Solutions to Linear Algebra Practice Problems 1

1. (a.)

[ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ −1 & 2 & −1 & 0 & 0 \\ 0 & −1 & 2 & −1 & 0 \\ 0 & 0 & −1 & 2 & 6 \end{matrix} ] { \atop r_{2} + \frac{1}{2}r_{1}} [ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ 0 & \frac{3}{2} & −1 & 0 & \frac{1}{2} \\ 0 & −1 & 2 & −1 & 0 \\ 0 & 0 & −1 & 2 & 6 \end{matrix} ]
˜\frac{2}{3}r_{2} [ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ 0 & 1 & −\frac{2}{3} & 0 & \frac{1}{3} \\ 0 & −1 & 2 & −1 & 0 \\ 0 & 0 & −1 & 2 & 6 \end{matrix} ]  { \atop r_{3} + r_{2}} [ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ 0 & 1 & −\frac{2}{3} & 0 & \frac{1}{2} \\ 0 & 0 & \frac{4}{3} & −1 & \frac{1}{3} \\ 0 & 0 & −1 & 2 & 6 \end{matrix} ]
˜\frac{3}{4}r_{3} [ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ 0 & 1 & −\frac{2}{3} & 0 & \frac{1}{2} \\ 0 & 0 & 1 & −\frac{3}{4} & \frac{1}{4} \\ 0 & 0 & −1 & 2 & 6 \end{matrix} ]  { \atop r_{4} + r_{3}} [ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ 0 & 1 & −\frac{2}{3} & 0 & \frac{1}{2} \\ 0 & 0 & 1 & −\frac{3}{4} & \frac{1}{4} \\ 0 & 0 & 0 & \frac{5}{4} & \frac{25}{4} \end{matrix} ]
˜\frac{5}{4}r_{4} [ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ 0 & 1 & −\frac{2}{3} & 0 & \frac{1}{2} \\ 0 & 0 & 1 & −\frac{3}{4} & \frac{1}{4} \\ 0 & 0 & 0 & 1 & 5 \end{matrix} ] { \atop r_{3} + \frac{3}{4}r_{4}} [ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ 0 & 1 & −\frac{2}{3} & 0 & \frac{1}{2} \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \end{matrix} ]
{ \atop r_{2} + \frac{3}{2}r_{3}} [ \begin{matrix}2 & −1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \end{matrix} ]  { \atop r_{1} + r_{2}} [ \begin{matrix}2 & 0 & 0 & 0 & 4 \\ 0 & 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \end{matrix} ]
˜\frac{1}{2}r_{1} [ \begin{matrix}1 & 0 & 0 & 0 & 2 \\ 0 & 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & 0 & 4 \\ 0 & 0 & 0 & 1 & 5 \end{matrix} ]

Thus (x_{1}, x_{2}, x_{3}, x_{4}) = (2, 3, 4, 5).

(b) The solution in part (a) implies that

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

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

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

So the reduced row echelon form of M is is [\begin{matrix}1 & 3 & 0 & 3 \\ 0 & 0 & 1 & −1 \\ 0 & 0 & 0 & 0 \end{matrix}]. The solu-

tion is x_{1} + 3x_{2} + 3x_{4} = 0 and x_{3} − x_{4} = 0. Thus x_{1} = −3x_{2} − 3x_{4}, x_{3} = x_{4}.

Any solution is of the form

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

3. (a)

\begin{aligned}& [ \begin{matrix}1 & 1 & 0 & 2 \\ 1 & 2 & 0 & 1 \\ 3 & 5 & a & b \end{matrix} ] & { \atop r_{2} + (−1)r_{1}, r_{3} + (−3)r_{1}} & [ \begin{matrix}1 & 1 & 0 & 2 \\ 0 & 1 & 0 & −1 \\ 0 & 2 & a & b − 6 \end{matrix} ] \\ { \atop r_{3} + (−2)r_{2}} & [ \begin{matrix}1 & 1 & 0 & 2 \\ 0 & 1 & 0 & −1 \\ 0 & 0 & a & b − 4 \end{matrix} ] & { \atop r_{1} + (−1)r_{2}} & [ \begin{matrix}1 & 0 & 0 & 3 \\ 0 & 1 & 0 & −1 \\ 0 & 0 & a & b − 4 \end{matrix} ] \end{aligned}

The system will have a unique solution when a ≠ 0. The solution is (3, −1, \frac{b−4}{a}).

(b) The system will have infinitely many solutions if a = 0 and b = 4.

(c) The system will be inconsistent if a = 0 and b ≠ 4 4.

A = [\begin{matrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \\ 3 & 4 & 1 & 2 \\ 4 & 1 & 2 & 3 \end{matrix}] and B = [\begin{matrix}1 & 1 & −1 \\ 1 & −1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}] .

AB = [\begin{matrix}1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 1 \\ 3 & 4 & 1 & 2 \\ 4 & 1 & 2 & 3 \end{matrix}] [\begin{matrix}1 & 1 & −1 \\ 1 & −1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}] = [\begin{matrix}10 & 6 & 10 \\ 10 & 4 & 10 \\ 10 & 2 & 10 \\ 10 & 8 & 10 \end{matrix}] .

B^{T} = [\begin{matrix}1 & 1 & 1 & 1 \\ 1 & −1 & 1 & 1 \\ −1 & 1 & 1 & 1 \end{matrix}] .

B^{T} A^{T} = (AB)^{T} = ([\begin{matrix}10 & 6 & 10 \\ 10 & 4 & 10 \\ 10 & 2 & 10 \\ 10 & 8 & 10 \end{matrix}])^{T} = [\begin{matrix}10 & 10 & 10 & 10 \\ 6 & 4 & 2 & 8 \\ 10 & 10 & 10 & 10 \end{matrix}] .

5.

[\begin{matrix}0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 1 \\ 2 & 0 & 0 & 0 \end{matrix}] { \atop move the last row to the first row} [\begin{matrix}2 & 0 & 0 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 1 & 4 \\ 0 & 0 & 0 & 1 \end{matrix}]

This matrix has four pivot vectors. So the columns of the matrix form a linearly independent set.

[\begin{matrix}−4 & −3 & 0 \\ 0 & −1 & 4 \\ 1 & 0 & 3 \\ 5 & 4 & 6 \end{matrix}] { \atop interchange first and third row} [\begin{matrix}1 & 0 & 3 \\ 0 & −1 & 4 \\ −4 & −3 & 0 \\ 5 & 4 & 6 \end{matrix}]

{ \atop r_{3} + 4r_{1}, r_{4} + (−5)r_{1}} [\begin{matrix}1 & 0 & 3 \\ 0 & −1 & 4 \\ 0 & −3 & 12 \\ 0 & 4 & −9 \end{matrix}]  { \atop (−1)r_{2}} [\begin{matrix}1 & 0 & 3 \\ 0 & 1 & −4 \\ 0 & −3 & 12 \\ 0 & 4 & −9 \end{matrix}]

{ \atop r_{3} + 3r_{2}, r_{4} + (−4)r_{2}} [\begin{matrix}1 & 0 & 3 \\ 0 & 1 & −4 \\ 0 & 0 & 0 \\ 0 & 0 & 7 \end{matrix}] { \atop interchange 3rd and 4th row, \frac{1}{7}r_{4}} [\begin{matrix}1 & 0 & 3 \\ 0 & 1 & −4 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}]

This matrix has three pivot vectors. So the columns of the matrix form a linearly independent set.

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}.

6.

[\begin{matrix}0 & 1 & 0 & t \\ −1 & 0 & t & 0 \\ 0 & −t & 0 & 1 \\ −t & 0 & −1 & 0 \end{matrix}] { \atop interchange 1st and 2nd row, (−1)r_{1}} [\begin{matrix}1 & 0 & −t & 0 \\ 0 & 1 & 0 & t \\ 0 & −t & 0 & 1 \\ −t & 0 & −1 & 0 \end{matrix}]

{ \atop r_{4} + t · r_{1}} [\begin{matrix}1 & 0 & −t & 0 \\ 0 & 1 & 0 & t \\ 0 & −t & 0 & 1 \\ 0 & 0 & −1 − t^{2} & 0 \end{matrix}] { \atop r_{3} + t · r_{2}} [\begin{matrix}1 & 0 & −t & 0 \\ 0 & 1 & 0 & t \\ 0 & 0 & 0 & 1 + t^{2} \\ 0 & 0 & −1 − t^{2} & 0 \end{matrix}]

{ \atop \frac{1}{1 + t^{2}} r_{3}, \frac{−1}{1 + t^{2}} r_{4}, r_{3} ↔ r_{4}} [\begin{matrix}1 & 0 & −t & 0 \\ 0 & 1 & 0 & t \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]

M has four pivot vectors so M is invertible. Note that we have used the fact that 1 + t^{2} ≠ 0 in the computation.

\begin{matrix}A = [\begin{matrix}1 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & −1 & 0 \\ 1 & 1 & 1 & 1 \end{matrix}] { \atop r_{2} + (−1)r_{1}, r_{4} + (−1)r_{1}} [\begin{matrix}1 & 1 & 0 & 0 \\ 0 & −1 & 0 & 0 \\ 0 & 1 & −1 & 0 \\ 0 & 0 & 1 & 1 \end{matrix}] \\ { \atop −(1)r_{2}, r_{3} + (−1)r_{2}} [\begin{matrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & −1 & 0 \\ 0 & 0 & 1 & 1 \end{matrix}] { \atop −(1)r_{3}, r_{4} + (−1)r_{3}} [\begin{matrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}] . \end{matrix}

A has four pivot vectors so A is invertible.

B = [\begin{matrix}1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{matrix}] [\begin{matrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix}]. B has four pivot vectors so B is

invertible.

7. (a) Let A = [\begin{matrix}3 & 7 & −3 \\ 3 & 8 & −1 \\ −6 & −14 & 7 \end{matrix}].

[ \begin{matrix}3 & 7 & −3 & 1 & 0 & 0 \\ 3 & 8 & −1 & 0 & 1 & 0 \\ −6 & −14 & 7 & 0 & 0 & 1 \end{matrix} ] { \atop r_{2} + (−1)r_{1}, r_{3} + 2r_{1}} [ \begin{matrix}3 & 7 & −3 & 1 & 0 & 0 \\ 0 & 1 & 2 & −1 & 1 & 0 \\ 0 & 0 & 1 & 2 & 0 & 1 \end{matrix} ]

{ \atop r_{2} + (−2)r_{3}, r_{1} + 3r_{3}} [ \begin{matrix}3 & 7 & 0 & 7 & 0 & 3 \\ 0 & 1 & 0 & −5 & 1 & −2 \\ 0 & 0 & 1 & 2 & 0 & 1 \end{matrix} ] { \atop r_{1} + (−7)r_{2}} [ \begin{matrix}3 & 0 & 0 & 42 & −7 & 17 \\ 0 & 1 & 0 & −5 & 1 & −2 \\ 0 & 0 & 1 & 2 & 0 & 1 \end{matrix} ]

˜\frac{1}{3}r_{1} [ \begin{matrix}1 & 0 & 0 & 14 & \frac{−7}{3} & \frac{17}{3} \\ 0 & 1 & 0 & −5 & 1 & −2 \\ 0 & 0 & 1 & 2 & 0 & 1 \end{matrix} ]

So A^{−1} = [ \begin{matrix}14 & \frac{−7}{3} & \frac{17}{3} \\ −5 & 1 & −2 \\ 2 & 0 & 1 \end{matrix} ] .

(b)Since A^{−1} exists, the equation Ax = b has a solution x = A^{−1}b. So the vector b = [\begin{matrix}1 \\ 2 \\ 3 \end{matrix}] lies in the span of the column vectors of A.

8. (a)

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

Thus M is invertible if a ≠ 1.

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

9. 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}

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

and T (e_{1}) = [1^{\frac{3}{2}}]. Similarly, T (e_{1} + e_{2}) − T (e_{1} − e_{2}) = [\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} +e_{2} +e_{3})−T (e_{1} +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}}] .