Question #114086

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


1
Expert's answer
2020-05-07T15:47:59-0400

A=B=[004322111]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(32)=4|A|=|B| = 4(3-2) = 4.

Now, adj(A)=adj(B)=[0114408120]T=[0481412100]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,

A1=B1=1Aadj(A)=[0121/4131/400]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)

A1+3BT=[0121/4131/400]+3[004322111]T=[0121/4131/400]+[0930631263]=[0851/47047/463]-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)

AT+A1=[004322111]+[0121/4131/400]=[01211/4155/411]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]




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