Problems 15-25
15
(a)) When
$$
\phantom{.} \
\mathbf{I} = \begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\ ,
$$
the matrix multiplication
$$
\phantom{.} \
\mathbf{Ix}
=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
x \begin{bmatrix}
1 \\ 0
\end{bmatrix}
+
y \begin{bmatrix}
0 \\ 1
\end{bmatrix}
=
\begin{bmatrix}
x + 0 \\
0 + y
\end{bmatrix}
=
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
\mathbf{b}
\ .
$$
\begin{array}{c c c}
\phantom{=} &
\mathbf{Ix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
& \phantom{=} \\[1em] = &
x \begin{bmatrix}
1 \\ 0
\end{bmatrix}
+
y \begin{bmatrix}
0 \\ 1
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
x + 0 \\
0 + y
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
x \\ y
\end{bmatrix}
& \phantom{=} \\[1em] = &
\phantom{.} \ \mathbf{b} \ .
\end{array}
(b) When
$$
\phantom{.} \
\mathbf{P} = \begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
\ ,
$$
the matrix multiplication
$$
\phantom{.} \
\mathbf{Px}
=
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
x \begin{bmatrix}
0 \\ 1
\end{bmatrix}
+
y \begin{bmatrix}
1 \\ 0
\end{bmatrix}
=
\begin{bmatrix}
0 + y \\
x + 0
\end{bmatrix}
=
\begin{bmatrix}
y \\ x
\end{bmatrix}
=
\mathbf{b}
\ .
$$
\begin{array}{c c c}
\phantom{=} &
\mathbf{Px}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
& \phantom{=} \\[1em] = &
x \begin{bmatrix}
0 \\ 1
\end{bmatrix}
+
y \begin{bmatrix}
1 \\ 0
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
0 + y \\
x + 0
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
x \\ y
\end{bmatrix}
& \phantom{=} \\[1em] = &
\phantom{.} \ \mathbf{b} \ .
\end{array}
16
(a) When
$$
\phantom{.} \
\mathbf{R} = \begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\ ,
$$
the matrix multiplication
$$
\mathbf{Rx}
=
\begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
x \begin{bmatrix}
\phantom{-}0 \\
-1
\end{bmatrix}
+
y \begin{bmatrix}
1 \\
0
\end{bmatrix}
=
\begin{bmatrix}
\phantom{-}0 + y \\
-x + 0
\end{bmatrix}
=
\begin{bmatrix}
y \\ -x
\end{bmatrix}
$$
$$
\begin{array}{c c c}
\phantom{=} &
\mathbf{Rx}
\phantom{=} \\[1em] = &
\begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
\phantom{=} \\[1em] = &
x \begin{bmatrix}
\phantom{-}0 \\
-1
\end{bmatrix}
y \begin{bmatrix}
1 \\
0
\end{bmatrix}
\phantom{=} \\[1em] = &
\begin{bmatrix}
\phantom{-}0 + y \\
-x + 0
\end{bmatrix}
\phantom{=} \\[1em] = &
\begin{bmatrix}
\phantom{-}y \\ -x
\end{bmatrix}
\ .
\end{array}
$$
As an example, let
$$
\mathbf{x}_1 = \begin{bmatrix}
1 \\ 1
\end{bmatrix}
, \quad
\mathbf{x}_2 = \begin{bmatrix}
-1 \\ 1
\end{bmatrix}
, \quad
\mathbf{x}_3 = \begin{bmatrix}
-1 \\ -1
\end{bmatrix}
, \quad
\mathbf{x}_4 = \begin{bmatrix}
1 \\ -1
\end{bmatrix}
, \quad
$$
\begin{array}{c}
\mathbf{x}_1 = \begin{bmatrix}
\phantom{-}1 \\ \phantom{-}1
\end{bmatrix}
, \quad \\[1em]
\mathbf{x}_2 = \begin{bmatrix}
-1 \\ \phantom{-}1
\end{bmatrix}
, \quad \\[1em]
\mathbf{x}_3 = \begin{bmatrix}
-1 \\ -1
\end{bmatrix}
, \quad \\[1em]
\mathbf{x}_4 = \begin{bmatrix}
\phantom{-}1 \\ -1
\end{bmatrix}
, \quad
\end{array}
then
$$
\begin{array}{c c c}
\mathbf{Rx}_1 & = & \begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
& = &
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
\ , \\[1em]
\mathbf{Rx}_2 & = & \begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
-1 \\ \phantom{-}1
\end{bmatrix}
& = &
\begin{bmatrix}
1 \\ 1
\end{bmatrix}
\ , \\[1em]
\mathbf{Rx}_3 & = & \begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
-1 \\ -1
\end{bmatrix}
& = &
\begin{bmatrix}
-1 \\ \phantom{-}1
\end{bmatrix}
\ , \\[1em]
\mathbf{Rx}_4 & = & \begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
\phantom{-}1 \\ -1
\end{bmatrix}
& = &
\begin{bmatrix}
1 \\ -1
\end{bmatrix}
\ .
\end{array}
$$
From where it’s also possible to see that
$$
\begin{array}{r c l}
\phantom{.} \ \mathbf{Rx}_1 & = & \mathbf{x}_4 \ , \\[0.5em]
\phantom{.} \ \mathbf{Rx}_2 & = & \mathbf{x}_1 \ , \\[0.5em]
\phantom{.} \ \mathbf{Rx}_3 & = & \mathbf{x}_2 \ , \\[0.5em]
\phantom{.} \ \mathbf{Rx}_4 & = & \mathbf{x}_3 \ .
\end{array}
$$
(b) A matrix that rotates a vector $\mathbf{x} = (x,y)^\top$ full $\pi$
radians is equal to matrix that rotates the $\mathbf{x}$ vector twice
$\pi/2$ radians. That is
$$
\phantom{.} \
\mathbf{R}(\mathbf{Rx})
=
\mathbf{R} \left( \begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
\right)
=
\mathbf{R}
\begin{bmatrix}
y \\ -x
\end{bmatrix}
=
\begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
y \\ -x
\end{bmatrix}
=
\begin{bmatrix}
-x \\ -y
\end{bmatrix}
\ .
$$
$$
\begin{array}{ c c c }
& \mathbf{R}(\mathbf{Rx})
& \phantom{=} \\[1em] = &
\mathbf{R} \left( \begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
\right)
& \phantom{=} \\[1em] = &
\mathbf{R}
\begin{bmatrix}
y \\ -x
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
\phantom{-}0 & 1 \\
-1 & 0
\end{bmatrix}
\begin{bmatrix}
y \\ -x
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
-x \\ -y
\end{bmatrix} \ .
\end{array}
$$
17
Find the matrix $\mathbf{P}$ that multiplies $[x,y,z]^\top$ to give
$[y,z,x]^\top$ is a kind of matrix that multiplies $x$ and $y$ by $0$
and $z$ by $1$ at the first column, $y$ and $z$ by $0$ and $y$ with $1$ at
the second column, and $x$ and $z$ by $0$ and $z$ by $1$ in the third
column.
Such a matrix is a matrix
$$
\phantom{.} \
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{bmatrix}
\ ,
$$
because
$$
\phantom{.} \
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
x \begin{bmatrix}
0 \\ 0 \\ 1
\end{bmatrix}
+
y \begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix}
+
z \begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}
=
\begin{bmatrix}
0x + \phantom{0}y + 0z \\
0x + 0y + \phantom{0}z \\
\phantom{0}x + 0y + 0z
\end{bmatrix}
=
\begin{bmatrix}
y \\ z \\ x
\end{bmatrix}
\ .
$$
$$
\begin{array}{ c c c }
&
\begin{bmatrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
& \\[1em] = &
x \begin{bmatrix}
0 \\ 0 \\ 1
\end{bmatrix}
+
y \begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix}
+
z \begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}
& \\[1em] = &
\begin{bmatrix}
0x + \phantom{0}y + 0z \\
0x + 0y + \phantom{0}z \\
\phantom{0}x + 0y + 0z
\end{bmatrix}
& \\[1em] = & \phantom{.} \
\begin{bmatrix}
y \\ z \\ x
\end{bmatrix} \ .
\end{array}
$$
A matrix $\mathbf{Q}$ that brings back the $[y,x,z]^\top$ to $[x,y,z]^\top$
is a matrix
$$
\phantom{.} \
\begin{bmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{bmatrix}
\ .
$$
because
$$
\phantom{.} \
\begin{bmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
y \\ z \\ x
\end{bmatrix}
=
y \begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}
+
z \begin{bmatrix}
0 \\ 0 \\ 1
\end{bmatrix}
+
x \begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix}
=
\begin{bmatrix}
0y + 0z + \phantom{0}x \\
\phantom{0}y + 0z + 0x \\
0y + \phantom{0}z + 0x
\end{bmatrix}
=
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
\ .
$$
$$
\begin{array}{c c c}
\phantom{=} &
\begin{bmatrix}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{bmatrix}
\begin{bmatrix}
y \\ z \\ x
\end{bmatrix}
& \\[1em] = &
y \begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix}
+
z \begin{bmatrix}
0 \\ 0 \\ 1
\end{bmatrix}
+
x \begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix}
& \\[1em] = &
\begin{bmatrix}
0y + 0z + \phantom{0}x \\
\phantom{0}y + 0z + 0x \\
0y + \phantom{0}z + 0x
\end{bmatrix}
& \\[1em] = & \phantom{.} \
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
\ .
\end{array}
$$
18
The situation is comparable to situation of two equations where
$$
(1 \cdot 3) + (0 \cdot 5) = 3
$$
and
$$
\phantom{.} \ ((-1) \cdot 3) + (1 \cdot 5) = 2 \ .
$$
Observing the coefficients in the above two equations, and rewriting the
computations in a matrix form leads to matrix multiplication
$$
\phantom{.} \
\begin{bmatrix}
\phantom{-}1 & 0 \\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
3 \\
5
\end{bmatrix}
=
3 \begin{bmatrix}
\phantom{-}1 \\
-1
\end{bmatrix}
+
5 \begin{bmatrix}
0 \\
1
\end{bmatrix}
=
\begin{bmatrix}
3(1) + 5(0) \\
3(-1) + 5(1) \\
\end{bmatrix}
=
\begin{bmatrix}
\phantom{-}3 + 0 \\
-3 + 5
\end{bmatrix}
=
\begin{bmatrix}
3 \\
2
\end{bmatrix}
\ .
$$
$$
\begin{array}{ c c c }
\phantom{=} &
\begin{bmatrix}
\phantom{-}1 & 0 \\
-1 & 1
\end{bmatrix}
\begin{bmatrix}
3 \\
5
\end{bmatrix}
& \\[1em] = &
3 \begin{bmatrix}
\phantom{-}1 \\
-1
\end{bmatrix}
+
5 \begin{bmatrix}
0 \\
1
\end{bmatrix}
& \\[1em] = &
\begin{bmatrix}
3(1) + 5(0) \\
3(-1) + 5(1) \\
\end{bmatrix}
& \\[1em] = &
\begin{bmatrix}
\phantom{-}3 + 0 \\
-3 + 5
\end{bmatrix}
& \\[1em] = & \phantom{.} \
\begin{bmatrix}
3 \\
2
\end{bmatrix}
\ .
\end{array}
$$
The situation is similar to the second equation. Written in anohter way,
the equations are
$$
\phantom{,} \ \phantom{-}1(3) + 0(5) + 0(7) = 3 \ ,
$$$$
\phantom{,} \ -1(3) + 1(5) + 0(7) = 3 \ ,
$$
and
$$
\phantom{,} \ 0(3) + 0(5) + 1(7) = 3 \ ,
$$
so the subtracting matrix computation becomes
$$
\phantom{.} \
\begin{bmatrix}
\phantom{-}1 & 0 & 0 \\
-1 & 1 & 0 \\
\phantom{-}0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
3 \\ 5 \\ 7
\end{bmatrix}
=
3 \begin{bmatrix}
\phantom{-}1 \\
-1 \\
0
\end{bmatrix}
+
5 \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}
+
7 \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}
=
\begin{bmatrix}
\phantom{-}3 + 0 + 0 \\
-3 + 5 + 0 \\
\phantom{-}0 + 0 + 7
\end{bmatrix}
=
\begin{bmatrix}
3 \\
2 \\
7
\end{bmatrix}
\ .
$$
$$
\begin{array}{ c c c }
\phantom{=} &
\begin{bmatrix}
\phantom{-}1 & 0 & 0 \\
-1 & 1 & 0 \\
\phantom{-}0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
3 \\ 5 \\ 7
\end{bmatrix}
\phantom{=} & \\[1em] = &
3 \begin{bmatrix}
\phantom{-}1 \\
-1 \\
0
\end{bmatrix}
+
5 \begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix}
+
7 \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}
\phantom{=} & \\[1em] = &
\begin{bmatrix}
\phantom{-}3 + 0 + 0 \\
-3 + 5 + 0 \\
\phantom{-}0 + 0 + 7
\end{bmatrix}
\phantom{=} & \\[1em] = & \phantom{.} \
\begin{bmatrix}
3 \\
2 \\
7
\end{bmatrix}
\ .
\end{array}
$$
19
When
$$
\phantom{,} \
\mathbf{E}^{-1} = \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix}
\ ,
$$
$$
\phantom{.} \
\mathbf{E}^{-1} \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
x + 0 + 0 \\
0 + y + 0 \\
x + 0 + z
\end{bmatrix}
=
\begin{bmatrix}
x \\ y \\ z + x
\end{bmatrix}
\ ,
$$
$$
\begin{array}{ c c c }
\phantom{=} &
\mathbf{E}^{-1} \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
x + 0 + 0 \\
0 + y + 0 \\
x + 0 + z
\end{bmatrix}
& \phantom{=} \\[1em] = & \phantom{,} \
\begin{bmatrix}
x \\ y \\ z + x
\end{bmatrix}
\ ,
\end{array}
$$
and when
$$
\phantom{,} \
\mathbf{E}^{-1} = \begin{bmatrix}
\phantom{-}1 & 0 & 0 \\
\phantom{-}0 & 1 & 0 \\
-1 & 0 & 1
\end{bmatrix}
\ ,
$$
$$
\phantom{.} \
\mathbf{E}^{-1} \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
\phantom{-}1 & 0 & 0 \\
\phantom{-}0 & 1 & 0 \\
-1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
=
\begin{bmatrix}
\phantom{-}x + 0 + 0 \\
\phantom{-}0 + y + 0 \\
-x + 0 + z
\end{bmatrix}
=
\begin{bmatrix}
x \\ y \\ z - x
\end{bmatrix}
\ .
$$
$$
\begin{array}{ c c c }
\phantom{=} &
\mathbf{E}^{-1} \begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
\phantom{-}1 & 0 & 0 \\
\phantom{-}0 & 1 & 0 \\
-1 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z
\end{bmatrix}
& \phantom{=} \\[1em] = &
\begin{bmatrix}
\phantom{-}x + 0 + 0 \\
\phantom{-}0 + y + 0 \\
-x + 0 + z
\end{bmatrix}
& \phantom{=} \\[1em] = & \phantom{,} \
\begin{bmatrix}
x \\ y \\ z + x
\end{bmatrix}
\ .
\end{array}
$$
The next multiplication brings the original vector back, because
$$
\phantom{.} \
\mathbf{E} \begin{bmatrix}
3 \\ 4 \\ 5
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
3 \\ 4 \\ 5
\end{bmatrix}
=
\begin{bmatrix}
3 \\
4 \\
5 + 3
\end{bmatrix}
=
\begin{bmatrix}
3 \\
4 \\
8
\end{bmatrix}
\ ,
$$
$$
\begin{array}{ c c c }
\phantom{=} &
\mathbf{E} \begin{bmatrix}
3 \\ 4 \\ 5
\end{bmatrix}
\phantom{=} \\[1em] = &
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
3 \\ 4 \\ 5
\end{bmatrix}
\phantom{=} \\[1em] = &
\begin{bmatrix}
3 \\
4 \\
5 + 3
\end{bmatrix}
\phantom{=} \\[1em] = & \phantom{.} \
\begin{bmatrix}
3 \\
4 \\
8
\end{bmatrix}
\ ,
\end{array}
$$
and
$$
\phantom{.} \
\mathbf{E}^{-1} \begin{bmatrix}
3 \\ 4 \\ 8
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
3 \\ 4 \\ 8
\end{bmatrix}
=
\begin{bmatrix}
3 \\
4 \\
8 - 3
\end{bmatrix}
=
\begin{bmatrix}
3 \\
4 \\
5
\end{bmatrix}
\ .
$$
$$
\begin{array}{ c c c }
\phantom{=} &
\mathbf{E}^{-1} \begin{bmatrix}
3 \\ 4 \\ 8
\end{bmatrix}
\phantom{=} \\[1em] = &
\begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
3 \\ 4 \\ 8
\end{bmatrix}
\phantom{=} \\[1em] = &
\begin{bmatrix}
3 \\
4 \\
8 + 3
\end{bmatrix}
\phantom{=} \\[1em] = & \phantom{.} \
\begin{bmatrix}
3 \\
4 \\
5
\end{bmatrix}
\ .
\end{array}
$$
20
The matrix multiplication
$$
\phantom{,} \
\mathbf{P}_1 \begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
x \\ y
\end{bmatrix}
=
\begin{bmatrix}
x \\ 0
\end{bmatrix}
\ ,
$$
which is a $[x,y]^{-t}$ vector projection onto the $x$-axis.
When
$$
\phantom{,} \
\mathbf{P}_2 = \begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}
\ ,
$$$$
\phantom{.} \
\mathbf{P}_2 \begin{bmatrix}
x \\ y
\end{bmatrix}
=
\begin{bmatrix}
0 \\ y
\end{bmatrix}
\ .
$$
which in turn is a $[x,y]^{-t}$ vector projection onto the $y$-axis.
Moreover,
$$
\phantom{.} \
\mathbf{P}_2 (\mathbf{P}_1 \mathbf{x})
=
\mathbf{P}_2 \begin{bmatrix}
x \\ 0
\end{bmatrix}
=
\mathbf{0}
\ ,
$$
i.e. $\mathbf{P}_2(\mathbf{P}_1 \mathbf{x})$ is equal to the zero matrix
$\mathbf{0}$.
21
–
22
The equation written as a linear combinarion is
$$
\phantom{.} \
\mathbf{Ax} = \begin{bmatrix}
1 & 4 & 5
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
=
x + 4y + 5 z
=
0
=
\mathbf{0}
\ .
$$
$$
\begin{array}{ c c c }
\phantom{=} & \mathbf{Ax} & \phantom{=}
\\[0.5em] = &
\begin{bmatrix}
1 & 4 & 5
\end{bmatrix}
\begin{bmatrix}
x \\ y \\ z
\end{bmatrix}
& \\[0.5em] = &
x + 4y + 5 z
& \\[1em] = &
0
& \\[1em] = &
\phantom{.} \ \mathbf{0} \ .
\end{array}
$$
23
The MATLAB/Octave command for the matrix $\mathbf{A}$ is
for the vector $\mathbf{x}$ the command is
and for the solution vector $\mathbf{b}$ the command is
A command that would test whether or not $\mathbf{Ax} = \mathbf{b}$ is
Running these command in the order of appearance says reveals that
$\mathbf{Ax} = \mathbf{b}$ is true.
24
The output of command
is
and the output of the command
is
Trying to run the command
produces an error, because trying to multpy a $3 \times 1$ and $3 \times 3$
is not defined in general.
25
The output of the command
is
and the output of the second command is
is