Question

Linear algebra,


Can someone help me solve part (b) of this ?

https://www.dropbox.com/s/15xddpvdsa78wck/Screenshot%202016-03-06%2015.45.10.png?dl=0

Answer 1

@Kainui

Answer 2

by definition, the adjoint Matrix, is the Matrix whose element of place \((i,j)\), is:
\[\huge {\left( {Adj\left( A \right)} \right)_{i,j}} = {\left( { - 1} \right)^{i + j}}\det \left( {{A_{ji}}} \right)\]
wherein
\[\huge {{A_{ji}}}\]
is the Matrix \(A\) without its \(j-\)th row, and without its \(i-\)th column

Answer 3

b) starting from the equality, if I multiply both sides to the left by \(A^{-1}\), I get:
\[\Large \left( {{A^{ - 1}} \cdot A} \right) \cdot Adj\left( A \right) = \left( {\det A} \right)\left( {{A^{ - 1}} \cdot {I_n}} \right)\]
please simplify

Answer 4

part c)
let's suppose that does exist another Matrix \(B\), such that:
\[\huge AB = I\]
if I multiply the above equation, by \(A^{-1}\) to the left, I get:
\[\huge \left( {{A^{ - 1}} \cdot A} \right) \cdot B = {A^{ - 1}} \cdot I\]
please simplify, what can you conclude?
similarly, if we start from te equation:
\[\huge CA = I\]