An algebra problem

Algebra Level 5

Let $D=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}$ and $P=\begin{pmatrix} 7 & 0 & 2 \\ 0 & 1 & 0 \\ 2 & 0 & 5 \end{pmatrix}$ .

Consider $A={ P }^{ -1 }DP$ . Find $\det({ A }^{ 2 }+A)$ .

Notation: $\det(\cdot)$ denotes the determinant function.

Source: Technological Institute of Aeronautics (ITA) admission exam

The answer is 144.

1 solution

Tom Engelsman
Dec 22, 2016

det $(A^2 + A)$ = det $[(P^{-1}DP)(P^{-1}DP) + P^{-1}DP];$

or det $(P^{-1}D^2P + P^{-1}DP);$

or det $(P^{-1}(D^2 + D)P);$

or det $(P^{-1}P) \cdot$ det $(D^2 + D);$

or det $(I)\cdot$ det $(D)\cdot$ det $(D + I);$

or $1^{3}\cdot(1\cdot2\cdot3) \cdot (2\cdot3\cdot4) = 6\cdot24 = \boxed{144}.$

Is Matrix multiplication distributive? I don't understand step no. 3 ..

Sriram Vudayagiri - 4 years, 5 months ago

That is correct, Sriram. The distributive law applies to matrices, as well as scalars! Scalars can actually be thought as a 1x1 matrix such as the example => a(b + c)d = (ab + ac)d = abd + acd. In this case, just equate a = P^-1, b = D^2, c = D, and d = P.........hope this helps!

tom engelsman - 4 years, 5 months ago

@Tom Engelsman Thank you very much!! ☺