MATHEMATICS 2 (MA10193) EXAMPLES SHEET 1: SOLUTIONS

1. We have

AA^{T} = ( \begin{matrix}cos w & p \\ sin w & q \end{matrix} ) ( \begin{matrix}cos w & sin w \\ p & q \end{matrix} ) = ( \begin{matrix}cos^{2} w + p^{2} & cos w sin w + pq \\ sin w cos w + qp & sin^{2} w + q^{2} \end{matrix} ) .

This is supposed to be equal to ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ), so cos^{2} w+p^{2} = 1 = cos^{2} w+sin^{2} w, sin^{2} w+q^{2} = 1, and cos w sin w + pq = 0. Therefore p = ± sin w, q = ± cos w and the signs have to be opposite. So we get two possible solutions: p = sin w, q = − cos w or p = − sin w, q = cos w.

2. We have

A = ( \begin{matrix}1 & 1 & 2 & −1 \\ 3 & 2 & 1 & −2 \\ 0 & 2 & 1 & 1 \end{matrix} ) and B = ( \begin{matrix}2 & 3 \\ −1 & 2 \\ 3 & 0 \\ 1 & 1 \end{matrix} ) .

so

A^{T} = ( \begin{matrix}1 & 3 & 0 \\ 1 & 2 & 2 \\ 2 & 1 & 1 \\ −1 & −2 & 1 \end{matrix} ) and B^{T} = ( \begin{matrix}2 & −1 & 3 & 1 \\ 3 & 2 & 0 & 1 \end{matrix} ) .

This gives

AB = ( \begin{matrix}6 & 4 \\ 5 & 11 \\ 2 & 5 \end{matrix} ) and B^{T} A^{T} = ( \begin{matrix}6 & 5 & 2 \\ 4 & 11 & 5 \end{matrix} ) .

3. (A + B)^{2} = A^{2} + AB + BA + B^{2} (not A^{2} + 2AB + B^{2}). If

A = ( \begin{matrix}2 & −1 \\ 3 & 2 \end{matrix} ) and B = ( \begin{matrix}2 & 3 \\ 1 & 1 \end{matrix} ) ,

then

A + B = ( \begin{matrix}4 & 2 \\ 4 & 3 \end{matrix} ) , (A + B)^{2} = ( \begin{matrix}4 & 2 \\ 4 & 3 \end{matrix} ) ( \begin{matrix}4 & 2 \\ 4 & 3 \end{matrix} ) = ( \begin{matrix}24 & 14 \\ 28 & 17 \end{matrix} ) ,

and

A^{2} = ( \begin{matrix}1 & −4 \\ 12 & 1 \end{matrix} ) , AB = ( \begin{matrix}3 & 5 \\ 8 & 11 \end{matrix} ) , BA = ( \begin{matrix}13 & 4 \\ 5 & 1 \end{matrix} ) , B^{2} = ( \begin{matrix}7 & 0 \\ 3 & 4 \end{matrix} )

and adding up all of these gives ( \begin{matrix}24 & 14 \\ 28 & 17 \end{matrix} ).

4. In this case, A and B are symmetric, so A^{T} = A and B^{T} = B. If AB is symmetric as well, then we have BA = B^{T} A^{T} = (AB)^{T} = AB, so it is true that AB = BA if A, B and AB are all symmetric. In this particular case we multiply the matrices and get

AB = ( \begin{matrix}3 − 3x + z & −6 + 2x & z − 3x + 1 \\ 3x − 6 + yz & 4 − 3x − 3y & xz − 6 + y \\ 3 − 3y + 3z & −12 + 2y & z − 3y + 3 \end{matrix} ) ,

so if this is symmetric we have

\begin{aligned}3x − 6 + yz & = −6 + 2x \\ z − 3x + 1 & = 3 − 3y + 3z \\ xz − 6 + y & = −12 + 2y. \end{aligned}

So x = −yz from the first equation, and y −xz = 6 from the third equation so y +yz^{2} = 6. Hence y = \frac{6}{z^{2}+1} and x = \frac{−6z}{z^{2}+1}. The second equation says 3x − 3y + 2z = −2, so we substitute the expressions we have fouund for x and y and then multiply by z^{2} + 1:

3 (\frac{−6z}{z^{2} + 1}) − 3 (\frac{6}{z^{2} + 1}) + 2z = −2

so

−18z − 18 + 2z(z^{2} + 1) = −2(z^{2} + 1),

i.e. 2z^{3}+2z^{2}−16z−16 = 0. One solution to this is z = −1: in fact it says (2(z+1)(z^{2}−8) = 0 so the solutions are z = −1 or z = ±\sqrt{8} = ±2\sqrt{2}. So there are three possible solutions: z = −1, x = 3 and y = 3; or z = 2\sqrt{2}, x = 4\sqrt{2}/3 and y = 2/3; or z = −2\sqrt{2}, x = −4\sqrt{2}/3 and y = 2/3.

5. We can write B = (B + B^{T})/2 + (B − B^{T})/2, i.e.

B = ( \begin{matrix}4 & \frac{3}{2} & \frac{1}{2} \\ \frac{3}{2} & −1 & \frac{5}{2} \\ \frac{1}{2} & \frac{5}{2} & 1 \end{matrix} ) + ( \begin{matrix}0 & \frac{−1}{2} & \frac{−5}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{5}{2} & \frac{−1}{2} & 0 \end{matrix} ) .

6. B^{2}, C^{2}, C^{T} AB and (AB)^{T} C are undefined. The others are

A^{2} = ( \begin{matrix}9 & 7 & 15 \\ 1 & 10 & 5 \\ 2 & 4 & 13 \end{matrix} ) ,  AA^{T} = ( \begin{matrix}30 & −1 & 13 \\ −1 & 6 & 4 \\ 13 & 4 & 13 \end{matrix} ) ,  BB^{T} = ( \begin{matrix}23 & 12 \\ 12 & 21 \end{matrix} ) ,

and

CC^{T} = ( \begin{matrix}4 & 2 & 6 \\ 2 & 1 & 3 \\ 6 & 3 & 9 \end{matrix} ) = C^{T} C.

7. AB = BA = I by multiplying them out. A^{T} B^{T} = (BA)^{T} = I^{T} = I. The first set of equations is

B ( \begin{matrix}x \\ y \\ z \end{matrix} ) = ( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} )

so the solution is

( \begin{matrix}x \\ y \\ z \end{matrix} ) = B^{−1} ( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} ) = A ( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} ) = ( \begin{matrix}7 \\ 9 \\ 30 \end{matrix} )

since B^{−1} = A. The second set is

B^{T} ( \begin{matrix}x \\ y \\ z \end{matrix} ) = ( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} )

and since B^{T}^{−1} = A^{T} the solution is

( \begin{matrix}x \\ y \\ z \end{matrix} ) = B^{T}^{−1} ( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} ) = A^{T} ( \begin{matrix}1 \\ 2 \\ 3 \end{matrix} ) = ( \begin{matrix}17 \\ 1 \\ 32 \end{matrix} )

GKS, 9/05/05