4)
Let
Z = ( a 11 a 12 . . . a 1 m a 21 a 22 . . . a 2 m ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 . . . a n m ) Z=\begin{pmatrix}
a_{11} & a_{12} & ... & a_{1m} \\
a_{21} & a_{22} & ... & a_{2m} \\
\vdots & \vdots & \vdots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nm} \\
\end{pmatrix} Z = ⎝ ⎛ a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ... ... ⋮ ... a 1 m a 2 m ⋮ a nm ⎠ ⎞ Then
3 Z = ( 3 a 11 3 a 12 . . . 3 a 1 m 3 a 21 3 a 22 . . . 3 a 2 m ⋮ ⋮ ⋮ ⋮ 3 a n 1 3 a n 2 . . . 3 a n m ) 3Z=\begin{pmatrix}
3a_{11} & 3a_{12} & ... & 3a_{1m} \\
3a_{21} & 3a_{22} & ... & 3a_{2m} \\
\vdots & \vdots & \vdots & \vdots \\
3a_{n1} & 3a_{n2} & ... & 3a_{nm} \\
\end{pmatrix} 3 Z = ⎝ ⎛ 3 a 11 3 a 21 ⋮ 3 a n 1 3 a 12 3 a 22 ⋮ 3 a n 2 ... ... ⋮ ... 3 a 1 m 3 a 2 m ⋮ 3 a nm ⎠ ⎞
Z + Z + Z = ( a 11 a 12 . . . a 1 m a 21 a 22 . . . a 2 m ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 . . . a n m ) Z+Z+Z=\begin{pmatrix}
a_{11} & a_{12} & ... & a_{1m} \\
a_{21} & a_{22} & ... & a_{2m} \\
\vdots & \vdots & \vdots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nm} \\
\end{pmatrix} Z + Z + Z = ⎝ ⎛ a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ... ... ⋮ ... a 1 m a 2 m ⋮ a nm ⎠ ⎞
+ ( a 11 a 12 . . . a 1 m a 21 a 22 . . . a 2 m ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 . . . a n m ) + ( a 11 a 12 . . . a 1 m a 21 a 22 . . . a 2 m ⋮ ⋮ ⋮ ⋮ a n 1 a n 2 . . . a n m ) +\begin{pmatrix}
a_{11} & a_{12} & ... & a_{1m} \\
a_{21} & a_{22} & ... & a_{2m} \\
\vdots & \vdots & \vdots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nm} \\
\end{pmatrix}+\begin{pmatrix}
a_{11} & a_{12} & ... & a_{1m} \\
a_{21} & a_{22} & ... & a_{2m} \\
\vdots & \vdots & \vdots & \vdots \\
a_{n1} & a_{n2} & ... & a_{nm} \\
\end{pmatrix} + ⎝ ⎛ a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ... ... ⋮ ... a 1 m a 2 m ⋮ a nm ⎠ ⎞ + ⎝ ⎛ a 11 a 21 ⋮ a n 1 a 12 a 22 ⋮ a n 2 ... ... ⋮ ... a 1 m a 2 m ⋮ a nm ⎠ ⎞
= ( a 11 + a 11 + a 11 a 12 + a 12 + a 12 . . . a 1 m + a 1 m + a 1 m a 21 + a 21 + a 21 a 22 + a 22 + a 22 . . . a 2 m + a 2 m + a 2 m ⋮ ⋮ ⋮ ⋮ a n 1 + a n 1 + a n 1 a n 2 + a n 2 + a n 2 . . . a n m + a n m + a n m ) =\begin{pmatrix}
a_{11}+a_{11}+a_{11} & a_{12}+a_{12}+a_{12} & ... & a_{1m}+a_{1m}+a_{1m} \\
a_{21}+a_{21}+a_{21} & a_{22}+a_{22}+a_{22} & ... & a_{2m} +a_{2m}+a_{2m}\\
\vdots & \vdots & \vdots & \vdots \\
a_{n1}+a_{n1}+a_{n1} & a_{n2}+a_{n2}+a_{n2} & ... & a_{nm}+a_{nm}+a_{nm} \\
\end{pmatrix} = ⎝ ⎛ a 11 + a 11 + a 11 a 21 + a 21 + a 21 ⋮ a n 1 + a n 1 + a n 1 a 12 + a 12 + a 12 a 22 + a 22 + a 22 ⋮ a n 2 + a n 2 + a n 2 ... ... ⋮ ... a 1 m + a 1 m + a 1 m a 2 m + a 2 m + a 2 m ⋮ a nm + a nm + a nm ⎠ ⎞
= ( 3 a 11 3 a 12 . . . 3 a 1 m 3 a 21 3 a 22 . . . 3 a 2 m ⋮ ⋮ ⋮ ⋮ 3 a n 1 3 a n 2 . . . 3 a n m ) = 3 Z =\begin{pmatrix}
3a_{11} & 3a_{12} & ... & 3a_{1m} \\
3a_{21} & 3a_{22} & ... & 3a_{2m} \\
\vdots & \vdots & \vdots & \vdots \\
3a_{n1} & 3a_{n2} & ... & 3a_{nm} \\
\end{pmatrix}=3Z = ⎝ ⎛ 3 a 11 3 a 21 ⋮ 3 a n 1 3 a 12 3 a 22 ⋮ 3 a n 2 ... ... ⋮ ... 3 a 1 m 3 a 2 m ⋮ 3 a nm ⎠ ⎞ = 3 Z
3 Z = Z + Z + Z 3Z = Z + Z + Z 3 Z = Z + Z + Z when Z Z Z is a matrix is True.
5)
X = ( 1 2 3 4 ) , E = ( a b ) X=\begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}, E=\begin{pmatrix}
a \\
b
\end{pmatrix} X = ( 1 3 2 4 ) , E = ( a b ) a)
X E = ( 1 2 3 4 ) ( a b ) = ( a + 2 b 3 a + 4 b ) XE=\begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}\begin{pmatrix}
a \\
b
\end{pmatrix}=\begin{pmatrix}
a +2b \\
3a+4b
\end{pmatrix} XE = ( 1 3 2 4 ) ( a b ) = ( a + 2 b 3 a + 4 b )
b)
The matrix E E E is 2 × 1 2\times1 2 × 1 matrix, the matrix X X X is 2 × 2 2\times2 2 × 2 matrix.
Since 1 ≠ 2 , 1\not=2, 1 = 2 , then
E X = ( a b ) ( 1 2 3 4 ) = d o e s n o t e x i s t EX=\begin{pmatrix}
a \\
b
\end{pmatrix}\begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}=does\ not\ exist EX = ( a b ) ( 1 3 2 4 ) = d oes n o t e x i s t c)
X T = ( 1 3 2 4 ) X^T=\begin{pmatrix}
1 & 3 \\
2 & 4
\end{pmatrix} X T = ( 1 2 3 4 )
X T X = ( 1 3 2 4 ) ( 1 2 3 4 ) = ( 1 + 9 2 + 12 2 + 12 4 + 16 ) X^TX=\begin{pmatrix}
1 & 3 \\
2 & 4
\end{pmatrix}\begin{pmatrix}
1 & 2 \\
3 & 4
\end{pmatrix}=\begin{pmatrix}
1+9 & 2+12 \\
2+12 & 4+16
\end{pmatrix} X T X = ( 1 2 3 4 ) ( 1 3 2 4 ) = ( 1 + 9 2 + 12 2 + 12 4 + 16 )
= ( 10 14 14 20 ) =\begin{pmatrix}
10 & 14 \\
14 & 20
\end{pmatrix} = ( 10 14 14 20 )
10.) Consider the linear equation
2 a + 3 b = 4 2a + 3b = 4 2 a + 3 b = 4 If ( a , b ) = ( 1 2 , 1 ) , (a, b)=(\dfrac{1}{2}, 1), ( a , b ) = ( 2 1 , 1 ) , then substitute
2 ( 1 2 ) + 3 ( 1 ) = 4 2(\dfrac{1}{2}) + 3(1) = 4 2 ( 2 1 ) + 3 ( 1 ) = 4
4 = 4 , T r u e 4=4, True 4 = 4 , T r u e Therefore ( a , b ) = ( 1 2 , 1 ) (a, b)=(\dfrac{1}{2}, 1) ( a , b ) = ( 2 1 , 1 ) is a solution to the equation 2 a + 3 b = 1. 2a+3b=1. 2 a + 3 b = 1.
11)
a) Two lines are parallel lines or skew lines.
b) Two lines are intersecting lines.
c) Two lines are coincident lines.
12) Since a homogeneous system always has a solution (the trivial solution), it can never be inconsistent.
Therefore a homogeneous linear system can have:
(a) A unique solution.
Or
(c) Infinite solutions.
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