Answer to Question #114086 in Linear Algebra for Cross

Question

Given: A= 004322111− B= 004322111− Find: a) –A^{-1}+ 3B^T b) B^{-1}+ ( A^T+A^{-1})

Answer 1

$A= B = \left[ {\begin{array}{cc} 0 & 0 & 4 \\ 3 & 2 & 2 \\ 1 & 1 & 1 \end{array} } \right]$ .

By expanding along 1st row, $|A|=|B| = 4(3-2) = 4$.

Now, $adj(A)=adj(B) = \left[ {\begin{array}{cc} 0 & -1 & 1 \\ 4 & -4 & 0 \\ -8 & 12 & 0 \\ \end{array} } \right]^T = \left[ {\begin{array}{cc} 0 & 4& -8 \\ -1& -4 & 12 \\ 1 & 0& 0 \\ \end{array} } \right]$ .

So,

$A^{-1} =B^{-1} = \frac{1}{|A|} adj(A) = \left[ {\begin{array}{cc} 0 & 1 & -2\\ -1/4 & -1& 3\\ 1/4 & 0& 0 \\ \end{array} } \right]$ .

a)

$-A^{-1} + 3 B^T = - \left[ {\begin{array}{cc} 0 & 1 & -2\\ -1/4 & -1& 3\\ 1/4 & 0& 0 \\ \end{array} } \right]+ 3 \left[ {\begin{array}{cc} 0 & 0 & 4 \\ 3 & 2 & 2 \\ 1 & 1 & 1 \end{array} } \right]^T \\ = \left[ {\begin{array}{cc} 0 & -1 & 2\\ 1/4 & 1& -3\\ -1/4 & 0& 0 \\ \end{array} } \right] + \left[ {\begin{array}{cc} 0 & 9& 3\\ 0& 6& 3\\ 12 & 6 & 3 \end{array} } \right] = \left[ {\begin{array}{cc} 0 & 8& 5\\ 1/4& 7& 0\\ 47/4& 6 & 3 \end{array} } \right]$

b)

$A^T + A^{-1} = \left[ {\begin{array}{cc} 0 & 0& 4\\ 3& 2& 2\\ 1 & 1 & 1 \end{array} } \right] + \left[ {\begin{array}{cc} 0 & 1 & -2\\ -1/4 & -1& 3\\ 1/4 & 0& 0 \\ \end{array} } \right] = \left[ {\begin{array}{cc} 0 & 1 & 2\\ 11/4 & 1& 5\\ 5/4 & 1& 1\\ \end{array} } \right]$