- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
3.11. $\quad$ Exercises
1. $\quad$ For each of the following statements about square matrices, give a proof or exhibit a counter example.
(a) $\quad$ $\det (A + B) = \det A + \det B.$
(b) $\quad$ $\det \{(A + B)^2\} = \{\det (A + B)\}^2$
(c) $\quad$ $\det \{(A + B)^2\} = \det (A^2 + 2AB + B^2)$
(d) $\quad$ $\det \{(A + B)^3\} = \det (A^3 + B^3).$
Solution. $\quad$
(a) $\quad$ $\det (A + B) = \det A + \det B.$
Counterexample. $\quad$ Let $A = I_n$ and $B = -I_n,$ where $I_n$ is the $n \times n$ identity matrix $n$ and $n$ is an even number. Then, $\det (A + B) = 0$ but $\det A + \det B = 2. \quad \blacksquare$
(b) $\quad$ $\det \{(A + B)^2\} = \{\det (A + B)\}^2$
Proof. $\quad$ By the product formula for determinants, we find that $$\det\{(A+B)^2\} = \det\{(A+B)(A+B)\} = \det (A+B) \det(A+B) = \{\det(A+B)\}^2. \quad \blacksquare$$
(c) $\quad$ $\det \{(A + B)^2\} = \det (A^2 + 2AB + B^2)$
Counterexample. $\quad$ Let $A = \begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix}1 & 0 \\ 1 & 1 \end{bmatrix}.$ We have
\begin{align*}
(A + B)^2 &=
\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}
\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \\
&= \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix} \\ \\
A^2 + 2AB + B^2 &=
\begin{bmatrix}
1 & 2 \\ 0 & 1
\end{bmatrix}
+
2 \begin{bmatrix}
2 & 1 \\ 1 & 1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 \\ 2 & 1
\end{bmatrix} \\
&= \begin{bmatrix}
6 & 4 \\ 4 & 4
\end{bmatrix}
\end{align*}
From this, we can see that $$\det \{(A + B)^2\} \neq \det (A^2 + 2AB + B^2). \quad \blacksquare$$
(d) $\quad$ $\det \{(A + B)^3\} = \det (A^3 + B^3).$
Counterexample. $\quad$ Let $A = \begin{bmatrix}1 & 1 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix}1 & 0 \\ 1 & 1 \end{bmatrix}.$ We have
\begin{align*}
(A + B)^3 &=
\begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}
\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \\
&= \begin{bmatrix} 14 & 13 \\ 13 & 14 \end{bmatrix} \\ \\
A^3 + B^3 &=
\begin{bmatrix}1 & 2 \\ 0 & 1\end{bmatrix}
\begin{bmatrix}
1 & 1 \\ 0 & 1
\end{bmatrix}
+
\begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix}
\begin{bmatrix}
1 & 0 \\ 1 & 1
\end{bmatrix} \\
&=
\begin{bmatrix}
1 & 3 \\ 0 & 1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 \\ 3 & 1
\end{bmatrix} \\
&=
\begin{bmatrix}
2 & 3 \\ 3 & 2
\end{bmatrix}
\end{align*}
As we can see, $\det\{(A + B)^3\} \neq \det(A^3 + B^3). \quad \blacksquare$