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Section 3.1 #22: Write the given system of linear equations as a matrix equation of the form $A\vec{x}=\vec{b}$:
$$\left\{ \begin{array}{llll}
-x_1 & & + 2x_3 & = 1 \\
x_1 & -x_2 & & =-2 \\
& x_2 &+ x_3 &= -1
\end{array} \right\}$$
Solution: Let $A = \begin{bmatrix} -1 & 0 & 2 \\ 1 & -1 & 0 \\ 0 & 1 & 1 \end{bmatrix}$, $\vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}$, and $\vec{b}=\begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}$, giving us
$$\begin{bmatrix} -1 & 0 & 2 \\ 1 & -1 & 0 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 \\ -2 \\ -1 \end{bmatrix}.$$
Section 3.1 #36: Let $B = \left[ \begin{array}{ll}
\dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} \\
\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}}
\end{array} \right]$. Find, with justification, $B^{2015}$.
Solution: By matrix multiplication, we see that
$$B^2 = \left[ \begin{array}{ll}
\dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} \\
\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}}
\end{array} \right] \left[ \begin{array}{ll}
\dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} \\
\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}}
\end{array} \right] = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix},$$
$$B^4 = (B^2)^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix},$$
$$B^6 = B^4 B^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix},$$
$$B^7 = B^6 B = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \left[ \begin{array}{ll}
\dfrac{1}{\sqrt{2}} & -\dfrac{1}{\sqrt{2}} \\
\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}}
\end{array} \right] = \begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} \\ -\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} \end{bmatrix},$$
and
$$B^8 = (B^4)^2 = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=I_2.$$
Therefore, since $2015=8 \cdot 251 + 7$, we compute
$$B^{2015} = B^{8 \cdot 251 + 7} = (B^8)^{251} B^7 = (I_2)^{251} B^7 = B^7 = \begin{bmatrix} \dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} \\ -\dfrac{1}{\sqrt{2}} & \dfrac{1}{\sqrt{2}} \end{bmatrix}.$$
Section 3.2 #4: Solve for $X$, given that $A = \left[ \begin{array}{ll}
1 & 2 \\
3 & 4
\end{array} \right]$ and $B = \left[ \begin{array}{rr}
-1 & 0 \\
1 & 1
\end{array} \right]$:
$$2(A-B+X) = 3(X-A).$$
Solution: First distribute the scalars on each side:
$$2A - 2B + 2X = 3X - 3A.$$
To both sides add $3A-2X$ to arrive at
$$5A-2B=X.$$
Now finish the problem by calculating
$$\begin{array}{ll}
X &= 5A - 2B \\
&=5 \begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix} - 2 \begin{bmatrix}
-1 & 0 \\
1 & 1
\end{bmatrix} \\
&= \begin{bmatrix}
7 & 10 \\
13 & 18
\end{bmatrix}
\end{array}$$
Section 3.2 #6: Write $B = \left[ \begin{array}{rr}
2 & 3 \\
-4 & 2
\end{array} \right]$ as a linear combination of $A_1 = \left[ \begin{array}{rr}
1 & 0 \\
0 & 1
\end{array} \right]$, $A_2 = \left[ \begin{array}{rr}
0 & -1 \\
1 & 0
\end{array} \right]$, and $A_3 = \begin{bmatrix}
1 & 1 \\
0 & 1
\end{bmatrix}$ if possible.
Solution: Consider the matrix equation $c_1 A_1 + c_2 A_2 + c_3 A_3= B$ and use algebra to simplify the left-hand side to get
$$\begin{bmatrix} c_1+c_3 & -c_2+c_3 \\ c_2 & c_1+c_3 \end{bmatrix} = \begin{bmatrix} 2 & 3 \\ -4 & 2 \end{bmatrix}.$$
We solve this system by writing the associated augmented matrix and putting it into reduced echelon form: calculate
$$\begin{array}{ll}
\begin{bmatrix}
1 & 0 & 1 & 2 \\
0 & -1 & 1 & 3 \\
0 & 1 & 0 & -4 \\
1 & 0 & 1 & 2
\end{bmatrix} &\stackrel{r_4^*=r_4-r_1}{\sim} \begin{bmatrix}
1&0&1&2 \\
0&-1&1&3\\
0&1&0&-4\\
0&0&0&0
\end{bmatrix} \\
&\stackrel{r_3^*=r_3+r_2}{\sim} \begin{bmatrix}
1&0&1&2 \\
0&-1&1&3 \\
0&0&1&-1 \\
0&0&0&0
\end{bmatrix} \\
&\stackrel{r_2^*=r_2-r_3, r_1^*=r_1-r_3}{\sim} \begin{bmatrix}
1&0&0&3 \\
0&-1&0&4 \\
0&0&1&-1 \\
0&0&0&0
\end{bmatrix} \\
&\stackrel{r_2^*=-r_2}{\sim} \begin{bmatrix}
1&0&0&3 \\
0&1&0&-4 \\
0&0&1&-1 \\
0&0&0&0
\end{bmatrix}
\end{array}$$
This corresponds to the solution $\begin{bmatrix} c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 4 \\ -1 \end{bmatrix}$, proving we can write
$$B=3A_1-4A_2-A_3.$$
Section 3.2 #14: Determine whether the given set of matrices is linearly independent or not:
$$\left\{ \left[ \begin{array}{rr}
1 & 2 \\
4 & 3
\end{array} \right], \left[ \begin{array}{rr}
2 & 1 \\
-1 & 0
\end{array} \right], \left[ \begin{array}{rr}
1 & 1 \\
1 & 1
\end{array} \right] \right\}.$$
Solution: Consider the matrix equation
$$c_1 \left[ \begin{array}{rr}
1 & 2 \\
4 & 3
\end{array} \right] + c_2 \left[ \begin{array}{rr}
2 & 1 \\
-1 & 0
\end{array} \right] + c_3 \left[ \begin{array}{rr}
1 & 1 \\
1 & 1
\end{array} \right] = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$
Simplify the left-hand side to get the equation
$$\begin{bmatrix} c_1+2c_2+c_3 & 2c_1 + c_2 + c_3 \\ 4c_1 - c_2 + c_3 & 3c_1 + c_3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$
Of course this is a way to encode the following system of equations:
$$\left\{ \begin{array}{rrrr}
c_1 & + 2c_2 &+c_3 &= 0 \\
2c_1 & +c_2 &+c_3 &= 0 \\
4c_1 &-c_2 & +c_3 &= 0 \\
3c_1 & &+c_3 &= 0.
\end{array} \right.$$
To solve the system, we will find the reduced echelon form of the associated augmented matrix: compute
$$\begin{bmatrix}
1 & 2 & 1 & 0 \\
2 & 1 & 1 & 0 \\
4 & -1 & 1 & 0 \\
3 & 0 & 1 & 0
\end{bmatrix} \sim \ldots \sim \begin{bmatrix} 1 & 0 & \frac{1}{3} & 0 \\ 0 & 1 & \frac{1}{3} & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}.$$
This encodes the pair of equations $c_1 + \frac{1}{3}c_3 = 0$ and $c_2 + \frac{1}{3}c_3 = 0$ and so we see that $c_3$ may be regarded as a free variable. Choosing, e.g. $c_3=1$ yields $c_1=-\frac{1}{3}$ and $c_2=-\frac{1}{3}$, a nontrivial solution of the matrix equation. Since we have found a nontrivial solution, we must conclude that the matrices are not independent.
Section 3.2 #44(b): Prove that for any scalar $k$,
$$\mathrm{tr}(kA) = k\mathrm{tr}(A).$$
Solution: Suppose that $A \in \mathbb{F}^{n \times n}$ for some $n$ and some field $\mathbb{F}$. Represent $A$ by its entries: $A=\{a_{ij}\}$. Then due to the laws of scalar multiplication of matrices, $kA=\{k \cdot a_{ij}\}$. From this we see that
$$\mathrm{tr}(kA) = ka_{11} + ka_{22} + \ldots + ka_{nn}.$$
On the other hand,
$$k \mathrm{tr}(A) = k(a_{11} + a_{22} + \ldots + a_{nn})=ka_{11} + ka_{22} + \ldots + ka_{nn}.$$
It is clear that these are equal, therefore
$$\mathrm{tr}(kA) = k\mathrm{tr}(A).$$
Problem A: If $A \in \mathbb{Z}_5^{2 \times 2}$ and $B \in \mathbb{Z}_5^{2 \times 2}$ with
$$A = \left[ \begin{array}{ll}
1 & 2 \\
3 & 4
\end{array} \right]$$
and
$$B = \left[ \begin{array}{ll}
1 & 3 \\
0 & 1
\end{array} \right],$$
compute $AB, BA, A^TB^T, B^TA^T,$ and $(BA)^T$. Are any of these matrices related?
Solution: First calculate
$$AB = \left[ \begin{array}{ll}
1 & 2 \\
3 & 4
\end{array} \right] \left[ \begin{array}{ll}
1 & 3 \\
0 & 1
\end{array} \right] = \begin{bmatrix} 1 & 5 \\ 3 & 13 \end{bmatrix},$$
now calculate
$$BA = \left[ \begin{array}{ll}
1 & 3 \\
0 & 1
\end{array} \right] \left[ \begin{array}{ll}
1 & 2 \\
3 & 4
\end{array} \right] = \begin{bmatrix} 10 & 14 \\ 3 & 4 \end{bmatrix},$$
now calculate
$$A^TB^T = \left[ \begin{array}{ll}
1 & 2 \\
3 & 4
\end{array} \right]^T \left[ \begin{array}{ll}
1 & 3 \\
0 & 1
\end{array} \right]^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 3 \\ 14 &4 \end{bmatrix},$$
now calculate
$$B^T A^T = \left[ \begin{array}{ll}
1 & 3 \\
0 & 1
\end{array} \right]^T \left[ \begin{array}{ll}
1 & 2 \\
3 & 4
\end{array} \right]^T = \begin{bmatrix} 1&0 \\ 3&1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 2&4 \end{bmatrix} = \begin{bmatrix} 1 & 3 \\ 5 & 13 \end{bmatrix},$$
and finally calculate
$$(BA)^T = \begin{bmatrix} 10 & 14 \\ 3 &4 \end{bmatrix} = \begin{bmatrix} 10 & 3 \\ 14 & 4 \end{bmatrix}.$$
The matrix $A^TB^T$ equals the matrix $(BA)^T$. This is a general formula that always holds (when the oprations are defined)! Notice the order of multiplication switches when the transposes are applied. This is similar to the formula for the inverse of a product in group theory: see here.
Problem B: If $A, B, C \in \mathbb{C}^{2 \times 2}$ with
$$A = \left[ \begin{array}{ll}
1 & 1 \\
i & 0
\end{array} \right],$$
$$B = \left[ \begin{array}{ll}
0 & 2+i \\
1 & 1
\end{array} \right],$$
and
$$C = \left[ \begin{array}{ll}
-1 & 1+i \\
1-i & 1
\end{array} \right].$$
Use the definition of linear independence (i.e. the equation) and solve it to determine whether $\{A,B,C\}$ is linearly independent or linearly dependent set of matrices.
Solution: We must solve the equation $c_1 A + c_2 B + c_3 C = 0_{2 \times 2}$. Simplifying the left-hand side yields
$$\begin{bmatrix}
c_1-c_3 & c_1 + (2+i)c_2 + (1+i)c_3 \\
i c_1 + c_2 + (1-i)c_3 & c_2 + c_3
\end{bmatrix}= \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}.$$
This matrix equation encodes the system of equations
$$\left\{ \begin{array}{rrrr}
c_1 & &-c_3 &= 0 \\
c_1 &+(2+i)c_2 &+(1+i)c_3 &= 0 \\
ic_1 &+c_2 &+(1-i)c_3 &= 0 \\
& c_2 &+ c_3 &= 0.
\end{array} \right.$$
Now we solve this system by putting the associated augmented matrix into reduced echelon form: compute
$$\begin{array}{ll}
\begin{bmatrix}
1 & 0 & -1 & 0 \\
1 & 2+i & 1+i & 0 \\
i & 1 & 1-i & 0 \\
0 & 1 & 1 & 0
\end{bmatrix} &\stackrel{r_2^*=r_2-r_1, r_3^*=r_3-ir_1}{\sim} \begin{bmatrix}
1 & 0 & -1 & 0 \\
0 & 2+i & 2+i & 0 \\
0 & 1 & 1 & 0 \\
0 & 1 & 1 & 0
\end{bmatrix} \sim \begin{bmatrix} 1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 &0 &0 \\ 0 & 0 & 0 & 0 \end{bmatrix} \\
\end{array}$$
This encodes the solution $c_1 - c_3 = 0$ and $c_2+c_3=0$. Here $c_3$ may be regarded as a free variable with $c_1=c_3$ and $c_2=-c_3$. Therefore we may find a nontrivial solution by setting $c_3=1$ yielding $c_1=1$ and $c_2=-1$. Therefore the set of matrices is not independent.
