_3 = 0 \\ -4x_1 + 5x_3 = 0 \\ -3x_1 + 6x_2 - 13x_3 = 0 \\ -\frac{7}{3} x_1 + 2x_2 - \frac{8}{3} x_3 = 0 \end{cases}\)

    After some steps, we have \(\left(\begin{array}{ccc|c}1 & 0 & 1 & \frac12\\ 0 & 1 & -1 & -\frac12\end{array}\right)\)​

    Therefore, the solutions are \(\left( \frac{1}{2} - x_3, -\frac{1}{2} + x_3, x_3 \right)\) for \(x_3 \in \mathbb{R}\).

  7. Consider the system of linear equations given by
    ​\(\begin{cases} x_1 - x_2 + 2x_3 = 1 \\ 2x_1 + 2x_3 = 1 \\ x_1 - 3x_2 + 4x_3 = 2 \end{cases}\)

    Write the augmented coefficient matrix and row-reduce it. Does this system have a solution? If so, describe all solutions explicitly.

    \(\begin{pmatrix}\begin{array}{ccc|c}1 & -1 & 2 & 1\\ 2 & 0 & 2 & 1\\ 1 & -3 & 4 & 2\end{array}\end{pmatrix}\Rightarrow\begin{pmatrix}\begin{array}{ccc|c}0 & 0 & 0 & 0\\ 0 & 1 & -1 & -\frac12\\ 1 & 0 & 1 & \frac12\end{array}\end{pmatrix}\Rightarrow\begin{cases}x+z=\frac12\\y-z=-\frac12\end{cases}\)​

    The solutions are \((\frac12-z,z-\frac12,z)\)

  8. Consider the matrix \(A\) with real coefficients given by
    ​\(A = \begin{pmatrix} 3 & -1 & 2 \\ 2 & 1 & 1 \\ 1 & -3 & 0 \end{pmatrix}\).

    By row-reducing:
    (a) Find all solutions in \(\mathbb{R}\) and in \(\mathbb{C}\) of \(Ax = 0\).

    After some steps, we have \(A=\begin{pmatrix}3 & -1 & 0\\ 0 & 5 & 0\\ 0 & 0 & 1\end{pmatrix}\)

    image​

    \(\Rightarrow x_3 = 0\), \(5x_2 = 0 \Rightarrow x_2 = 0\). And \(3x_1 = x_2 = 0\).

    Therefore, we only have the trivial solution \(x = 0\).

    (b) Find all solutions in \(\mathbb{R}\) and in \(\mathbb{C}\) of \(Ax = \begin{pmatrix} 1 \\ i \\ 0 \end{pmatrix}\).

    We need to do the same operations on \(\begin{pmatrix}1\\ i\\ 0\end{pmatrix}\), which is \(\begin{pmatrix}i-\frac12\\ \frac{i}{3}-\frac16\\ -\frac{4i}{3}+\frac76\end{pmatrix}\in\mathbb{C}^3\)​

    Therefore, we have no solution in \(\R\) and only one in \(\mathbb{C}\)

  9. For each of the following matrices, use elementary row operations to discover whether it is invertible or not. If invertible, find its inverse.

    (a) \(A = \begin{pmatrix} 2 & 5 & -1 \\ 4 & -1 & 2 \\ 6 & 4 & 1 \end{pmatrix}\)

    Remember that if we have a square matrix \(A \in M_{n \times n}(\mathbb{F})\) and it is possible to get \((A \mid \text{Id}) \rightarrow (\text{Id} \mid B)\) through row-reducing, then \(A\) is invertible and \(A^{-1} = B\).

    Given \(\begin{pmatrix}\begin{array}{ccc|ccc}2 & 5 & -1 & 1 & 0 & 0\\ 4 & -1 & 2 & 0 & 1 & 0\\ 6 & 4 & 1 & 0 & 0 & 1\end{array}\end{pmatrix}\)

    \(R_2 - 2 R_1\);\(R_3 - 3 R_1\) .Then we have \(\begin{pmatrix}\begin{array}{ccc|ccc}2 & 5 & -1 & 1 & 0 & 0\\ 0 & -11 & 4 & -2 & 1 & 0\\ 0 & -11 & 4 & -3 & 0 & 1\end{array}\end{pmatrix}\)

    Notice that we now have a row of zeros: \(\begin{pmatrix}\begin{array}{ccc|ccc}2 & 5 & -1 & 1 & 0 & 0\\ 0 & -11 & 4 & -2 & 1 & 0\\ 0 & 0 & 0 & -1 & -1 & 1\end{array}\end{pmatrix}\)

    So \(A\) is not invertible.

    (b) \(B = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 2 & 4 \\ 0 & 1 & -2 \end{pmatrix}\)

    We have\(\begin{pmatrix}\begin{array}{ccc|ccc}1 & -1 & 2 & 1 & 0 & 0 \\ 3 & 2 & 4 & 0 & 1 & 0 \\ 0 & 1 & -2 & 0 & 0 & 1\end{array}\end{pmatrix}\)

    Thus \(A^{-1} = \begin{pmatrix}\begin{array}{ccc}1 & 0 & -1 \\ -3/4 & 1/4 & -1/4 \\ -3/8 & 1/8 & -5/8\end{array}\end{pmatrix}\).

  10. Let \(A = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \in M_{2 \times 1}(\mathbb{F})\) and \(B = (b_1 \ b_2) \in M_{1 \times 2}(\mathbb{F})\). Show that \(C = AB\) is not invertible.

    \(C=\begin{pmatrix}a_1\\ a_2\end{pmatrix}(b_1\ b_2)=\begin{pmatrix}a_1b_1 & a_1b_2\\ a_2b_1 & a_2b_2\end{pmatrix}\)

    Since \(a_1b_1a_2b_2=a_1b_1a_2b_2\), by det, it is not invertible

    Or

    Let \(A = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \in M_{2 \times 1}(\mathbb{F})\) and \(B = \begin{pmatrix} b_1 & b_2 \end{pmatrix} \in M_{1 \times 2}(\mathbb{F})\).

    Show that \(C = A B\) is not invertible.

    We have that \(AB = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix} \begin{pmatrix} b_1 & b_2 \end{pmatrix} = \begin{pmatrix} a_1 b_1 & a_1 b_2 \\ a_2 b_1 & a_2 b_2 \end{pmatrix}\).

    Let's consider cases:

    • If \(a_1\) or \(a_2\) or \(b_1\) or \(b_2\) are zero \(\implies\) we will have either a row or a column of zeros.

    \(\implies C\) is not invertible.

    • If \(a_1 \neq 0\), \(a_2 \neq 0\), \(b_1 \neq 0\), and \(b_2 \neq 0\):

    Then \(C = \begin{pmatrix} a_1 b_1 & a_1 b_2 \\ a_2 b_1 & a_2 b_2 \end{pmatrix}\). Performing the row operation \(R_2 - \frac{a_2}{a_1} R_1\) yields \(\begin{pmatrix} a_1 b_1 & a_1 b_2 \\ 0 & 0 \end{pmatrix}\), and this is not invertible.

    \(\implies C\) is not invertible.

    In all cases, \(C\) is not invertible.

  11. Let \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). Prove, using elementary row operations, that \(A\) is invertible if and only if \(ad - bc \neq 0\).

    \(\Rightarrow\))

    Suppose \(B = \begin{pmatrix} x & y \\ z & w \end{pmatrix}\), then \(AB = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x & y \\ z & w \end{pmatrix} = \begin{pmatrix} ax + bz & ay + bw \\ cx + dz & cy + dw \end{pmatrix}\).

    Therefore \(AB=I=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}\Leftrightarrow\begin{cases}ax+bz=1\\ ay+bw=0\\ cx+dz=0\\ cy+dw=1\end{cases}\)​

    So, in the matrix form, we have:
    ​\(\left(\begin{array}{cccc|c}a & 0 & b & 0 & 1\\ 0 & a & 0 & b & 0\\ c & 0 & d & 0 & 0\\ 0 & c & 0 & d & 1\end{array}\right)\).

    Then let \(aR_1-bR_3,aR_2-bR_4\) we have \(\left(\begin{array}{cccc|c}ad-bc & 0 & 0 & 0 & d\\ 0 & ad-bc & 0 & 0 & -b\\ c & 0 & d & 0 & 0\\ 0 & c & 0 & d & 1\end{array}\right)\)

    Thus \(ad-bc\neq 0\)

    \(\Leftarrow\)) Suppose \(ad - bc \neq 0\)

    \(R_1 \rightarrow R_1 / (ad - bc)\) and \(R_2 \rightarrow R_2 / (ad - bc)\):\(\left(\begin{array}{cccc|c}1 & 0 & 0 & 0 & d/(ad-bc)\\ 0 & 1 & 0 & 0 & -b/(ad-bc)\\ c & 0 & d & 0 & 0\\ 0 & c & 0 & d & 1\end{array}\right)\)

    \(R_3 - c R_1\) and \(R_4 - c R_2\):\(\left(\begin{array}{cccc|c}1 & 0 & 0 & 0 & d/(ad-bc)\\ 0 & 1 & 0 & 0 & -b/(ad-bc)\\ 0 & 0 & d & 0 & -cd/(ad-bc)\\ 0 & 0 & 0 & d & 1+bc/(ad-bc)\end{array}\right)\)

    Suppose \(d\neq 0\), then let \(R_3/d,R_4/d\)

    Thus \(B\) exists and by definition \(B=A^{-1}\Rightarrow A\) is invertible

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