- #1
dcl
- 55
- 0
[tex]
c = \left[ {\begin{array}{*{20}c}
{2 - x} & 5 & 1 \\
{ - 3} & 0 & x \\
{ - 2} & 1 & 2 \\
\end{array}} \right][/tex]
a) Calculate det(C).
My answer was [tex]x^2 - 12x + 27[/tex].
b) Calculate det(2C).
Umm, would this just be 2*det(C)?
Couldn't find anything more helpful in my notes.
c) State the values for 'x' for which C is not invertible.
I believe the value for 'x' that would make this non invertable would be the solution that det(c) = 0. (A matrix has no inverse when the determinant = 0 yeh?)
which would be x = 9 or 3
is this correct?
c = \left[ {\begin{array}{*{20}c}
{2 - x} & 5 & 1 \\
{ - 3} & 0 & x \\
{ - 2} & 1 & 2 \\
\end{array}} \right][/tex]
a) Calculate det(C).
My answer was [tex]x^2 - 12x + 27[/tex].
b) Calculate det(2C).
Umm, would this just be 2*det(C)?
Couldn't find anything more helpful in my notes.
c) State the values for 'x' for which C is not invertible.
I believe the value for 'x' that would make this non invertable would be the solution that det(c) = 0. (A matrix has no inverse when the determinant = 0 yeh?)
which would be x = 9 or 3
is this correct?