- 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
4. $\quad$ State and prove a generalization of Exercise 3 for $n \times n$ matrices.
Solution. $\quad$ For any $n \times n$ matrix of the form: $ \begin{bmatrix} I & O \\ B & A \end{bmatrix} $ or $ \begin{bmatrix} A & B \\ O & I \end{bmatrix}, $ where $I$ is an $m \times m$ identity matrix for $m \lt n,$ $A$ is any $k \times k$ matrix where $m + k = n,$ $B$ is a block of arbitrary values, and $O$ is a block of zeros, we have \begin{align*} \det \begin{bmatrix} I & O \\ B & A \end{bmatrix} &= \det A \qquad \text{and} \qquad \det \begin{bmatrix} A & B \\ O & I \end{bmatrix} = \det A. \end{align*} Proof. $\quad$ As in Exercise 3, we can use the Gauss-Jordan process to zero the elements of block $B$ without changing the determinant of the matrix. As such, \begin{align*} \det \begin{bmatrix} I & O \\ B & A \end{bmatrix} = \det \begin{bmatrix} I & O \\ O & A \end{bmatrix} \end{align*} and \begin{align*} \det \begin{bmatrix} A & B \\ O & I \end{bmatrix} = \det \begin{bmatrix} A & O \\ O & I \end{bmatrix} \end{align*} But as we showed in Theorem 3.7, the determinant of a block-diagonal matrix is the product of the determinants of its blocks: \begin{align*} \det \begin{bmatrix} I & O \\ O & A \end{bmatrix} &= (\det I)(\det A) = \det A \\ \\ \text{and} \\ \\ \det \begin{bmatrix} A & O \\ O & I \end{bmatrix} &= (\det A)(\det I) = \det A. \quad \blacksquare \end{align*}