0.1 Solutions for Note 3

1.

If \(A\) is invertible, then \(A^{-1} A x = A^{-1} b\) and \(x = A^{-1} b\). Therefore, the system \(Ax = b\) is always consistent.

2.

Consider the following representation of the system. \begin{align*} x + 2y &= 1 \\ 2x + 4y &= k \end{align*}

Notice that if \(k = 2\), then there exist infinitely many solutions while \(k \neq 2\) gives no solutions. Thus the system never has a unique solution.

3.

First and foremost, consider the following augmented matrix. \[ \begin {bNiceArray}{ccc|c} 1 & 1 & 1 & 10 \\ 3 & 4 & -1 & 2 \\ -1 & -3 & 1 & 4 \end {bNiceArray} \] Using elementary row operations, the augmented matrix can be transformed as following. \begin{align*} \begin {bNiceArray}{ccc|c} 1 & 1 & 1 & 10 \\ 3 & 4 & -1 & 2 \\ -1 & -3 & 1 & 4 \end {bNiceArray} &\rightarrow \begin {bNiceArray}{ccc|c} 1 & 1 & 1 & 10 \\ 0 & 1 & -4 & -28 \\ -1 & -3 & 1 & 4 \end {bNiceArray} \\ &\rightarrow \begin {bNiceArray}{ccc|c} 1 & 1 & 1 & 10 \\ 0 & 1 & -4 & -28 \\ 0 & -2 & 2 & 14 \end {bNiceArray} \\ &\rightarrow \begin {bNiceArray}{ccc|c} 1 & 1 & 1 & 10 \\ 0 & 1 & -4 & -28 \\ 0 & 0 & -6 & -42 \end {bNiceArray} \end{align*}

Therefore, \(-6z = -42\) and \(z = 7\). Moreover, \(y - 4z = -28\) and \(y = 0\). Finally, \(x + y + z = 10\) and \(x = 3\). Therefore, the solution to the system is \((3, 0, 7)\).

4.

First and foremost, let \(A = \begin {bmatrix} 1 & 1 & 1 \\ 3 & 4 & -1 \\ -1 & -3 & 1 \end {bmatrix}\). Using elementary row operations, \(A\) can be transformed as following. \begin{align*} \begin {bmatrix} 1 & 1 & 1 \\ 3 & 4 & -1 \\ -1 & -3 & 1 \end {bmatrix} &\rightarrow \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & -4 \\ -1 & -3 & 1 \end {bmatrix} \rightarrow \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & -4 \\ 0 & -2 & 2 \end {bmatrix} \\ &\rightarrow \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & -4 \\ 0 & 0 & -6 \end {bmatrix} \end{align*}

In other words, \(A\) can be represented as the matrix multiplication below. \begin{align*} &\quad \ \begin {bmatrix} 1 & 1 & 1 \\ 3 & 4 & -1 \\ -1 & -3 & 1 \end {bmatrix} \\ &= \begin {bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix} \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 0 & 1 \end {bmatrix} \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end {bmatrix} \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & -4 \\ 0 & 0 & -6 \end {bmatrix} \\ &= \begin {bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ 0 & 0 & 1 \end {bmatrix} \begin {bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & -2 & 1 \end {bmatrix} \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & -4 \\ 0 & 0 & -6 \end {bmatrix} \\ &= \begin {bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -1 & -2 & 1 \end {bmatrix} \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & -4 \\ 0 & 0 & -6 \end {bmatrix} \end{align*}

Consider the following system for equation for \(Ly = b\). \[ \begin {bmatrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -1 & -2 & 1 \end {bmatrix} \begin {bmatrix} x' \\ y' \\ z' \end {bmatrix} = \begin {bmatrix} 10 \\ 2 \\ 4 \end {bmatrix} \] It is evident that \(x' = 10\), \(y' = 2 - 3x' = -28\), and \(z' = 4 + x' + 2y' = -42\). Continuing, the system for \(Ux = y\) can be solved. \[ \begin {bmatrix} 1 & 1 & 1 \\ 0 & 1 & -4 \\ 0 & 0 & -6 \end {bmatrix} \begin {bmatrix} x \\ y \\ z \end {bmatrix} = \begin {bmatrix} 10 \\ -28 \\ -42 \end {bmatrix} \] Therefore, \(z = 7\), \(y = 0\), and \(x = 3\) holds and the solution for the system is \((3, 0, 7)\) as shown in the previous problem.

5.

To find the inverse of \(A = \begin {bmatrix} 1 & 1 & 1 \\ 3 & 4 & -1 \\ -1 & -3 & 1 \end {bmatrix}\), consider the following augmentation and transformations. \begin{align*} \begin {bNiceArray}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & 0 \\ 3 & 4 & -1 & 0 & 1 & 0 \\ -1 & -3 & 1 & 0 & 0 & 1 \end {bNiceArray} &\rightarrow \begin {bNiceArray}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & -4 & -3 & 1 & 0 \\ -1 & -3 & 1 & 0 & 0 & 1 \end {bNiceArray} \\ &\rightarrow \begin {bNiceArray}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & -4 & -3 & 1 & 0 \\ 0 & -2 & 2 & 1 & 0 & 1 \end {bNiceArray} \\ &\rightarrow \begin {bNiceArray}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & -4 & -3 & 1 & 0 \\ 0 & 0 & -6 & -5 & 2 & 1 \end {bNiceArray} \\ &\rightarrow \begin {bNiceArray}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & -4 & -3 & 1 & 0 \\ 0 & 0 & 1 & \frac {5}{6} & -\frac {1}{3} & -\frac {1}{6} \end {bNiceArray} \\ &\rightarrow \begin {bNiceArray}{ccc|ccc} 1 & 1 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & \frac {1}{3} & -\frac {1}{3} & -\frac {2}{3} \\ 0 & 0 & 1 & \frac {5}{6} & -\frac {1}{3} & -\frac {1}{6} \end {bNiceArray} \\ &\rightarrow \begin {bNiceArray}{ccc|ccc} 1 & 0 & 0 & -\frac {1}{6} & \frac {2}{3} & \frac {5}{6} \\ 0 & 1 & 0 & \frac {1}{3} & -\frac {1}{3} & -\frac {2}{3} \\ 0 & 0 & 1 & \frac {5}{6} & -\frac {1}{3} & -\frac {1}{6} \end {bNiceArray} \\ \end{align*}

Therefore, \(A^{-1} = \begin {bmatrix} -\frac {1}{6} & \frac {2}{3} & \frac {5}{6} \\ \frac {1}{3} & -\frac {1}{3} & -\frac {2}{3} \\ \frac {5}{6} & -\frac {1}{3} & -\frac {1}{6} \end {bmatrix}\) and the system could be solved by multiplying \(A^{-1}\) and \(b\). \[ \begin {bmatrix} -\frac {1}{6} & \frac {2}{3} & \frac {5}{6} \\ \frac {1}{3} & -\frac {1}{3} & -\frac {2}{3} \\ \frac {5}{6} & -\frac {1}{3} & -\frac {1}{6} \end {bmatrix} \begin {bmatrix} 10 \\ 2 \\ 4 \end {bmatrix} = \begin {bmatrix} 3 \\ 0 \\ 7 \end {bmatrix} \] Thus, the solution is \((3, 0, 7)\), which is consistent with the previous problems.