A = ( 3 0 − 1 2 1 1 ) , B = ( 1 5 4 − 1 0 2 ) , C = ( 2 0 1 1 4 2 3 1 5 ) , D = ( − 1 0 1 3 2 4 ) , E = ( − 1 1 2 4 1 3 ) A=\begin{pmatrix}
3 & 0 \\
-1 & 2\\
1&1
\end{pmatrix}, B=\begin{pmatrix}
1 & 5 \\
4 & -1\\
0&2
\end{pmatrix}, \\
C=\begin{pmatrix}
2 & 0&1 \\
1 & 4&2\\
3&1&5
\end{pmatrix}, D=\begin{pmatrix}
-1 &0&1 \\
3 & 2&4
\end{pmatrix}, \\
E=\begin{pmatrix}
-1 &1&2 \\
4 & 1&3
\end{pmatrix} A = ⎝ ⎛ 3 − 1 1 0 2 1 ⎠ ⎞ , B = ⎝ ⎛ 1 4 0 5 − 1 2 ⎠ ⎞ , C = ⎝ ⎛ 2 1 3 0 4 1 1 2 5 ⎠ ⎞ , D = ( − 1 3 0 2 1 4 ) , E = ( − 1 4 1 1 2 3 )
− 4 C 2 = − 4 ( 2 0 1 1 4 2 3 1 5 ) ⋅ ( 2 0 1 1 4 2 3 1 5 ) = = − 4 ( 4 + 0 + 3 0 + 0 + 1 2 + 0 + 5 2 + 4 + 6 0 + 16 + 2 1 + 8 + 10 6 + 1 + 15 0 + 4 + 5 3 + 2 + 25 ) = = − 4 ( 7 1 7 12 18 19 22 9 30 ) = = ( − 28 − 4 − 28 − 48 − 72 − 76 − 88 − 36 − 120 ) -4C^2=-4\begin{pmatrix}
2 & 0&1 \\
1 & 4&2\\
3&1&5
\end{pmatrix}\cdot\begin{pmatrix}
2 & 0&1 \\
1 & 4&2\\
3&1&5
\end{pmatrix}=\\
=-4\begin{pmatrix}
4+0+3 & 0+0+1&2+0+5 \\
2+4+6 & 0+16+2&1+8+10\\
6+1+15&0+4+5&3+2+25
\end{pmatrix}=\\
=-4\begin{pmatrix}
7 & 1&7 \\
12 & 18&19\\
22&9&30
\end{pmatrix}=\\
=\begin{pmatrix}
- 28 & -4&-28 \\
-48 & -72&-76\\
-88&-36&-120
\end{pmatrix} − 4 C 2 = − 4 ⎝ ⎛ 2 1 3 0 4 1 1 2 5 ⎠ ⎞ ⋅ ⎝ ⎛ 2 1 3 0 4 1 1 2 5 ⎠ ⎞ = = − 4 ⎝ ⎛ 4 + 0 + 3 2 + 4 + 6 6 + 1 + 15 0 + 0 + 1 0 + 16 + 2 0 + 4 + 5 2 + 0 + 5 1 + 8 + 10 3 + 2 + 25 ⎠ ⎞ = = − 4 ⎝ ⎛ 7 12 22 1 18 9 7 19 30 ⎠ ⎞ = = ⎝ ⎛ − 28 − 48 − 88 − 4 − 72 − 36 − 28 − 76 − 120 ⎠ ⎞
2.
( E − D ) T = = ( ( − 1 1 2 4 1 3 ) − ( − 1 0 1 3 2 4 ) ) T = = ( ( 0 1 1 1 − 1 − 1 ) ) T = = ( 0 1 1 − 1 1 − 1 ) (E-D)^T=\\=\left(\begin{pmatrix}
-1 &1&2 \\
4 & 1&3
\end{pmatrix}-\begin{pmatrix}
-1 &0&1 \\
3 & 2&4
\end{pmatrix}\right)^T=\\
=\left(\begin{pmatrix}
0 &1&1 \\
1 & -1&-1
\end{pmatrix}\right)^T=\\
=\begin{pmatrix}
0 & 1 \\
1& -1\\
1&-1
\end{pmatrix} ( E − D ) T = = ( ( − 1 4 1 1 2 3 ) − ( − 1 3 0 2 1 4 ) ) T = = ( ( 0 1 1 − 1 1 − 1 ) ) T = = ⎝ ⎛ 0 1 1 1 − 1 − 1 ⎠ ⎞
3.
A ( B C ) A(BC) A ( BC )
B : 3 × 2 , C : 3 × 3 B:3\times2,
C:3\times3\\ B : 3 × 2 , C : 3 × 3
The product B C BC BC does not exist.
4.
( B T B ) C (B^TB)C ( B T B ) C
B T = ( 1 4 0 5 − 1 2 ) , B = ( 1 5 4 − 1 0 2 ) B T ⋅ B = ( 1 4 0 5 − 1 2 ) ⋅ ( 1 5 4 − 1 0 2 ) = = ( 1 + 16 + 0 5 − 4 + 0 5 − 4 + 0 25 + 1 + 4 ) = ( 17 1 1 30 ) B^T=\begin{pmatrix}
1 & 4&0 \\
5 & -1&2
\end{pmatrix},B=\begin{pmatrix}
1 & 5 \\
4 & -1\\
0&2
\end{pmatrix}\\
B^T\cdot B=\begin{pmatrix}
1 & 4&0 \\
5 & -1&2
\end{pmatrix}\cdot\begin{pmatrix}
1 & 5 \\
4 & -1\\
0&2
\end{pmatrix}=\\
=
\begin{pmatrix}
1+16+0 & 5-4+0 \\
5-4+0 & 25+1+4
\end{pmatrix}=\begin{pmatrix}
17 & 1 \\
1 & 30
\end{pmatrix} B T = ( 1 5 4 − 1 0 2 ) , B = ⎝ ⎛ 1 4 0 5 − 1 2 ⎠ ⎞ B T ⋅ B = ( 1 5 4 − 1 0 2 ) ⋅ ⎝ ⎛ 1 4 0 5 − 1 2 ⎠ ⎞ = = ( 1 + 16 + 0 5 − 4 + 0 5 − 4 + 0 25 + 1 + 4 ) = ( 17 1 1 30 )
B T ⋅ B : 2 × 2 , C : 3 × 3 B^T\cdot B:2\times2, C:3\times3 B T ⋅ B : 2 × 2 , C : 3 × 3
The product ( B T B ) C (B^TB)C ( B T B ) C does not exist.
5.
3 B − A B 3 B = 3 ( 1 5 4 − 1 0 2 ) = ( 3 15 12 − 3 0 6 ) 3B-AB\\
3B=3\begin{pmatrix}
1 & 5 \\
4 & -1\\
0&2
\end{pmatrix}=\begin{pmatrix}
3 & 15 \\
12 & -3\\
0&6
\end{pmatrix} 3 B − A B 3 B = 3 ⎝ ⎛ 1 4 0 5 − 1 2 ⎠ ⎞ = ⎝ ⎛ 3 12 0 15 − 3 6 ⎠ ⎞
A : 3 × 2 , B : 3 × 2 A:3\times2, B:3\times2 A : 3 × 2 , B : 3 × 2
The product A B AB A B does not exist.
Then 3B-AB does not exist.
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