- 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
3. $\quad$ Let $A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ a & b & c & d \\ e & f & g & h \end{bmatrix},$ $B = \begin{bmatrix} a & b & c & d \\ e & f & g & h \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}.$ Prove that $\det A = \det \begin{bmatrix} c & d \\ g & h \end{bmatrix}$ and that $\det B = \det \begin{bmatrix} a & b \\ e & f \end{bmatrix}.$
Proof. $\quad$ We recall from Section 4 that adding one row to a multiple of another in a square matrix leaves the determinant unchanged. As such, we have \begin{align*} \det \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ a & b & c & d \\ e & f & g & h \end{bmatrix} &= \det \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & c & d \\ 0 & 0 & g & h \end{bmatrix} \\ \\ &= \left(\det \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right) \left( \det \begin{bmatrix}c & d \\ g & h\end{bmatrix}\right) \\ &= \det \begin{bmatrix}c & d \\ g & h\end{bmatrix} \end{align*} and \begin{align*} \det \begin{bmatrix} a & b & c & d \\ e & f & g & h \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} &= \det \begin{bmatrix} a & b & 0 & 0 \\ e & f & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \\ &= \left( \det \begin{bmatrix}a & b \\ e & f\end{bmatrix}\right) \left( \det \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}\right)\\ &= \det \begin{bmatrix}a & b \\ e & f\end{bmatrix}. \quad \blacksquare \end{align*}