Find the det(-2A) and compare it to det(A) for A={1 1}{3 -1}
Solution:-
solving det(A)
we get
"\\left|\\begin{array}{cc}1 & 1\\\\3 & -1\\end{array}\\right| = \\left(1\\right)\\cdot \\left(-1\\right) - \\left(1\\right)\\cdot \\left(3\\right) = -4" ................(1)
we know the property of determinant
"det(xA)=" "x^ndet(A)" where n is the order .
here
"det(-2A)=" "(-2)^2det(A)"
"=4(-4)" ...............(from 1)
"=-16"
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