A = B = [ 0 0 4 3 2 2 1 1 1 ] A= B =
\left[ {\begin{array}{cc}
0 & 0 & 4 \\
3 & 2 & 2 \\
1 & 1 & 1
\end{array} } \right] A = B = ⎣ ⎡ 0 3 1 0 2 1 4 2 1 ⎦ ⎤ .
By expanding along 1st row, ∣ A ∣ = ∣ B ∣ = 4 ( 3 − 2 ) = 4 |A|=|B| = 4(3-2) = 4 ∣ A ∣ = ∣ B ∣ = 4 ( 3 − 2 ) = 4 .
Now, a d j ( A ) = a d j ( B ) = [ 0 − 1 1 4 − 4 0 − 8 12 0 ] T = [ 0 4 − 8 − 1 − 4 12 1 0 0 ] adj(A)=adj(B) =
\left[ {\begin{array}{cc}
0 & -1 & 1 \\
4 & -4 & 0 \\
-8 & 12 & 0 \\
\end{array} } \right]^T =
\left[ {\begin{array}{cc}
0 & 4& -8 \\
-1& -4 & 12 \\
1 & 0& 0 \\
\end{array} } \right] a d j ( A ) = a d j ( B ) = ⎣ ⎡ 0 4 − 8 − 1 − 4 12 1 0 0 ⎦ ⎤ T = ⎣ ⎡ 0 − 1 1 4 − 4 0 − 8 12 0 ⎦ ⎤ .
So,
A − 1 = B − 1 = 1 ∣ A ∣ a d j ( A ) = [ 0 1 − 2 − 1 / 4 − 1 3 1 / 4 0 0 ] A^{-1} =B^{-1} = \frac{1}{|A|} adj(A) = \left[ {\begin{array}{cc}
0 & 1 & -2\\
-1/4 & -1& 3\\
1/4 & 0& 0 \\
\end{array} } \right] A − 1 = B − 1 = ∣ A ∣ 1 a d j ( A ) = ⎣ ⎡ 0 − 1/4 1/4 1 − 1 0 − 2 3 0 ⎦ ⎤ .
a)
− A − 1 + 3 B T = − [ 0 1 − 2 − 1 / 4 − 1 3 1 / 4 0 0 ] + 3 [ 0 0 4 3 2 2 1 1 1 ] T = [ 0 − 1 2 1 / 4 1 − 3 − 1 / 4 0 0 ] + [ 0 9 3 0 6 3 12 6 3 ] = [ 0 8 5 1 / 4 7 0 47 / 4 6 3 ] -A^{-1} + 3 B^T = - \left[ {\begin{array}{cc}
0 & 1 & -2\\
-1/4 & -1& 3\\
1/4 & 0& 0 \\
\end{array} } \right]+ 3 \left[ {\begin{array}{cc}
0 & 0 & 4 \\
3 & 2 & 2 \\
1 & 1 & 1
\end{array} } \right]^T
\\
= \left[ {\begin{array}{cc}
0 & -1 & 2\\
1/4 & 1& -3\\
-1/4 & 0& 0 \\
\end{array} } \right] + \left[ {\begin{array}{cc}
0 & 9& 3\\
0& 6& 3\\
12 & 6 & 3
\end{array} } \right] = \left[ {\begin{array}{cc}
0 & 8& 5\\
1/4& 7& 0\\
47/4& 6 & 3
\end{array} } \right] − A − 1 + 3 B T = − ⎣ ⎡ 0 − 1/4 1/4 1 − 1 0 − 2 3 0 ⎦ ⎤ + 3 ⎣ ⎡ 0 3 1 0 2 1 4 2 1 ⎦ ⎤ T = ⎣ ⎡ 0 1/4 − 1/4 − 1 1 0 2 − 3 0 ⎦ ⎤ + ⎣ ⎡ 0 0 12 9 6 6 3 3 3 ⎦ ⎤ = ⎣ ⎡ 0 1/4 47/4 8 7 6 5 0 3 ⎦ ⎤
b)
A T + A − 1 = [ 0 0 4 3 2 2 1 1 1 ] + [ 0 1 − 2 − 1 / 4 − 1 3 1 / 4 0 0 ] = [ 0 1 2 11 / 4 1 5 5 / 4 1 1 ] A^T + A^{-1} = \left[ {\begin{array}{cc}
0 & 0& 4\\
3& 2& 2\\
1 & 1 & 1
\end{array} } \right] + \left[ {\begin{array}{cc}
0 & 1 & -2\\
-1/4 & -1& 3\\
1/4 & 0& 0 \\
\end{array} } \right] = \left[ {\begin{array}{cc}
0 & 1 & 2\\
11/4 & 1& 5\\
5/4 & 1& 1\\
\end{array} } \right] A T + A − 1 = ⎣ ⎡ 0 3 1 0 2 1 4 2 1 ⎦ ⎤ + ⎣ ⎡ 0 − 1/4 1/4 1 − 1 0 − 2 3 0 ⎦ ⎤ = ⎣ ⎡ 0 11/4 5/4 1 1 1 2 5 1 ⎦ ⎤
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