Exercises – Short Day

Exercises – Short Day#

Exercise 1: Transpose of Matrices#

Calculate

\[\begin{split}\left[ \begin{array}{cc} 1 & 2\\ 3 & 4\\ 5 & 6\\ \end{array} \right]^T.\end{split}\]

Answer

The transpose of the matrix is

\[\begin{split}\left[ \begin{array}{ccc} 1 & 3 & 5\\ 2 & 4 & 6\\ \end{array} \right].\end{split}\]

Exercise 2: Determinants#

We are given the following matrix:

\[\begin{split} {\mathbf A}=\left[ \begin{array}{ccc} 1 & 3 & 5\\ 2 & 4 & 6\\ 3 & 0 & 0\\ \end{array} \right]. \end{split}\]

The intention with this Exercise is to determine the determinant of this matrix by expanding the determinant by a row or column.

Question a#

Calculate the matrices \({\mathbf A}(1;1), {\mathbf A}(2;1),\) and \({\mathbf A}(3;1)\) as well as there determinants.

Hint

In Definition 8.1.1 in the textbook the notation \({\mathbf A}(i;j)\) is explained.

Answer

\[\begin{split} {\mathbf A}(1;1)=\left[ \begin{array}{ccc} 4 & 6\\ 0 & 0\\ \end{array} \right] \quad {\mathbf A}(2;1)=\left[ \begin{array}{ccc} 3 & 5\\ 0 & 0\\ \end{array} \right] \quad {\mathbf A}(3;1)=\left[ \begin{array}{ccc} 3 & 5\\ 4 & 6\\ \end{array} \right]. \end{split}\]

You can achieve the determinants directly by using the method from Eksempel 8.1.1 in the textbook:

\[\mathrm{det}({\mathbf A(1;1)})=0 \quad \mathrm{det}({\mathbf A(2;1)})=0 \quad \mathrm{det}({\mathbf A(3;1)})=-2.\]

Question b#

Calculate \(\mathrm{det}({\mathbf A})\) using Definition 8.1.2 in the textbook. In other words, calculate \(\mathrm{det}({\mathbf A})\) by expanding the determinant by the first column.

Answer

By expanding by the first column, we get directly from Definition 8.1.2 that

\[\mathrm{det}({\mathbf A})=1\cdot \mathrm{det}({\mathbf A}(1;1))-2\cdot \mathrm{det}({\mathbf A}(2;1))+3\cdot \mathrm{det}({\mathbf A}(3;1))=-6.\]

Question c#

For the given matrix \({\mathbf A}\) it is more convenient to expand the determinant by the third row of the matrix. Why?

Hint

In the beginning of Section 8.2 it is explained what it means to expand a determinant by a row or column.

Answer

Due to the many zeroes in the third row of the matrix we have that:

\[ \mathrm{det}({\mathbf A})=3\cdot \mathrm{det}({\mathbf A}(3;1))=3 \cdot (3 \cdot 6-4 \cdot 5)=-6. \]

Question d#

To practice the expansion of determinants by a row, we will calculate \(\mathrm{det}({\mathbf A})\) one last time.

Calculate the matrices \({\mathbf A}(2;1), {\mathbf A}(2;2),\) and \({\mathbf A}(2;3)\) as well as their determinants. Now calculate \(\mathrm{det}({\mathbf A})\) by expanding the determinant by the second row.

Answer

\[\begin{split} {\mathbf A}(2;1)=\left[ \begin{array}{ccc} 3 & 5\\ 0 & 0\\ \end{array} \right] \quad {\mathbf A}(2;2)=\left[ \begin{array}{ccc} 1 & 5\\ 3 & 0\\ \end{array} \right] \quad {\mathbf A}(2;3)=\left[ \begin{array}{ccc} 1 & 3\\ 3 & 0\\ \end{array} \right]. \end{split}\]

and

\[\mathrm{det}({\mathbf A(2;1)})=0 \quad \mathrm{det}({\mathbf A(2;2)})=-15 \quad \mathrm{det}({\mathbf A(2;3)})=-9.\]

We now get from expanding by the second row that

\[\mathrm{det}({\mathbf A})=-2\cdot \mathrm{det}({\mathbf A}(2;1))+4\cdot \mathrm{det}({\mathbf A}(2;2))-6\cdot \mathrm{det}({\mathbf A}(2;3))=-6.\]

Exercise 3: Determinants and Linear Independence#

We are given the following three vectors in \(\mathbb{R}^3\):

\[\begin{split} {\mathbf v_1}=\left[ \begin{array}{c} 1\\ 0\\ 4\\ \end{array} \right], \quad {\mathbf v_2}=\left[ \begin{array}{c} 1\\ 1\\ 7\\ \end{array} \right], \quad {\mathbf v_3}=\left[ \begin{array}{c} 1\\ -1\\ 1\\ \end{array} \right]. \end{split}\]

Use Corollary 8.2.5 in the textbook to determine whether the vectors \({\mathbf v}_1, {\mathbf v}_2,\) and \({\mathbf v}_3\) are linearly independent or not.

Hint

First, calculate the determinant of a matrix whose columns are \({\mathbf v}_1, {\mathbf v}_2,\) and \({\mathbf v}_3\). Then expand the determinant by a column or row of you own choice - a smart choice is the one with the most zeroes in order to reduce your work load.

Hint

In Corollary 8.2.5, the determining factor is whether the determinant of the matrix mentioned in the previous hint is zero or not.

Answer

The three given vectors are linearly dependent.

Exercise 4: Linear Dependence#

We are given the vectors

\[\begin{split} {\mathbf u}_1=\left[ \begin{array}{c} 1\\ 1\\ -6\\ \end{array} \right] \quad {\mathbf u}_2=\left[ \begin{array}{c} 1\\ 1\\ a\\ \end{array} \right] \quad {\mathbf u}_3=\left[ \begin{array}{c} a\\ -2\\ 0\\ \end{array} \right], \end{split}\]

where \(a \in \mathbb{R}\) is a constant. Determine all values of the constant \(a\) that will make the vectors \({\mathbf u}_1, {\mathbf u}_2,\) and \({\mathbf u}_3\) linearly dependent.

Hint

Corollary 8.2.5 in the textbook can again be of use.

Hint

The determinant of the matrix

\[\begin{split} {\mathbf A}=\left[ \begin{array}{ccc} 1 & 1 & a\\ 1 & 1 & -2\\ -6 & a & 0\\ \end{array} \right] \end{split}\]

will after a few intermediate calculations become \(a^2+8a+12\). For which calues of \(a\) is the determinant equal to zero?

Answer

Since \(a^2+8a+12=0\) if and only if \(a=-2 \, \vee a=-6\), then according to Corollary 8.2.5 the three given vectors are linearly dependent if and only if \(a \in \{-2,-6\}.\)

Exercise 5: Transpose of Square Matrices#

We konw from Theorem 7.2.2 in the textbook that \(({\mathbf A} \cdot {\mathbf B})^T={\mathbf B}^T \cdot {\mathbf A}^T\) for all \({\mathbf A} \in \mathbb{F}^{m \times n}\) and \({\mathbf B} \in \mathbb{F}^{n \times \ell}\). The symbol \(\mathbb{F}\) is in this course a placeholder for \(\mathbb{R}\) or \(\mathbb{C}.\)

Question a#

Conclude that \(({\mathbf A} \cdot {\mathbf B})^T={\mathbf B}^T \cdot {\mathbf A}^T\) for all \({\mathbf A},{\mathbf B} \in \mathbb{F}^{n \times n}\).

Hint

One can let \(m\) and \(\ell\) be equal to \(n\).

Question b#

Assume that the matrix \(\mathbf A \in \mathbb{F}^{n \times n}\) and that \(\rho({\mathbf A})=n\). Use the result from Question a to realize that \(({\mathbf A}^{T})^{-1}=({\mathbf A}^{-1})^T.\)

Hint

Begin by figuring out which properties the inverse matrix of \({\mathbf A}^{T}\) should have according to Definition 7.3.1.

Hint

\({\mathbf A}^{-1}\) exists because \(\rho({\mathbf A})=n\). Hence we can define the matrix \(\mathbf B=({\mathbf A}^{-1})^T\).

Now use the result from Question a to realize that the matrix \(\mathbf B\) has precisely the wanted proporties that the inverse of \({\mathbf A}^T\) must have according to Definition 7.3.1. If those properties are achieved then we can conclude that \({\mathbf B}=({\mathbf A}^T)^{-1}\).

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