Let $A$ and $B$ be two square matrices of order $2\times 2$, where $\det(A)=1$ and $\det(B)=2$ then value of $\det(A+\alpha B)-\det(\alpha A+B)$

Question

Let $A$ and $B$ be two square matrices of order $2\times 2$, where $\det(A)=1$ and $\det(B)=2$ then value of $$\det(A+\alpha B)-\det(\alpha A+B)$$ where $\alpha \in \mathbb{R}$ is

(A)$\alpha^2$

(B)$0$

(C)$\alpha^2-1$

(D)None of these

My Attempt

Let A be the matrix \begin{bmatrix}1&0\\0&1\end{bmatrix} and $B$ be the matrix \begin{bmatrix}2&1\\2&2\end{bmatrix} Substituting in the expression $\det(A+\alpha B)-\det(\alpha A+B)$, I am able to get $\alpha^2-1$.

When I took $B$ to be the matrix \begin{bmatrix}2&0\\0&1\end{bmatrix} I again ended up getting $\alpha^2-1$.

But is there some general way to go about it.

Answer 1

Use the property for $2\times 2$ matrix $M, N$

$$|M+xN|=|M|+x^2|N|+x\left[\text{Tr}(M)\text{Tr}(N)-\text{Tr}(MN)\right]$$

We have: $$|A+\alpha B|=|A|+\alpha^2|B|+\alpha\left[\text{Tr}(A)\text{Tr}(B)-\text{Tr}(AB)\right]$$

$$|B+\alpha A|=|B|+\alpha^2|A|+\alpha\left[\text{Tr}(B)\text{Tr}(A)-\text{Tr}(BA)\right]$$

Note $\text{Tr}(AB)=\text{Tr}(BA)$, then we have:

$$|A+\alpha B|-|B+\alpha A|=|A|+\alpha^2|B|-|B|-\alpha^2|A|=\alpha^2-1$$

Answer 2

Let $\lambda$ and $\mu$ be the two eigenvalues of $X=A^{-1}B$. Then $\lambda\mu=\det X=2$ and $$ \begin{aligned} \det(A+\alpha B)-\det(\alpha A+B) &=\det(A)\left[\det(I+\alpha A^{-1}B)-\det(\alpha I+A^{-1}B)\right]\\ &=\det(I+\alpha X)-\det(\alpha I+X)\\ &=(1+\alpha\lambda)(1+\alpha\mu)-(\alpha+\lambda)(\alpha+\mu)\\ &=(1+\alpha\lambda+\alpha\mu+2\alpha^2)-(\alpha^2+\alpha\lambda+\alpha\mu+2)\\ &=\alpha^2-1.\\ \end{aligned} $$

Answer 3

Let $A=\begin{bmatrix}a &b\\c&d \end{bmatrix}$ and $B=\begin{bmatrix}e &f\\g&h \end{bmatrix}$. Since Det$(A)=1$, we have that $ad-bc=1$ and since Det$(B)=2$, we have that $eh-fg=2$. Now $$A+\alpha B=\begin{bmatrix}a+\alpha e & b+\alpha f\\c+\alpha g&d+\alpha h\end{bmatrix}, ~ \alpha A+B=\begin{bmatrix}\alpha a+e & \alpha b+f \\\alpha c+g&\alpha d+h\end{bmatrix}.$$ Therefore $$Det(A+\alpha B)-Det(\alpha A+B)\\=\left((a+\alpha e)(d+\alpha h)-(b+\alpha f)(c+\alpha g)\right)-\left((\alpha a+e)(\alpha d+h)-(\alpha b+f)(\alpha c+g)\right)\\ =((ad-bc)+\alpha^2(eh-fg))-(\alpha^2(ad-bc)+(eh-fg))\\ =(1+2\alpha^2)-(\alpha^2+2)=\alpha^2-1.$$