Exercises – Long Day

Exercises – Long Day#

Exercise 1: Linear Combinations#

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

\[\begin{split} {\mathbf u}=\left[ \begin{array}{c} 4\\ -2\\ 0\\ \end{array} \right], \quad {\mathbf v}=\left[ \begin{array}{c} 7\\ 4\\ 3\\ \end{array} \right], \quad {\mathbf w}=\left[ \begin{array}{c} 1\\ 7\\ 3\\ \end{array} \right]. \end{split}\]

Question a#

Calculate the following linear kombinations of the three vectors \({\mathbf u}, {\mathbf v},\) and \({\mathbf w}.\)

  1. \(4{\mathbf v}.\)

  2. \(2{\mathbf u}-3{\mathbf w}\).

  3. \(3{\mathbf u}-2{\mathbf v}+2{\mathbf w}\).

Answer

  1. \[\begin{split}4{\mathbf v}=\left[ \begin{array}{c} 28\\ 16\\ 12\\ \end{array} \right]. \end{split}\]
\[\begin{split} 2{\mathbf u}-3{\mathbf w}=\left[ \begin{array}{c} 5\\ -25\\ -9\\ \end{array} \right]. \end{split}\]
\[\begin{split}3{\mathbf u}-2{\mathbf v}+2{\mathbf w}=\left[ \begin{array}{c} 0\\ 0\\ 0\\ \end{array} \right].\end{split}\]

Question b#

Determine whether the vectors \({\mathbf u}, {\mathbf v},\) and \({\mathbf w}\) are linearly independent based on your answers to Question a.

Hint

In Definition 7.1.1 in the textbook you can read more about what it means for vectors to be linearly independent/dependent. Equations (7-3) and (7-4) may also be useful to have a look at.

Answer

The vectors \({\mathbf u}, {\mathbf v},\) and \({\mathbf w}\) are not linearly independent due to what we saw in part 3 of Question along with Equation (7-4) in the textbook.

Question c#

Show that the vectors \({\mathbf u}\) and \({\mathbf v}\) are linearly independent.

Hint

First, consider Equation (7-3) in the textbook. If \(c_1,c_2 \in \mathbb{R}\) fulfill that

\[\begin{split}c_1 \cdot {\mathbf u}+c_2 \cdot {\mathbf v}=\left[ \begin{array}{c} 0\\ 0\\ 0\\ \end{array} \right],\end{split}\]

then it can be shown that \(c_1=c_2=0\)?

Answer

The vectors \({\mathbf u}\) and \({\mathbf v}\) are linearly independent.

Exercise 2: Linearly Independent Vectors in \(\mathbb{R}^4\)#

We are given the following three vectors in \(\mathbb{R}^4\):

\[\begin{split} {\mathbf u}=\left[ \begin{array}{c} 1\\ 1\\ 0\\ 0\\ \end{array} \right], \quad {\mathbf v}=\left[ \begin{array}{c} 0\\ 1\\ 1\\ 0\\ \end{array} \right], \quad {\mathbf w}=\left[ \begin{array}{c} 0\\ 0\\ 1\\ 1\\ \end{array} \right]. \end{split}\]

Question a#

Show that the vectors \({\mathbf u}, {\mathbf v},\) and \({\mathbf w}\) are linearly independent.

Hint

Assume that

\[\begin{split} c_1\cdot\left[ \begin{array}{c} 1\\ 1\\ 0\\ 0\\ \end{array} \right]+c_2\cdot\left[ \begin{array}{c} 0\\ 1\\ 1\\ 0\\ \end{array} \right]+c_3\cdot\left[ \begin{array}{c} 0\\ 0\\ 1\\ 1\\ \end{array} \right]=\left[ \begin{array}{c} 0\\ 0\\ 0\\ 0\\ \end{array} \right]. \end{split}\]

What can we now say about \(c_1, c_2,\) and \(c_3\)?

Question b#

Assume that \({\mathbf c} \in \mathbb{R}^4\) fulfills that the vectors \({\mathbf u}, {\mathbf v}, {\mathbf w},\) and \({\mathbf c}\) are linearly independent. Show that then the vector \(\mathbf c\) cannot be written as a linear combination of \({\mathbf u}, {\mathbf v},\) and \({\mathbf w}\).

Hint

Using contraposition (Equation (1.22) in the textbook) we see that what we wish to prove is logically equivalent with the proposition: if \({\mathbf c}\) can be written as a linear combination of \({\mathbf u}, {\mathbf v},\) and \({\mathbf w}\), then the vectors \({\mathbf u}, {\mathbf v}, {\mathbf w},\) and \({\mathbf c}\) are linearly dependent. Try proving this proposition instead.

Hint

If \({\mathbf c}=c_1\cdot {\mathbf u}+c_2\cdot {\mathbf v}+c_3\cdot {\mathbf w}\) for some \(c_1,c_2,c_3 \in \mathbb{R}\), then

\[\begin{split}{\mathbf c}-c_1\cdot {\mathbf u}-c_2\cdot {\mathbf v}-c_3\cdot {\mathbf w}=\left[ \begin{array}{c} 0\\ 0\\ 0\\ 0\\ \end{array} \right].\end{split}\]

Question c#

Find a vector \({\mathbf b} \in \mathbb{R}^4\) such that the vectors \({\mathbf u}, {\mathbf v}, {\mathbf w},\) and \({\mathbf b}\) are linearly independent.

Hint

What we learned in Question b can be used here: try to find a vector \({\mathbf b} \in \mathbb{R}^4\) that cannot be written as a linear combination of \({\mathbf u}, {\mathbf v},\) and \({\mathbf w}\).

Hint

Can the vector

\[\begin{split}\left[ \begin{array}{c} 1\\ 0\\ 0\\ 0\\ \end{array} \right]\end{split}\]

be written as a linear combination of \({\mathbf u}, {\mathbf v},\) and \({\mathbf w}\)?

Exercise 3: Linear Independence and Reduced Row-Echelon Form#

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

\[\begin{split} {\mathbf v_1}=\left[ \begin{array}{c} 1\\ 0\\ 4\\ \end{array} \right], \quad {\mathbf v_2}=\left[ \begin{array}{c} 1\\ 1\\ 7\\ \end{array} \right], \quad {\mathbf v_3}=\left[ \begin{array}{c} 1\\ 2\\ 3\\ \end{array} \right]. \end{split}\]

Question a#

Let \(\mathbf A\) be the \(3 \times 3\) matrix that has \({\mathbf v}_1, {\mathbf v}_2,\) and \({\mathbf v}_3\) as columns. Calculate the reduced row-echelon form of the matrix and use this form to determine the rank of \({\mathbf A}\).

Answer

The reduced row-echelon form of \(\mathbf A\) is

\[\begin{split} \left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right]. \end{split}\]

In other words, the reduced row-echelon form of \(\mathbf A\) is \({\mathbf I}_3\), the \(3 \times 3\) identity matrix. It contains three pivot element, hence \(\rho(\mathbf A)=3.\)

Question b#

Are the given vectors \({\mathbf v}_1, {\mathbf v}_2,\) and \({\mathbf v}_3\) linearly independent?

Hint

Theorem 7.1.3 in the textbook is useful here.

Answer

Using Theorem 7.1.3 we conclude that the three given vectors are linearly independent.

Question c#

Can four vectors \({\mathbf w}_1, {\mathbf w}_2, {\mathbf w}_3,\) and \({\mathbf w}_4\) in \(\mathbb{R}^3\) be linearly independent?

Hint

Let \(\mathbf B\) be the \(3 \times 4\) matrix whose columns are \({\mathbf w}_1, {\mathbf w}_2, {\mathbf w}_3,\) and \({\mathbf w}_4\). Consider why the rank of \(\mathbf B\) cannot be greater than three.

Answer

Since \(\mathbf B\) has three rows, its reduced row-echelon form can maximally have three pivots. Hence, \(\rho(\mathbf B) \le 3\). Using Theorem 7.1.3 we conclude that four vectors \({\mathbf w}_1, {\mathbf w}_2, {\mathbf w}_3,\) and \({\mathbf w}_4\) in \(\mathbb{R}^3\) always will be linearly dependent. In conclusion, the answer is no.

Exercise 4: Matrix Products#

We are given the following three matrices:

\[\begin{split} {\mathbf A}=\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right] \quad {\mathbf B}=\left[ \begin{array}{cc} 1 & 4\\ 2 & 1\\ 3 & 0\\ \end{array} \right] \quad {\mathbf C}=\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 2 & 0 & 0 \\ \end{array} \right]. \end{split}\]

Question a#

Which of the nine combinations of matrix products \({\mathbf A}\cdot{\mathbf A}\), \(\, {\mathbf A}\cdot{\mathbf B}\), \(\, {\mathbf A}\cdot{\mathbf C}\), \(\, {\mathbf B}\cdot{\mathbf A}\), \(\, {\mathbf B}\cdot{\mathbf B}\), \(\, {\mathbf B}\cdot{\mathbf C}\), \(\, {\mathbf C}\cdot{\mathbf A}\), \(\, {\mathbf C}\cdot{\mathbf B}\), \(\, {\mathbf C}\cdot{\mathbf C}\) are defined?

Hint

According to Definition 7.2.2 in the textbook two matrices can only be multiplied together if the number of columns in the first matrix equals the number of rows in the second matrix.

Answer

Only these matrix products are possible: \({\mathbf A}\cdot{\mathbf B}, \, {\mathbf A}\cdot{\mathbf C}, \, {\mathbf B}\cdot{\mathbf A}, \, {\mathbf C}\cdot{\mathbf B}, \, {\mathbf C}\cdot{\mathbf C}\).

Question b#

Calculate the matrix products \({\mathbf A}\cdot{\mathbf B}\) and \({\mathbf B}\cdot{\mathbf A}\). Does it hold true that \({\mathbf A}\cdot{\mathbf B}={\mathbf B}\cdot{\mathbf A}\)?

Answer

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

It is clear that \({\mathbf A}\cdot{\mathbf B} \neq {\mathbf B}\cdot{\mathbf A}\). The matrices \({\mathbf A}\cdot{\mathbf B}\) and \({\mathbf B}\cdot{\mathbf A}\) do not even have the same size!

Question c#

Calculate the product of the scalar \(-1/2\) with the matrix \(\mathbf C\). In other words, calculate \((-1/2) \cdot {\mathbf C}\).

Answer

\[\begin{split}(-1/2) \cdot {\mathbf C}=\left[ \begin{array}{ccc} -1/2 & 0 & 0 \\ 0 & 1/2 & 0 \\ -1 & 0 & 0 \\ \end{array} \right]\end{split}\]

Question d#

Determine whether the following sums of matrices are defined, and calculate those that are: \((-1/2) \cdot {\mathbf C}+{\mathbf A} \cdot {\mathbf B}\) and \((-1/2) \cdot {\mathbf C}+{\mathbf B} \cdot {\mathbf A}.\)

Answer

The sum \((-1/2) \cdot {\mathbf C}+{\mathbf A} \cdot {\mathbf B}\) is not defined, since \((-1/2) \cdot {\mathbf C}\) is a \(3 \times 3\) matrix while \({\mathbf A} \cdot {\mathbf B}\) is a \(2 \times 2\) matrix.

The sum \((-1/2) \cdot {\mathbf C}+{\mathbf B} \cdot {\mathbf A}\) is defined, since the matrices \((-1/2) \cdot {\mathbf C}\) and \({\mathbf A} \cdot {\mathbf B}\) have the same size. The result is:

\[\begin{split} (-1/2) \cdot {\mathbf C}+{\mathbf B} \cdot {\mathbf A}=\left[ \begin{array}{ccc} 1/2 & 4 & 0 \\ 2 & 3/2 & 0 \\ 2 & 0 & 0 \\ \end{array} \right]. \end{split}\]

Exercise 5: Matrix-Vector Product and Systems of Linear Equations#

We are given the following system of linear equations

\[\begin{split} \left\{ \begin{array}{ccc} x_1 + x_4 & = & 0\\ -2x_1+x_2 +3x_4 & = & 1\\ 5x_1 +x_3 -4x_4 & = & 1\\ 4x_1 +x_2 +x_3 & = & 2\\ \end{array} \right. \end{split}\]

Question a#

Write the system in the same form as Equation (7-5) in the textbook.

Answer

\[\begin{split} \left[ \begin{array}{ccccc} 1 & 0 & 0 & 1 \\ -2 & 1 & 0 & 3 \\ 5 & 0 & 1 & -4 \\ 4 & 1 & 1 & 0 \\ \end{array} \right] \cdot \left[ \begin{array}{c} x_1 \\ x_2 \\ x_3 \\ x_4 \\ \end{array} \right]= \left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 2 \\ \end{array} \right]. \end{split}\]

Question b#

We define

\[\begin{split}{\mathbf A}=\left[ \begin{array}{ccccc} 1 & 0 & 0 & 1 \\ -2 & 1 & 0 & 3 \\ 5 & 0 & 1 & -4 \\ 4 & 1 & 1 & 0 \\ \end{array} \right], \quad {\mathbf b}=\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 2 \\ \end{array} \right].\end{split}\]

Check that the given linear equation system has the augmented matrix \([{\mathbf A}|{\mathbf b}]\). In Exercise 3b from Short Day in Week 6 we checked that the linear equation system whose augmented matrix is \([{\mathbf A}|{\mathbf b}]\) has the following particular solution:

\[\begin{split} {\mathbf v}=\left[ \begin{array}{c} 0 \\ 1 \\ 1 \\ 0 \\ \end{array} \right]. \end{split}\]

Check this again, but this time by calculating the product \({\mathbf A}\cdot{\mathbf v}\).

Exercise 6: Finding Inverse Matrices#

We are given the following square matrix:

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

Question a#

Determine whether the matrix \({\mathbf A}^{-1}\) exists, and if so then find it.

Hint

The procedure explained right before Example 7.3.1 in the textbook can be followed. Examples 7.3.1 and 7.3.2 can be useful as well if you wish to see some examples.

Answer

The reduced row-echelon form of the matrix \([{\mathbf A}|{\mathbf I}_3]\) is

\[\begin{split}\left[ \begin{array}{cccccc} 1 & 0 & 0 & 1/2 & -1/2 & 0 \\ 0 & 1 & 0 & 1/2 & 1/2 & 0 \\ 0 & 0 & 1 & -5/2 & 5/2 & 1 \\ \end{array} \right]. \end{split}\]

Hence, \({\mathbf A}^{-1}\) exists, and we have

\[\begin{split}{\mathbf A}^{-1}=\left[ \begin{array}{ccc} 1/2 & -1/2 & 0 \\ 1/2 & 1/2 & 0 \\ -5/2 & 5/2 & 1 \\ \end{array} \right].\end{split}\]

Exercise 7: Inverse Matrices and Systems of Linear Equations#

Let \({\mathbf A} \in \mathbb{C}^{n \times n}\) be a square matrix and assume that \({\mathbf A}\) is an invertible matrix. Furthermore, let \({\mathbf b} \in \mathbb{C}^n\) be a vector.

Question a#

Show that the vector \({\mathbf A}^{-1}\cdot {\mathbf b} \in \mathbb{C}^n\) is a solution to the linear equation system whose augmented matrix is \([{\mathbf A}|{\mathbf b}].\)

Hint

First, rewrite the equation system to matrix form as in Equation (7-5) in the textbook.

Hint

In matrix form the system will look as follows:

\[\begin{split} {\mathbf A} \cdot \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]={\mathbf b}.\end{split}\]

Is this fulfilled if

\[\begin{split} \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]={\mathbf A}^{-1} \cdot {\mathbf b}?\end{split}\]

Question b#

Show that the vector \({\mathbf A}^{-1} \cdot {\mathbf b}\) is the only solution to the linear equation system whose augmented matrix is \([{\mathbf A}|{\mathbf b}].\)

Hint

Multiply the equation

\[\begin{split} {\mathbf A} \cdot \left[ \begin{array}{c} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{array} \right]={\mathbf b}\end{split}\]

by \({\mathbf A}^{-1}\) from the left on both sides of the equation sign.

Exercise 8: Systems of Equations with Variable Coefficients#

For all real values of \(a\) we are given the following system of linear equations:

\[\begin{align*} ax_1 + x_2 + x_3 &= 1\\ x_1 + ax_2 + x_3 &= 1\\ x_1 + x_2 + ax_3 &= 1 \end{align*}\]

Question a#

State the augmented matrix of the equation system.

Question b#

What is the solution for \(a=-2\)?

Question c#

Determine the solution to the linear equation system.

Hint

Along the way in your calculations you might risk division by zero. Note these sepcial cases down and treat them separately afterwards.

Answer

For \(a\in\mathbb R\setminus\{-2,1\}\) the solution is \((x_1,x_2,x_3)=(\frac1{a+2},\frac1{a+2},\frac1{a+2})\), and for \(a=1\) the solution is \((x_1,x_2,x_3)=(1,0,0)+t_1(-1,1,0)+t_2(-1,0,1)\quad,t_1,t_2\in\mathbb R\). There is no solution for \(a=-2\).

Remark#

The Python test on the notebook from Python module 1 released last week will open in Möbius at 15:30 and close at 17:00.

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