Extra (optional) exercises – Long Day

These exercises are optional extra material for students who have finished the exercises of the day and have the time and interest for more.

Extra exercise 1: Uniqueness of the inverse matrix

Definition 7.3.1 in the textbook tells that the inverse matrix of a square matrix \(\mathbf A\) is denoted by \(\mathbf A^{-1}\), if it exists. It is implicit in this notation that a square matrix has only one inverse matrix. Let us show that this indeed is true.

For a given square matrix \(\mathbf A \in \mathbb F^{n \times n}\), where \(n\) is a natural number, assume that there exist two matrices \(\mathbf B\) and \(\mathbf C\) such that

\[\mathbf A\cdot\mathbf B=\mathbf B\cdot\mathbf A=\mathbf I_n\]

and

\[\mathbf A \cdot\mathbf C=\mathbf C\cdot\mathbf A=\mathbf I_n.\]

Show that \(\mathbf B=\mathbf C\) then follows. Conclude that if a matrix has an inverse, then this inverse is uniquely determined.

Extra exercise 2: Calculation of matrix products

Given an invertible matrix \(\mathbf A \in \mathbb F^{n \times n}\) and a matrix \(\mathbf B \in \mathbb F^{n \times m}\), we form an \(n \times (n+m)\) matrix \([\mathbf A | \mathbf B]\) by placing the columns of \(\mathbf B\) next to (to the right of) the columns of \(\mathbf A\). With the two example matrices

\[\begin{split}\mathbf A = \left[ \begin{array}{ccc} 1 & 1 & 2\\ 0 & 2 & 1 \\ 2 & 0 & 4\\ \end{array} \right] \quad \text{ and } \quad \mathbf B = \left[ \begin{array}{cccc} 2 & 0 & 0 & 2\\ -2 & 0 & 1 & 3\\ 0 & 3 & 1 & -1\\ \end{array} \right], \end{split}\]

this construction forms the new matrix

\[\begin{split}[\mathbf {A} | \mathbf{B}]=\left[ \begin{array}{ccccccc} 1 & 1 & 2 & 2 & 0 & 0 & 2\\ 0 & 2 & 1 & -2 & 0 & 1 & 3\\ 2 & 0 & 4 & 0 & 3 & 1 & -1\\ \end{array} \right].\end{split}\]

Question a

Explain generally why the reduced row-echelon form of the matrix \([\mathbf A | \mathbf B]\) has the form \([\mathbf I_n | \mathbf C]\) for some \(n \times m\) matrix \(\mathbf C\).

Hint

Remember that we were given an invertible matrix \(\mathbf A\). What does that mean for the reduced row-echelon form of \(\mathbf A\)?

Question b

Find a formula for the matrix \(\mathbf C\) from the previous question.

Hint

If \(\mathbf B\) consists of just one column (that is, if \(m=1\)), what is then the solution to the system of linear equations \(\mathbf A \cdot \mathbf x=\mathbf B\)?

Answer

\(\mathbf C={\mathbf A}^{-1}\cdot {\mathbf B}.\)

A special case is the following: if \(\mathbf B = \mathbf I_n\), then \(\mathbf C={\mathbf A}^{-1}\), which agrees with the procedure for calculating \(\mathbf A^{-1}\) in Section 7.3 in the textbook.

Question c

We are now given an invertible matrix \(\mathbf A \in \mathbb F^{n \times n}\) as before and matrices \(\mathbf B_1 \in \mathbb F^{n \times m_1}, \mathbf B_2 \in \mathbb F^{n \times m_2}, \dots, \mathbf B_r \in \mathbb F^{n \times m_r}\). Show that by finding the reduced row-echelon form of the matrix \([\mathbf A | \mathbf B_1 | \mathbf B_2 | \dots | \mathbf B_r]\), we can determine the matrices \(\mathbf A^{-1}\cdot \mathbf B_1, \mathbf A^{-1}\cdot \mathbf B_2, \dots, \mathbf A^{-1}\cdot \mathbf B_r\) in one go.