Solutions to Linear Algebra Practice Problems 1

1. Show that A = [ \begin{matrix}1 & 2 & −2 & 0 & 7 \\ −2 & −3 & 1 & −1 & −5 \\ −3 & −4 & 0 & −2 & −3 \\ 3 & 6 & −6 & 5 & 1 \end{matrix} ] is row equivalent to [ \begin{matrix}1 & 0 & 4 & 0 & −3 \\ 0 & 1 & −3 & 0 & 5 \\ 0 & 0 & 0 & 1 & −4 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} ].

Solution: A = [ \begin{matrix}1 & 2 & −2 & 0 & 7 \\ −2 & −3 & 1 & −1 & −5 \\ −3 & −4 & 0 & −2 & −3 \\ 3 & 6 & −6 & 5 & 1 \end{matrix} ] ∼ (r_{2} := r_{2}+2r_{1}) [ \begin{matrix}1 & 2 & −2 & 0 & 7 \\ 0 & 1 & −3 & −1 & 9 \\ −3 & −4 & 0 & −2 & −3 \\ 3 & 6 & −6 & 5 & 1 \end{matrix} ]

∼ (r_{3} := r_{3} + 3r_{1}) [ \begin{matrix}1 & 2 & −2 & 0 & 7 \\ 0 & 1 & −3 & −1 & 9 \\ 0 & 2 & −6 & −2 & 18 \\ 3 & 6 & −6 & 5 & 1 \end{matrix} ] ∼ (r_{4} := r_{4} − 3r_{1})

[ \begin{matrix}1 & 2 & −2 & 0 & 7 \\ 0 & 1 & −3 & −1 & 9 \\ 0 & 2 & −6 & −2 & 18 \\ 0 & 0 & 0 & 5 & −20 \end{matrix} ] ∼ (r_{3} := r_{3} −2r_{2}) [ \begin{matrix}1 & 2 & −2 & 0 & 7 \\ 0 & 1 & −3 & −1 & 9 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 5 & −20 \end{matrix} ]

∼ (r_{3} ↔ r_{4}, r_{3} := r_{3}/5) [ \begin{matrix}1 & 2 & −2 & 0 & 7 \\ 0 & 1 & −3 & −1 & 9 \\ 0 & 0 & 0 & 1 & −4 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} ] ∼ (r_{2} := r_{2} + r_{3})

[ \begin{matrix}1 & 2 & −2 & 0 & 7 \\ 0 & 1 & −3 & 0 & 5 \\ 0 & 0 & 0 & 1 & −4 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} ] ∼ (r_{1} := r_{1} − 2r_{2}) [ \begin{matrix}1 & 0 & 4 & 0 & −3 \\ 0 & 1 & −3 & 0 & 5 \\ 0 & 0 & 0 & 1 & −4 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} ]

2. Determine if the following systems are consistent and if so give all solutions in parametric vector form.

(a)

x_{1} −2x_{2} = 3
2x_{1} −7x_{2} = 0
−5x_{1} +8x_{2} = 5

Solution: The augmented matrix is [ \begin{matrix}1 & −2 & 3 \\ 2 & −7 & 0 \\ −5 & 8 & 5 \end{matrix} ] ∼ (r_{2} := r_{2} −2r_{1})

[ \begin{matrix}1 & −2 & 3 \\ 0 & −3 & −6 \\ −5 & 8 & 5 \end{matrix} ] ∼ (r_{3} := r_{3} + 5r_{1}) [ \begin{matrix}1 & −2 & 3 \\ 0 & −3 & −6 \\ 0 & −2 & 20 \end{matrix} ]

∼ (r_{2} := r_{2}/ − 3, r_{3} := r_{3}/ − 2) [ \begin{matrix}1 & −2 & 3 \\ 0 & 1 & 2 \\ 0 & 1 & −10 \end{matrix} ] ∼ (r_{3} := r_{3} − r_{2})

[ \begin{matrix}1 & −2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & −12 \end{matrix} ]. The last row implies that 0 · x_{2} = −12 which is impossible. So this system is inconsistent. (b)

\begin{aligned}x_{1} & +2x_{2} & −3x_{3} & +x_{4} & = 1 \\ −x_{1} & −2x_{2} & +4x_{3} & −x_{4} & = 6 \\ −2x_{1} & −4x_{2} & +7x_{3} & −x_{4} & = 1 \end{aligned}

The augmented matrix is [ \begin{matrix}1 & 2 & −3 & 1 & 1 \\ −1 & −2 & 4 & −1 & 6 \\ −2 & −4 & 7 & −1 & 1 \end{matrix} ] ∼ (r_{2} := r_{2} + r_{1})

[ \begin{matrix}1 & 2 & −3 & 1 & 1 \\ 0 & 0 & 1 & 0 & 7 \\ −2 & −4 & 7 & −1 & 1 \end{matrix} ] ∼ (r_{3} := r_{3} + 2r_{1}) [ \begin{matrix}1 & 2 & −3 & 1 & 1 \\ 0 & 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 1 & 3 \end{matrix} ]

∼ (r_{3} := r_{3}−r_{2}) [ \begin{matrix}1 & 2 & −3 & 1 & 1 \\ 0 & 0 & 1 & 0 & 7 \\ 0 & 0 & 0 & 1 & −4 \end{matrix} ] ∼ (r_{1} := r_{1}−r_{3}) [ \begin{matrix}1 & −2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & −12 \end{matrix} ]

∼ (r_{1} := r_{1}−r_{3}) [ \begin{matrix}1 & 2 & −3 & 0 & 5 \\ 0 & 0 & 1 & 0 & 7 \\ 0 & 0 & 0 & 1 & −4 \end{matrix} ]∼ (r_{1} := r_{1}+3r_{2}) [ \begin{matrix}1 & 2 & 0 & 0 & 26 \\ 0 & 0 & 1 & 0 & 7 \\ 0 & 0 & 0 & 1 & −4 \end{matrix} ].

So x_{2} is free. The solution is x_{1} = 26 − 2x_{2}, x_{3} = 7, x_{4} = −47. Its

parametric vector form is [\begin{matrix}x_{1} \\ x_{2} \\ x_{4} \\ x_{4} \end{matrix}] = [\begin{matrix}26 − 2x_{2} \\ x_{2} \\ 7 \\ −4 \end{matrix}] = [\begin{matrix}26 \\ 0 \\ 7 \\ −4 \end{matrix}] + x_{2} [\begin{matrix}−2 \\ 1 \\ 0 \\ 0 \end{matrix}].

3. Let A = [\begin{matrix}1 & 3 & −4 & 7 \\ 2 & 6 & 5 & 1 \\ 3 & 9 & 4 & 5 \end{matrix}].

(a) Find all the solutions of the non-homogeneous system Ax = b, and write them in parametric form, where b = [\begin{matrix}−1 \\ −2 \\ −3 \end{matrix}]. Solution: Consider the augmented matrix [A b] = [\begin{matrix}1 & 3 & −4 & 7 & −1 \\ 2 & 6 & 5 & 1 & −2 \\ 3 & 9 & 4 & 5 & −3 \end{matrix}]. Now we perform row operations on the augmented matrix.

[\begin{matrix}1 & 3 & −4 & 7 & −1 \\ 2 & 6 & 5 & 1 & −2 \\ 3 & 9 & 4 & 5 & −3 \end{matrix}] { \atop r_{2} := (−2)r_{1} + r_{2}, r_{3} := (−3)r_{1} + r_{3}} [\begin{matrix}1 & 3 & −4 & 7 & −1 \\ 0 & 0 & 13 & −13 & 0 \\ 0 & 0 & 16 & −16 & 0 \end{matrix}]
{ \atop r_{2} := \frac{1}{13}r_{2}, r_{3} := \frac{1}{16}r_{3}, r_{3} := r_{3} − r_{2}} [\begin{matrix}1 & 3 & −4 & 7 & −1 \\ 0 & 0 & 1 & −1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]
{ \atop r_{1} := 4r_{2} + r_{1}} [\begin{matrix}1 & 3 & 0 & 3 & −1 \\ 0 & 0 & 1 & −1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}]

So the solution is

\{ \begin{matrix}x_{1} + 3x_{2} + 3x_{4} & = −1 \\ x_{3} − x_{4} & = 0 \\ x_{2} and x_{4} are free. \end{matrix}  (1)

So

\{ \begin{matrix}x_{1} & = −1 − 3x_{2} − 3x_{4} \\ x_{2} & is free \\ x_{3} & = x_{4} \\ x_{4} & is free. \end{matrix}  (2)

Thus the solution of Ax = [\begin{matrix}−1 \\ −2 \\ −3 \end{matrix}] is

x = [\begin{matrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{matrix}] = [\begin{matrix}−1 − 3x_{2} − 3x_{4} \\ x_{2} \\ x_{4} \\ x_{4} \end{matrix}] = [\begin{matrix}−1 \\ 0 \\ 0 \\ 0 \end{matrix}] + x_{2} [\begin{matrix}−3 \\ 1 \\ 0 \\ 0 \end{matrix}] + x_{4} [\begin{matrix}−3 \\ 0 \\ 1 \\ 1 \end{matrix}].

where x_{2} and x_{4} are any numbers.

(b) Find all the solutions of the homogeneous system Ax = 0, and write them in parametric form. Solution:From previous example, we know that the solution of Ax = 0 is of the form

x = x_{2} [\begin{matrix}−3 \\ 1 \\ 0 \\ 0 \end{matrix}] + x_{4} [\begin{matrix}−3 \\ 0 \\ 1 \\ 1 \end{matrix}]

where x_{2} and x_{4} are any numbers.

(c) Are the columns of the matrix A linearly independent? Write down a linear relation between the columns of A if they are dependent. Solution: Since Ax = 0 has nontrivial solution, we know that the columns of the matrix A are linearly dependent. The solution is

x = x_{2} [\begin{matrix}−3 \\ 1 \\ 0 \\ 0 \end{matrix}] + x_{4} [\begin{matrix}−3 \\ 0 \\ 1 \\ 1 \end{matrix}]. Choosing x_{2} = 1 and x_{3} = 0, We have s x = [\begin{matrix}−3 \\ 1 \\ 0 \\ 0 \end{matrix}]. This implies that

−3 [\begin{matrix}1 \\ 2 \\ 3 \end{matrix}] + 1 · [\begin{matrix}3 \\ 6 \\ 9 \end{matrix}] + 0 · [\begin{matrix}−4 \\ 5 \\ 4 \end{matrix}] + 0 · [\begin{matrix}7 \\ 1 \\ 5 \end{matrix}] = [\begin{matrix}0 \\ 0 \\ 0 \end{matrix}].

4. Let S = Span\{ [ \begin{matrix}1 \\ −2 \\ 3 \\ 1 \end{matrix} ] , [ \begin{matrix}0 \\ 1 \\ 1 \\ −2 \end{matrix} ] , [ \begin{matrix}1 \\ −3 \\ 2 \\ 3 \end{matrix} ] , [ \begin{matrix}0 \\ 1 \\ 1 \\ −4 \end{matrix} ] \}.

(a) Find all the vectors u = [\begin{matrix}a \\ b \\ c \\ d \end{matrix}] such that the u is in S. Write these u in parametric form. Justify your answer. Solution: Note that u ∈ S iff the following system is consistent.

[ \begin{matrix}1 & 0 & 1 & 0 & a \\ −2 & 1 & −3 & 1 & b \\ 3 & 1 & 2 & 1 & c \\ 1 & −2 & 3 & −3 & d \end{matrix} ] ∼ (r_{2} := r_{2}+2r_{1}) [ \begin{matrix}1 & 0 & 1 & 0 & a \\ 0 & 1 & −1 & 1 & b + 2 a \\ 3 & 1 & 2 & 1 & c \\ 1 & −2 & 3 & −3 & d \end{matrix} ]

∼ (r_{3} := r_{3} − 3r_{1}) [ \begin{matrix}1 & 0 & 1 & 0 & a \\ 0 & 1 & −1 & 1 & b + 2 a \\ 0 & 1 & −1 & 1 & c − 3 a \\ 1 & −2 & 3 & −3 & d \end{matrix} ]

∼ (r_{4} := r_{4} − 3r_{1}) [ \begin{matrix}1 & 0 & 1 & 0 & a \\ 0 & 1 & −1 & 1 & b + 2 a \\ 0 & 1 & −1 & 1 & c − 3 a \\ 0 & −2 & 2 & −3 & d − a \end{matrix} ] ∼ (r_{3} := r_{3} − r_{2})

[ \begin{matrix}1 & 0 & 1 & 0 & a \\ 0 & 1 & −1 & 1 & b + 2 a \\ 0 & 0 & 0 & 0 & c − 5 a − b \\ 0 & −2 & 2 & −3 & d − a \end{matrix} ] ∼ (r_{4} := r_{4}+2r_{2}) [ \begin{matrix}1 & 0 & 1 & 0 & a \\ 0 & 1 & −1 & 1 & b + 2 a \\ 0 & 0 & 0 & 0 & c − 5 a − b \\ 0 & 0 & 0 & −1 & d + 3 a + 2 b \end{matrix} ]

∼ (r_{4} ↔ r_{3}, r_{3} := −r_{3}) [ \begin{matrix}1 & 0 & 1 & 0 & a \\ 0 & 1 & −1 & 1 & b + 2 a \\ 0 & 0 & 0 & 1 & −d − 3 a − 2 b \\ 0 & 0 & 0 & 0 & c − 5 a − b \end{matrix} ]. This sys-

tem is consistent if c − 5 a − b = 0. So c = 5a + b and u = [\begin{matrix}a \\ b \\ c \\ d \end{matrix}] =

[\begin{matrix}a \\ b \\ 5a + b \\ d \end{matrix}] = a [\begin{matrix}1 \\ 0 \\ 5 \\ 1 \end{matrix}] + b [\begin{matrix}0 \\ 1 \\ 1 \\ 0 \end{matrix}] + d [\begin{matrix}0 \\ 0 \\ 0 \\ 1 \end{matrix}].

(b) Is v = [ \begin{matrix}−1 \\ 3 \\ −2 \\ 1 \end{matrix} ] in S.

Solution: We have a = −1, b = 3 and c = −2. So c − 5a − b = −2 + 5 − 3 = 0 So v ∈ S.

(c) Is w = [ \begin{matrix}1 \\ 3 \\ −2 \\ 1 \end{matrix} ] in S.

Solution: We have a = 1, b = 3 and c = −2. So c−5a−b = −2−5−3 = −10 ≠ 0 So v is not in S.

5. Consider a linear system whose augmented matrix is of the form

[ \begin{matrix}1 & 1 & 1 & 2 \\ 1 & 2 & 1 & b \\ 3 & 5 & a & 1 \end{matrix} ]

(a) For what values of a will the system have a unique solution? What is the solution?(your answer may involve a and b)

(b) For what values of a and b will the system have infinitely many solutions?

(c) For what values of a and b will the system be inconsistent?

Answer:

[ \begin{matrix}1 & 1 & 1 & 2 \\ 1 & 2 & 1 & b \\ 3 & 5 & a & 1 \end{matrix} ] ∼ (r_{2} := r_{2} + (−1)r_{1}) [ \begin{matrix}1 & 1 & 1 & 2 \\ 0 & 1 & 0 & b − 2 \\ 3 & 5 & a & 1 \end{matrix} ]

∼ (r_{3} := r_{3} + (−3)r_{1}) [ \begin{matrix}1 & 1 & 1 & 2 \\ 0 & 1 & 0 & b − 2 \\ 0 & 2 & a − 3 & −5 \end{matrix} ]

∼ (r_{3} := r_{3} + (−2)r_{2}) [ \begin{matrix}1 & 1 & 1 & 2 \\ 0 & 1 & 0 & b − 2 \\ 0 & 0 & a − 3 & −1 − 2 b \end{matrix} ].

(a) If a − 3 ≠ 0 then previous augmented matrix is row equiva-

lent to ( r_{3} := \frac{1}{a−3}r_{3}) [ \begin{matrix}1 & 1 & 1 & 2 \\ 0 & 1 & 0 & b − 2 \\ 0 & 0 & 1 & \frac{−1−2 b}{a−3} \end{matrix} ] ∼ (r_{1} := r_{1} − r_{3})

[ \begin{matrix}1 & 1 & 0 & 2 + \frac{1+2 b}{a−3} \\ 0 & 1 & 0 & b − 2 \\ 0 & 0 & 1 & \frac{−1−2 b}{a−3} \end{matrix} ]

∼ (r_{1} := r_{1} − r_{2}) [ \begin{matrix}1 & 0 & 0 & 4 + \frac{1+2 b}{a−3} − b \\ 0 & 1 & 0 & b − 2 \\ 0 & 0 & 1 & \frac{−1−2 b}{a−3} \end{matrix} ]

The system will have a unique solution when a ≠ 3. The solution is x_{1} = 4 + \frac{1+2 b}{a−3} − b, x_{2} = b − 2 and x_{2} = \frac{−1−2 b}{a−3}.

(b) The system will have infinitely many solutions if a − 3 = 0 and −1 − 2b = 0, i.e a = 3 and b = −\frac{1}{2}.

(c) The system will be inconsistent if a − 3 = 0 and −1 − 2b ≠ 0, i.e a = 3 and b ≠ −\frac{1}{2}.

6. (a) [\begin{matrix}−2 & 1 \\ 4 & −2 \\ 0 & 0 \\ −6 & 3 \end{matrix}] = [v_{1} v_{2}]. We have v_{1} = −2v_{2}. So the set of column vectors is linearly dependent.

(b) [\begin{matrix}−2 & 1 \\ 4 & −2 \\ 2 & 2 \end{matrix}]. The first column vector is not a multiple of the second column vector. So the set of column vectors is linearly independent.

(c) [\begin{matrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{matrix}] { \atop r_{3} := r_{3} − r_{1}, r_{4} := r_{4} − r_{1}} [ \begin{matrix}1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & −1 & 1 \\ 0 & −1 & 0 \end{matrix} ]

∼ r_{2} := −r_{4}, r_{4} := r_{2} [ \begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & −1 & 1 \\ 0 & 1 & 1 \end{matrix} ] ∼ r_{3} := r_{3} + r_{2}, r_{4} := r_{4} − r_{2}

[ \begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \end{matrix} ] ∼ r_{4} := r_{4} − r_{3}, r_{1} := r_{1} − r_{2} [ \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} ]. So x_{1} = x_{2} =

x_{3} = 0. Thus the columns of the matrix is linearly independent. This matrix has three pivot vectors. So the columns of the matrix form a linearly independent set.

\begin{aligned}& [\begin{matrix}1 & 1 & 3 \\ 0 & 1 & 2 \\ 1 & −1 & −1 \\ 1 & 0 & 1 \end{matrix}] & ; \begin{matrix}r_{4} := r_{4} − r_{1} \\ r_{3} := r_{3} − r_{1} \end{matrix} & [ \begin{matrix}1 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & −2 & −4 \\ 0 & −1 & −2 \end{matrix} ] \\ ; \begin{matrix}r_{3} := r_{3} + 2r_{2} \\ r_{4} := r_{4} + r_{2} \end{matrix} & [ \begin{matrix}1 & 1 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ] & { \atop r_{1} := r_{1} − r_{2}} & [ \begin{matrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} ] \end{aligned}

This matrix has only two pivot vectors. So the columns of the matrix form a linearly dependent set.

(e)

The column vectors of

[\begin{matrix}−4 & −3 & 1 & 5 & 1 \\ 2 & −1 & 4 & −1 & 2 \\ 1 & 2 & 3 & 6 & −3 \\ 5 & 4 & 6 & −3 & 2 \end{matrix}]

form a dependent set since we have five column vectors in R^{4}. We will have at least one free variable for the solution of Ax = 0.

7. (a)

\begin{matrix}M = [\begin{matrix}1 & 1 & 2 \\ 1 & a + 1 & 3 \\ 1 & a & a + 1 \end{matrix}] { \atop r_{2} := r_{2} + (−1)r_{1}, r_{3} := r_{3} + (−1)r_{1}} [\begin{matrix}1 & 1 & 2 \\ 0 & a & 1 \\ 0 & a − 1 & a − 1 \end{matrix}] \\ { \atop r_{3} := \frac{1}{a − 1}r_{3} ifa − 1 ≠ 0, r_{2} ↔ r_{3}} & [\begin{matrix}1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & a & 1 \end{matrix}] & { \atop r_{3} := r_{3} + (−a)r_{2}} & [\begin{matrix}1 & 1 & 2 \\ 0 & 1 & 1 \\ 0 & 0 & 1 − a \end{matrix}] \end{matrix}

Thus the column vectors are independent if a ≠ 1.

(b) The column vectors are dependent if a = 1.

8. First, note that

\begin{matrix}T ([\begin{matrix}1 \\ 2 \\ 3 \end{matrix}]) = & T ([\begin{matrix}1 \\ 0 \\ 0 \end{matrix}] + 2 [\begin{matrix}0 \\ 1 \\ 0 \end{matrix}] + 3 [\begin{matrix}0 \\ 0 \\ 1 \end{matrix}] = T ([\begin{matrix}1 \\ 0 \\ 0 \end{matrix}]) + T (2 [\begin{matrix}0 \\ 1 \\ 0 \end{matrix}]) + T (3 [\begin{matrix}0 \\ 0 \\ 1 \end{matrix}]) \\ = & T ([\begin{matrix}1 \\ 0 \\ 0 \end{matrix}]) + 2T ([\begin{matrix}0 \\ 1 \\ 0 \end{matrix}]) + 3T ([\begin{matrix}0 \\ 0 \\ 1 \end{matrix}] = T (e_{1}) + 2T (e_{2}) + 3T (e_{3}). \end{matrix}

We need to find T (e_{1}), T (e_{2}) and T (e_{3}). Since T is linear, we have T (e_{1} + e_{2}) = T (e_{1}) + T (e_{2}), T (e_{1} − e_{2}) = T (e_{1})−T (e_{2}) and T (e_{1}+e_{2}+e_{3}) = T (e_{1})+T (e_{2})+T (e_{3}). The conditions

T (e_{1} + e_{2}) = [\begin{matrix}1 \\ −1 \end{matrix}] , T (e_{1} − e_{2}) = [\begin{matrix}2 \\ 3 \end{matrix}] and T (e_{1} + e_{2} + e_{3}) = [\begin{matrix}1 \\ −2 \end{matrix}] can

be written as

\{ \begin{matrix}T (e_{1}) + T (e_{2}) & = [\begin{matrix}1 \\ −1 \end{matrix}] \\ T (e_{1}) − T (e_{2}) & = [\begin{matrix}2 \\ 3 \end{matrix}] \\ T (e_{1}) + T (e_{2}) + T (e_{3}) & = [\begin{matrix}1 \\ −2 \end{matrix}] . \end{matrix}  (3)

Adding T (e_{1})+T (e_{2}) = [\begin{matrix}1 \\ −1 \end{matrix}] and T (e_{1})−T (e_{2}) = [\begin{matrix}2 \\ 3 \end{matrix}], we get 2T (e_{1}) = [\begin{matrix}3 \\ 2 \end{matrix}] and T (e_{1}) = [1^{\frac{3}{2}}]. Similarly, T (e_{1}) + T (e_{2}) − (T (e_{1}) − T (e_{2})) =

[\begin{matrix}1 \\ −1 \end{matrix}] − [\begin{matrix}2 \\ 3 \end{matrix}] = [\begin{matrix}−1 \\ −4 \end{matrix}]. So 2T (e_{2}) = [\begin{matrix}−1 \\ −4 \end{matrix}] and T (e_{2}) = [\begin{matrix}−\frac{1}{2} \\ −2 \end{matrix}]. From

T (e_{1})+T (e_{2})+T (e_{3})−(T (e_{1})+T (e_{2})) = [\begin{matrix}1 \\ −2 \end{matrix}]−[\begin{matrix}1 \\ −1 \end{matrix}], we get T (e_{3}) =

[\begin{matrix}0 \\ −1 \end{matrix}]. Hence T ([\begin{matrix}1 \\ 2 \\ 3 \end{matrix}]) = T (e_{1}) + 2T (e_{2}) + 3T (e_{3}) = [1^{\frac{3}{2}}] + 2 · [\begin{matrix}−\frac{1}{2} \\ −2 \end{matrix}] +

3 · [\begin{matrix}0 \\ −1 \end{matrix}] = [−6^{\frac{1}{2}}] .