Answer:
a. False
b. False
c. True
d. True
e. False
f. False
Step-by-step explanation:
Hi,
We have certain properties for matrices, (where A and B are nxn matrices and I is the identity matrix) :
[tex]{(A^{-1})}^{-1} = A[/tex]
[tex](AB)^{-1} = B^{-1}A^{-1}[/tex]
[tex](A')^{-1} = (A^{-1})'[/tex]
[tex](A^{n})^{-1} = (A^{-1})^{n} = A^{-n}[/tex]
[tex]AA^{-1} = A^{-1}A = I[/tex]
[tex]AI = IA = A[/tex]
Using these properties, we verify the provided statements:
A. False.
None of the properties help verify this statement. We ca use an example for counter:
Let [tex]A =\left[\begin{array}{cc}1&2\\2&0\\\end{array}\right][/tex] and [tex]B = \left[\begin{array}{cc}5&1\\3&2\end{array}\right][/tex] , we calculate the L.H.S:
[tex]A+B = \left[\begin{array}{cc}1+5&2+1\\2+3&0+2\end{array}\right]\\= \left[\begin{array}{cc}6&3\\5&2\end{array}\right][/tex]
The square of (A+B):
[tex](A + B)^{2} = \left[\begin{array}{cc}36&9\\25&4\end{array}\right][/tex]
Lets calculate the R.H.S:
[tex]A^{2} =\left[\begin{array}{cc}1&4\\4&0\\\end{array}\right]\\B^{2} = \left[\begin{array}{cc}25&1\\9&4\end{array}\right]\\2AB = \left[\begin{array}{cc} (1 \times 5) + (2 \times 3) &(1 \times 1) + (2 \times 2)\\(2 \times 5) + (0 \times 3)& (2 \times 1) + (0 \times 2)\end{array}\right]\\= \left[\begin{array}{cc} 11 &5\\10& 2\end{array}\right][/tex]
[tex]A^{2} + B^{2} + 2AB = \left[\begin{array}{ccc}1+25+11&4+1+5 \\4+9+10&0+4+2\\\end{array}\right] \\= \left[\begin{array}{ccc}37&10 \\23&6\\\end{array}\right][/tex]
This proves that: L.H.S ≠ R.H.S
Hence, A is false.
B. False
This can only hold when the eigenvalues for A are real.
[tex]trace (A^{2}) > 0, det (A^{2}) > 0 : \\(A + A^{-1}) = ( I + A^{2} ) A^ {- 1} = ( A ( I + A ^{2} )^ {-1})^ {-1}[/tex]
C. True
This is a simplification of the distribution property of matrices.
D. True
The property that inverse is possible for any "n" value of the matrix.
E. False
Similar to part A, we can show that this property is invalid for any nxn matrix. Let:
[tex]A = \left[\begin{array}{cc}1&2\\0&1\end{array}\right] \\A^{-1} = \left[\begin{array}{cc}1&-2\\0&1\end{array}\right][/tex]
L.H.S:
[tex]A + A^{-1} = \left[\begin{array}{cc}1+1&2-2\\0+0&1+1\end{array}\right] = \left[\begin{array}{cc}2&0\\0&2\end{array}\right][/tex]
[tex](A + A^{-1})^{9} = \left[\begin{array}{cc}512&0\\0&512\end{array}\right][/tex]
R.H.S:
[tex]A^{9} = \left[\begin{array}{cc}1&512\\0&1\end{array}\right] \\A^{-9}= \left[\begin{array}{cc}1&-0.001953125\\0&1\end{array}\right]\\\\\\A^{9} + A^{-9} = \left[\begin{array}{cc}2&512\\0&2\end{array}\right] \\[/tex]
Since, L.H.S ≠ R.H.S, the statement is false.
F. False
This is a basic matrix rule, that commutative property does not apply on matrices.
