Linear Algebra for Team-Based Inquiry Learning: 2023 Edition PREVIEW

1. \begin{equation*} x_1\left[\begin{array}{c} 3 \\ -1 \\ 0 \end{array}{}\right] + x_2 \left[\begin{array}{c}2 \\ -4 \\ 1 \end{array}{}\right]+ x_3 \left[\begin{array}{c} 1 \\ 1 \\ -1 \end{array}{}\right] + x_4 \left[\begin{array}{c} 1 \\ -7 \\ 0 \end{array}{}\right] = \left[\begin{array}{c} 1 \\ 0 \\ -2 \end{array}{}\right] \end{equation*}
2. \begin{equation*} \left[\begin{array}{cccc|c} 3 & 2 & 0 & 1 & 1 \\ -1 & -4 & 1 & -7 & 0 \\ 0 & 1 & -1 & 0 & -2 \end{array}\right] \end{equation*}

1. • \(A=\left[\begin{array}{ccc} -4 & 0 & 4 \\ 0 & 1 & -2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\) is not in reduced row echelon form because the pivots are not all \(1\text{.}\)

   • \(B=\left[\begin{array}{ccc} 0 & 1 & 2 \\ 1 & 0 & -3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\) is not in reduced row echelon form because the pivots are not descending to the right.

   • \(C=\left[\begin{array}{ccc} 1 & -4 & 4 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\) is not in reduced row echelon form because not every entry above and below each pivot is zero.

2. \begin{alignat*}{4} \left[\begin{array}{cccc} 0 & 3 & 1 & 2 \\ 1 & 2 & -1 & -3 \\ 2 & 4 & -1 & -1 \end{array}\right] &\sim& \left[\begin{array}{cccc} \circledNumber{1} & 2 & -1 & -3 \\ 0 & 3 & 1 & 2 \\ 2 & 4 & -1 & -1 \end{array}\right] &\hspace{0.2in} \text{Swap Rows 1 and 2}& \\ &\sim& \left[\begin{array}{cccc} \circledNumber{1} & 2 & -1 & -3 \\ 0 & 3 & 1 & 2 \\ 0 & 0 & 1 & 5 \end{array}\right] &\hspace{0.2in} \text{Add } -2 \text{ Row 1 to Row 3}& \\ &\sim& \left[\begin{array}{cccc} \circledNumber{1} & 2 & -1 & -3 \\ 0 & \circledNumber{1} & \frac{1}{3} & \frac{2}{3} \\ 0 & 0 & 1 & 5 \end{array}\right] &\hspace{0.2in} \text{Multiply Row 3 by } \frac{1}{3}& \\ &\sim& \left[\begin{array}{cccc} \circledNumber{1} & 0 & -\frac{5}{3} & -\frac{13}{3} \\ 0 & \circledNumber{1} & \frac{1}{3} & \frac{2}{3} \\ 0 & 0 & \circledNumber{1} & 5 \end{array}\right] &\hspace{0.2in} \text{Add } -2 \text{ Row 2 to Row 1}& \\ &\sim& \left[\begin{array}{cccc} \circledNumber{1} & 0 & -\frac{5}{3} & -\frac{13}{3} \\ 0 & \circledNumber{1} & 0 & -1 \\ 0 & 0 & \circledNumber{1} & 5 \end{array}\right] &\hspace{0.2in} \text{Add } -\frac{1}{3} \text{ Row 3 to Row 2}& \\ &\sim& \left[\begin{array}{cccc} \circledNumber{1} & 0 & 0 & 4 \\ 0 & \circledNumber{1} & 0 & -1 \\ 0 & 0 & \circledNumber{1} & 5 \end{array}\right] &\hspace{0.2in} \text{Add } \frac{5}{3} \text{ Row 3 to Row 1}& \end{alignat*}

1. \begin{equation*} \mathrm{RREF}\left[\begin{array}{ccc|c} -2 & 1 & 1 & -2 \\ -2 & -3 & -3 & 0 \\ 3 & 1 & 1 & 3 \end{array}\right]=\left[\begin{array}{ccc|c} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right] \end{equation*}

   This matrix corresponds to the simpler system

   \begin{equation*} \begin{matrix} x_{1} & & & & & = & 0 \\ & & x_{2} & + & x_{3} & = & 0 \\ & & & & 0 & = & 1 \\ \end{matrix} \end{equation*}

   The third equation \(0=1\) indicates that the system has no solutions. The solution set is \(\emptyset\text{.}\)

2. \begin{equation*} \mathrm{RREF}\left[\begin{array}{ccc|c} -5 & 3 & 14 & 1 \\ 3 & -2 & -9 & 0 \\ -1 & 2 & 7 & -4 \end{array}\right]=\left[\begin{array}{ccc|c} 1 & 0 & -1 & -2 \\ 0 & 1 & 3 & -3 \\ 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

   This matrix corresponds to the simpler system

   \begin{equation*} \begin{matrix} x_{1} & & & - & x_3 & = & -2 \\ & & x_{2} & + & 3\,x_{3} & = & -3 \\ & & & & 0 & = & 0 \\ \end{matrix}. \end{equation*}

   Since there are three variables and two nontrivial equations, the solution set has infinitely-many solutions.

3. \begin{equation*} \mathrm{RREF}\left[\begin{array}{ccc|c} 0 & 1 & 2 & -5 \\ -1 & -4 & -4 & 11 \\ -1 & -4 & -3 & 8 \end{array}\right]=\left[\begin{array}{ccc|c} 1 & 0 & 0 & -3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & -3 \end{array}\right] \end{equation*}

   This matrix corresponds to the simpler system

   \begin{equation*} \begin{matrix} x_{1} & & & & & = & -3 \\ & & x_{2} & & & = & 1 \\ & & & & x_{3} & = & -3 \\ \end{matrix}. \end{equation*}

   This system has one solution. The solution set is \(\left\{ \left[\begin{array}{c} -3 \\ 1 \\ -3 \end{array}\right] \right\}\text{.}\)

First, we compute

This corresponds to the simpler system

Since the second column is a non-pivot column, we let \(x_2=a\text{.}\) Making this substitution and then solving for \(x_1\text{,}\) \(x_3\text{,}\) and \(x_4\) produces the system

Thus, the solution set is \(\left\{ \left[\begin{array}{c} -a + 1 \\ a \\ 1 \\ 2 \end{array}\right] \,\middle|\, a \in\mathbb R \right\} \text{.}\)

1. We compute both sides:

   \begin{align*} c \odot \left((x_1,y_1) \oplus (x_2,y_2) \right) &= c \odot (2x_1+2x_2,2y_1+2y_2)\\ &= (c(2x_1+2x_2),c^2(2y_1+2y_2))\\ &= (2cx_1+2cx_2,2c^2y_1+2c^2y_2) \end{align*}

   and

   \begin{align*} c\odot (x_1,y_1) \oplus c \odot (x_2,y_2) &= (cx_1,c^2y_1) \oplus (cx_2,c^2y_2)\\ &= (2cx_1+2cx_2,2c^2y_1+2c^2y_2) \end{align*}

   Since these are the same, we have shown that the property holds.

2. To show \(V\) is not a vector space, we must show that it fails one of the 8 defining properties of vector spaces. We will show that scalar multiplication does not distribute over scalar addition, i.e., there are values such that

   \begin{equation*} (c+d)\odot(x,y) \neq c \odot(x,y) \oplus d\odot(x,y) \end{equation*}

   • (Solution method 1) First, we compute

     \begin{align*} (c+d)\odot(x,y) &= ((c+d)x,(c+d)^2y)\\ &= ( (c+d)x, (c^2+2cd+d^2)y). \end{align*}

     Then we compute

     \begin{align*} c\odot (x,y) \oplus d\odot(x,y) &= (cx,c^2y) \oplus (dx,d^2y)\\ &= ( 2cx+2dx, 2c^2y+2d^2y). \end{align*}

     Since \((c+d)x\not=2cx+2dy\) when \(c,d,x,y=1\text{,}\) the property fails to hold.

   • (Solution method 2) When we let \(c,d,x,y=1\text{,}\) we may simplify both sides as follows.

     \begin{align*} (c+d)\odot(x,y) &= 2\odot(1,1)\\ &= (2\cdot1,2^2\cdot1)\\ &=(2,4) \end{align*}
     \begin{align*} c\odot (x,y) \oplus d\odot(x,y) &= 1\odot(1,1)\oplus 1\odot(1,1)\\ &= (1\cdot1,1^2\cdot1)\oplus(1\cdot1,1^2\cdot1)\\ &= (1,1)\oplus(1,1)\\ &= (2\cdot1+2\cdot1,2\cdot1+2\cdot1)\\ &= (4,4) \end{align*}

     Since these ordered pairs are different, the property fails to hold.

• \(\left[\begin{array}{c} -13 \\ 3 \\ -13 \end{array}\right]\)is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 3 \end{array}\right] , \text{ and } \left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right]\) exactly when the vector equation

  \begin{equation*} x_1\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] +x_2 \left[\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right] +x_3 \left[\begin{array}{c} 3 \\ 0 \\ 3 \end{array}\right] +x_4 \left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right] = \left[\begin{array}{c} -13 \\ 3 \\ -13 \end{array}\right] \end{equation*}

  has a solution. To solve this vector equation, we compute

  \begin{equation*} \mathrm{RREF}\, \left[\begin{array}{cccc|c} 1 & 2 & 3 & -5 & -13 \\ 0 & 0 & 0 & 1 & 3 \\ 1 & 2 & 3 & -5 & -13 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 2 & 3 & 0 & 2 \\ 0 & 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]\text{.} \end{equation*}

  We see that this vector equation has solution set \(\left\{\left[\begin{array}{c}2-2a-3b \\ a \\ b \\ 3 \end{array}\right]\ \middle|\ a,b \in \mathbb{R}\right\}\text{,}\) so \(\left[\begin{array}{c} -13 \\ 3 \\ -13 \end{array}\right]\) is a linear combination; for example, \(2 \left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] + 3 \left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right] = \left[\begin{array}{c} -13 \\ 3 \\ -13 \end{array}\right]\)

• \(\left[\begin{array}{c} -13 \\ 3 \\ -15 \end{array}\right]\) is a linear combination of the vectors \(\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] , \left[\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right] , \left[\begin{array}{c} 3 \\ 0 \\ 3 \end{array}\right] , \text{ and } \left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right]\) exactly when the vector equation

  \begin{equation*} x_1\left[\begin{array}{c} 1 \\ 0 \\ 1 \end{array}\right] +x_2 \left[\begin{array}{c} 2 \\ 0 \\ 2 \end{array}\right] +x_3 \left[\begin{array}{c} 3 \\ 0 \\ 3 \end{array}\right] +x_4 \left[\begin{array}{c} -5 \\ 1 \\ -5 \end{array}\right] = \left[\begin{array}{c} -13 \\ 3 \\ -15 \end{array}\right] \end{equation*}

  has a solution. To solve this vector equation, we compute

  \begin{equation*} \mathrm{RREF}\, \left[\begin{array}{cccc|c} 1 & 2 & 3 & -5 & -13 \\ 0 & 0 & 0 & 1 & 3 \\ 1 & 2 & 3 & -5 & -15 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 2 & 3 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right]\text{.} \end{equation*}

  This vector equation has no solution, so \(\left[\begin{array}{c} -13 \\ 3 \\ -15 \end{array}\right]\) is not a linear combination.

The set of vectors \(\left\{ \left[\begin{array}{c} 1 \\ -1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 3 \\ -2 \\ 3 \\ 3 \end{array}\right] , \left[\begin{array}{c} 10 \\ -7 \\ 11 \\ 9 \end{array}\right] , \left[\begin{array}{c} -6 \\ 3 \\ -3 \\ -9 \end{array}\right] \right\}\) spans \(\mathbb{R}^4\) exactly when the vector equation

has a solution for all \(\vec{v} \in \mathbb{R}^4\text{.}\) If there is some vector \(\vec{v} \in \mathbb{R}^4\) for which this vector equation has no solution, then the set does not span \(\mathbb{R}^4\text{.}\) To answer this, we compute

We see that for some \(\vec{v} \in \mathbb{R}^4\text{,}\) this vector equation will not have a solution, so the set of vectors \(\left\{ \left[\begin{array}{c} 1 \\ -1 \\ 2 \\ 0 \end{array}\right] , \left[\begin{array}{c} 3 \\ -2 \\ 3 \\ 3 \end{array}\right] , \left[\begin{array}{c} 10 \\ -7 \\ 11 \\ 9 \end{array}\right] , \left[\begin{array}{c} -6 \\ 3 \\ -3 \\ -9 \end{array}\right] \right\}\) does not span \(\mathbb{R}^4\text{.}\)

To show that \(W\) is a subspace, let \(\vec v=\left[\begin{array}{c} x_1 \\y_1 \\ z_1 \\ w_1 \end{array}\right]\in W\) and \(\vec w=\left[\begin{array}{c} x_2 \\y_2 \\ z_2 \\ w_2 \end{array}\right] \in W \text{,}\) so we know that \(x_1+y_1=3z_1+2w_1\) and \(x_2+y_2=3z_2+2w_2\text{.}\)

Consider

To see if \(\vec{v}+\vec{w} \in W\text{,}\) we need to check if \((x_1+x_2)+(y_1+y_2) = 3(z_1+z_2)+2(w_1+w_2)\text{.}\) We compute

Thus \(\vec v+\vec w\in W\text{,}\) so \(W\) is closed under vector addition.

Now consider

Similarly, to check that \(c\vec{v} \in W\text{,}\) we need to check if \(cx_1+cy_1=3(cz_1)+2(cw_1)\text{,}\) so we compute

and we see that \(c\vec v\in W\text{,}\) so \(W\) is closed under scalar multiplication. Therefore \(W\) is a subspace of \(\IR^3\text{.}\)

Now, to show \(U\) is not a subspace, we will show that it is not closed under vector addition.

• (Solution Method 1) Now let \(\vec v=\left[\begin{array}{c} x_1 \\y_1 \\ z_1 \\ w_1 \end{array}\right]\in U\) and \(\vec w=\left[\begin{array}{c} x_2 \\y_2 \\ z_2 \\ w_2 \end{array}\right] \in U \text{,}\) so we know that \(x_1+y_1=3z_1+w_1^2\) and \(x_2+y_2=3z_2+w_2^2\text{.}\)

  Consider

  \begin{equation*} \vec{v}+\vec{w}= \left[\begin{array}{c} x_1 \\y_1 \\ z_1 \\ w_1\end{array}\right] +\left[\begin{array}{c} x_2 \\y_2 \\ z_2 \\ w_2 \end{array}\right] =\left[\begin{array}{c} x_1+x_2 \\y_1+y_2 \\ z_1+z_2 \\w_1+w_2 \end{array}\right] . \end{equation*}

  To see if \(\vec{v}+\vec{w} \in U\text{,}\) we need to check if \((x_1+x_2)+(y_1+y_2) = 3(z_1+z_2)+(w_1+w_2)^2\text{.}\) We compute

  \begin{align*} (x_1+x_2)+(y_1+y_2) &= (x_1+y_1)+(x_2+y_2) &\text{by regrouping}\\ &= (3z_1+w_1^2)+(3z_2+w_2^2) &\text{since }\\ &=3(z_1+z_2)+(w_1^2+w_2^2) &\text{by regrouping} \end{align*}

  and thus \(\vec v+\vec w\in U\) \textbf{only when} \(w_1^2+w_2^2=(w_1+w_2)^2\text{.}\) Since this is not true in general, \(U\) is not closed under vector addition, and thus cannot be a subspace.

• (Solution Method 2) Note that the vector \(\vec v=\left[\begin{array}{c} 0\\1\\0\\1\end{array}\right] \) belongs to \(U\) since \(0+1=3(0)+1^2\text{.}\) However, the vector \(2\vec v=\left[\begin{array}{c} 0\\2\\0\\2\end{array}\right] \) does not belong to \(U\) since \(0+2\not=3(0)+2^2\text{.}\) Therefore \(U\) is not closed under scalar multiplication, and thus is not a subspace.

The set of vectors \(\left\{ \left[\begin{array}{c} 1 \\ 3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -3 \end{array}\right] \right\}\) is linearly independent exactly when the vector equation

has no non-trivial (i.e. nonzero) solutions. The set is linearly dependent when there exists a nontrivial (i.e. nonzero) solution. We compute

Thus, this vector equation has a solution set \(\left\{ \left[\begin{array}{c}a \\ a \\ 0 \end{array}\right]\ \middle|\ a \in \mathbb{R}\right\}\text{.}\) Since there are nontrivial solutions, we conclude that the set of vectors \(\left\{ \left[\begin{array}{c} 1 \\ 3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} -1 \\ -3 \\ -4 \\ 4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -3 \end{array}\right] \right\}\) is linearly dependent.

The set of vectors \(\left\{ \left[\begin{array}{c} 1 \\ 3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ 11 \\ 18 \\ -18 \end{array}\right] , \left[\begin{array}{c} -2 \\ -7 \\ -11 \\ 11 \end{array}\right] \right\}\) is a basis of \(\mathbb{R}^4\) exactly when it is linearly independent and the set spans \(\mathbb{R}^4\text{.}\) If it is either linearly dependent, or the set does not span \(\mathbb{R}^4\text{,}\) then the set is not a basis.

To answer this, we compute

We see that this set of vectors is linearly dependent, so therefore the set of vectors \(\left\{ \left[\begin{array}{c} 1 \\ 3 \\ 4 \\ -4 \end{array}\right] , \left[\begin{array}{c} 0 \\ 1 \\ 3 \\ -3 \end{array}\right] , \left[\begin{array}{c} 3 \\ 11 \\ 18 \\ -18 \end{array}\right] , \left[\begin{array}{c} -2 \\ -7 \\ -11 \\ 11 \end{array}\right] \right\}\) is not a basis.

1. Observe that

   \begin{equation*} \RREF \left[\begin{array}{ccccc} 1 & 1 & 3 & 1 & 2 \\ -3 & 0 & -6 & 6 & 3 \\ -1 & 1 & -1 & 1 & 0 \\ 2 & -2 & 2 & -1 & 1 \end{array}\right] = \left[\begin{array}{ccccc} 1 & 0 & 2 & 0 & 1 \\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right] \end{equation*}

   If we remove the vectors yielding non-pivot columns, the resulting set will span the same vectors while being linearly independent. Therefore

   \begin{equation*} \left\{ \left[\begin{array}{c} 1 \\ -3 \\ -1 \\ 2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 0 \\ 1 \\ -2 \end{array}\right] , \left[\begin{array}{c} 1 \\ 6 \\ 1 \\ -1 \end{array}\right] \right\} \end{equation*}

   is a basis of \(W\text{.}\)

2. Since this (and thus every other) basis has three vectors in it, the dimension of \(W\) is \(3\text{.}\)

The set of polynomials

is linearly independent exactly when the polynomial equation

has no nontrivial (i.e. nonzero) solutions. The set is linearly dependent when this equation has a nontrivial (i.e. nonzero) solution.

To solve this equation, we distribute and then collect coefficients to obtain

These polynomials are equal precisely when their coefficients are equal, leading to the system

To solve this, we compute

The system has (infintely many) nontrivial solutions, so we that the set of polynomials is linearly dependent.

1. Observe that

   \begin{equation*} \RREF \left[\begin{array}{ccccc|c} 1 & 1 & 3 & 1 & 2 & 0\\ -3 & 0 & -6 & 6 & 3 & 0\\ -1 & 1 & -1 & 1 & 0 & 0\\ 2 & -2 & 2 & -1 & 1& 0 \end{array}\right] = \left[\begin{array}{ccccc|c} 1 & 0 & 2 & 0 & 1 &0\\ 0 & 1 & 1 & 0 & 0 &0\\ 0 & 0 & 0 & 1 & 1 &0\\ 0 & 0 & 0 & 0 & 0&0 \end{array}\right] \end{equation*}

   Letting \(x_3=a\) and \(x_5=b\) (since those correspond to the non-pivot columns), this is equivalent to the system

   \begin{alignat*}{6} x_1 &\,\,& &\,+\,& 2x_3 &\,\,& &\,+\,& x_5 &=& 0\\ &\,\,& x_2 &\,+\,& x_3 &\,\,& &\,\,& &=& 0\\ &\,\,& &\,\,& x_3 &\,\,& &\,\,& &=& a\\ &\,\,& &\,\,& &\,\,& x_4 &\,+\,& x_5 &=& 0\\ &\,\,& &\,\,& &\,\,& &\,\,& x_5 &=& b \end{alignat*}

   Thus, the solution set is

   \begin{equation*} \setBuilder{\left[\begin{array}{c} -2a-b \\ -a \\ a \\ -b \\ b \end{array}\right]}{a,b \in \IR} . \end{equation*}

2. Since we can write

   \begin{equation*} \left[\begin{array}{c} -2a-b \\ -a \\ a \\ -b \\ b \end{array}\right] = a \left[\begin{array}{c} -2 \\ -1 \\ 1 \\ 0 \\ 0 \end{array}\right] + b \left[\begin{array}{c} -1 \\ 0 \\ 0 \\ -1 \\ 1 \end{array}\right], \end{equation*}

   a basis for the solution space is

   \begin{equation*} \left \{ \left[\begin{array}{c} -2 \\ -1 \\ 1 \\ 0 \\ 0 \end{array}\right] , \left[\begin{array}{c} -1 \\ 0 \\ 0 \\ -1 \\ 1 \end{array}\right] \right\}. \end{equation*}

To show \(S\) is a linear transformation, we must show two things:

To show \(S\) respects addition, we compute

But note that \(S(f(x))=3xf(x)\) and \(S(g(x))=3xg(x)\text{,}\) so we have \(S(f(x)+g(x))=S(f(x))+S(g(x))\text{.}\)

For the second part, we compute

But note that \(cS(f(x))=c(3xf(x))=3cxf(x)\) as well, so we have \(S(cf(x))=cS(f(x))\text{.}\) Now, since \(S\) respects both addition and scalar multiplication, we can conclude \(S\) is a linear transformation.

• (Solution method 1) As for \(T\text{,}\) we compute

  \begin{align*} T(f(x)+g(x))& =3 (f(x)+g(x))'(f(x)+g(x)) &\text{by definition of }\\ &= 3(f'(x)+g'(x))(f(x)+g(x)) & \text{since the derivative is linear}\\ &= 3f(x)f'(x)+3f(x)g'(x)+3f'(x)g(x)+3g(x)g'(x) &\text{by distributing} \end{align*}

  However, note that \(T(f(x))+T(g(x))=3f'(x)f(x)+3g'(x)g(x)\text{,}\) which is not always the same polynomial (for example, when \(f(x)=g(x)=x\)). So we see that \(T(f(x)+g(x)) \neq T(f(x))+T(g(x))\text{,}\) so \(T\) does not respect addition and is therefore not a linear transformation.

• (Solution method 2) As for \(T\text{,}\) we may choose the polynomial \(f(x)=x\) and scalar \(c=2\text{.}\) Then

  \begin{equation*} T(cf(x))=T(2x)=3(2x)'(2x)=3(2)(2x)=12x. \end{equation*}

  But on the other hand,

  \begin{equation*} cT(f(x))=2T(x)=2(3)(x)'(x)=2(3)(1)(x)=6x. \end{equation*}

  Since this isn't the same polynomial, \(T\) does not preserve multiplication and is therefore not a linear transformation.

1. Since

   \begin{align*} T\left(\left[\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right]\right) &= \left[\begin{array}{c} -1 \\ -1 \\ 7 \\0\end{array}\right]\\ T\left(\left[\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\right]\right) &= \left[\begin{array}{c} 1 \\ 3 \\ 1 \\0 \end{array}\right]\\ T\left(\left[\begin{array}{c} 0 \\ 0 \\ 1 \end{array}\right]\right) &= \left[\begin{array}{c} 0 \\ -1 \\ 3 \\ 0 \end{array}\right], \end{align*}

   the standard matrix for \(T\) is \(\left[\begin{array}{ccc} -1 & 1 & 0 \\ -1 & 3 & -1 \\ 7 & 1 & 3 \\ 0 & 0 & 0 \end{array}\right] \text{.}\)

2. \begin{equation*} S\left(\left[\begin{array}{c} -2 \\ 1 \\ 3 \\ 2 \end{array}\right] \right) = -2S(\vec{e}_1)+S(\vec{e}_2)+3S(\vec{e}_3)+2S(\vec{e}_4) \end{equation*}
   \begin{equation*} = -2 \left[\begin{array}{c} 2 \\ 0 \\ 3 \end{array}\right] + \left[\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right] + 3 \left[\begin{array}{c} 4 \\ -1 \\ -2 \end{array}\right]+2\left[\begin{array}{c} 1 \\ -1 \\ 4 \end{array}\right] = \left[\begin{array}{c} 13 \\ -4 \\ -6\end{array}\right]. \end{equation*}

1. To find the image we compute

   \begin{equation*} \Im(T) = T\left(\vspan\left\{\vec{e}_1,\vec{e}_2,\vec{e}_3,\vec{e}_4\right\}\right) \end{equation*}
   \begin{equation*} = \vspan\left\{T(\vec{e}_1),T(\vec{e}_2),T(\vec{e}_3),T(\vec{e}_4)\right\} \end{equation*}
   \begin{equation*} = \vspan\left\{\left[\begin{array}{c}1 \\ 2 \\ 1 \end{array}\right], \left[\begin{array}{c} 3 \\ 4 \\ 6 \end{array}\right], \left[\begin{array}{c} 2 \\ 6 \\ -1 \end{array}\right], \left[\begin{array}{c} -3 \\ -10 \\ 3 \end{array}\right]\right\}. \end{equation*}

2. The kernel is the solution set of the corresponding homogeneous system of equations, i.e.

   \begin{alignat*}{5} x &+& 3y &+& 2z &-& 3w &=& 0\\ 2x &+& 4y &+& 6z &-& 10w &=& 0\\ x &+& 6y &-& z &+& 3w &=& 0 . \end{alignat*}

   So we compute

   \begin{equation*} \RREF\left[\begin{array}{cccc|c} 1 & 3 & 2 & -3 & 0 \\ 2 & 4 & 6 & -10 &0 \\ 1 & 6 & -1 & 3 & 0 \end{array}\right] = \left[\begin{array}{cccc|c} 1 & 0 & 5 & -9 & 0 \\ 0 & 1 & -1 & 2 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right]. \end{equation*}

   Then, letting \(z=a\) and \(w=b\) we have

   \begin{equation*} \ker T = \setBuilder{\left[\begin{array}{c}-5a+9b \\ a-2b \\ a \\ b \end{array}\right]}{a,b \in \IR}. \end{equation*}

3. Since \(\Im(T) = \vspan\left\{\left[\begin{array}{c}1 \\ 2 \\ 1 \end{array}\right], \left[\begin{array}{c} 3 \\ 4 \\ 6 \end{array}\right], \left[\begin{array}{c} 2 \\ 6 \\ -1 \end{array}\right], \left[\begin{array}{c} -3 \\ -10 \\ 3 \end{array}\right]\right\}\text{,}\) we simply need to find a linearly independent subset of these four spanning vectors. So we compute

   \begin{equation*} \RREF \left[\begin{array}{cccc}1 & 3 & 2 & -3 \\ 2 & 4 & 6 & -10 \\ 1 & 6 & -1 & 3 \end{array}\right] = \left[\begin{array}{cccc} 1 & 0 & 5 & -9 \\ 0 & 1 & -1 & 2 \\ 0 & 0 & 0 & 0\end{array}\right]. \end{equation*}

   Since the first two columns are pivot columns, they form a linearly independent spanning set, so a basis for \(\Im T\) is \(\setList{\left[\begin{array}{c}1\\2\\1 \end{array}\right], \left[\begin{array}{c}3\\4\\6 \end{array}\right]}.\)

   To find a basis for the kernel, note that

   \begin{equation*} \ker T = \setBuilder{\left[\begin{array}{c}-5a+9b \\ a-2b \\ a \\ b \end{array}\right]}{a,b \in \IR} \end{equation*}
   \begin{equation*} = \setBuilder{a \left[\begin{array}{c}-5 \\ 1 \\ 1 \\ 0 \end{array}\right]+b \left[\begin{array}{c} 9 \\ -2 \\ 0 \\ 1 \end{array}\right]}{a,b \in \IR} \end{equation*}
   \begin{equation*} = \vspan\left\{ \left[\begin{array}{c} -5 \\ 1 \\ 1 \\ 0 \end{array}\right], \left[\begin{array}{c} 9 \\ -2 \\ 0 \\ 1 \end{array}\right]\right\}. \end{equation*}

   so a basis for the kernel is

   \begin{equation*} \setList{\left[\begin{array}{c}-5 \\ 1 \\ 1 \\ 0 \end{array}\right], \left[\begin{array}{c}9 \\ -2 \\ 0 \\ 1 \end{array}\right]}. \end{equation*}

4. The dimension of the image (the rank) is \(2\text{,}\) the dimension of the kernel (the nullity) is \(2\text{,}\) and the dimension of the domain of \(T\) is \(4\text{,}\) so we see \(2+2=4\text{,}\) which verifies that the sum of the rank and nullity of \(T\) is the dimension of the domain of \(T\text{.}\)

Compute

1. Note that the third and fourth columns are non-pivot columns, which means \(\ker T\) contains infinitely many vectors, so \(T\) is not injective.

2. Since there are only two pivots, the image (i.e. the span of the columns) is a 2-dimensional subspace (and thus does not equal \(\IR^3\)), so \(T\) is not surjective.

\(AC\) is the only one that can be computed, since \(C\) corresponds to a linear transformation \(\mathbb{R}^3 \rightarrow \mathbb{R}^2\) and \(A\) corresponds to a linear transfromation \(\mathbb{R}^2 \rightarrow \mathbb{R}^2\text{.}\) Thus the composition \(AC\) corresponds to a linear transformation \(\mathbb{R}^3 \rightarrow \mathbb{R}^2\) with a \(2\times 3\) standard matrix. We compute

Thus

1. \(\displaystyle P=\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \\ 4 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right]\)

2. \(\displaystyle Q=\left[\begin{array}{cccc} -4 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right]\)

3. \(\displaystyle QPA\)

We compute

We see \(D\) is bijective, and therefore invertible. To compute the inverse, we solve \(D\vec{x}=\vec{e}_1\) by computing

Similarly, we solve \(D\vec{x}=\vec{e}_2\) by computing

Similarly, we solve \(D\vec{x}=\vec{e}_3\) by computing

Similarly, we solve \(D\vec{x}=\vec{e}_4\) by computing

Combining these, we obtain

We compute

We see \(N\) is not bijective and thus is not invertible.

1. Adding a multiple of one row to another row does not change the determinant, so \(\det(B)=\det(A)=-7\text{.}\)

2. Scaling a row scales the determinant by the same factor, so so \(\det(B)=-3\det(A)=-3(-7)=21\text{.}\)

3. Swaping rows changes the sign of the determinant, so \(\det(B)=-\det(A)=7\text{.}\)

Here is one possible solution, first applying a single row operation, and then performing Laplace/cofactor expansions to reduce the determinant to a linear combination of \(2\times 2\) determinants:

Here is another possible solution, using row and column operations to first reduce the determinant to a \(3\times 3\) matrix and then applying a formula:

Compute the characteristic polynomial:

The eigenvalues are the roots of the characteristic polynomial, namely \(2\) and \(3\text{.}\)

The eigenspace associated to \(3\) is the kernel of \(A-3I\text{,}\) so we compute

Thus we see the kernel is

which has a basis of \(\left\{ \left[\begin{array}{c} -1 \\ \frac{3}{2} \\ 1 \end{array}\right] \right\}\text{.}\)