- 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
5. $\quad$ Let $A = \begin{bmatrix} a & b & 0 & 0 \\ c & d & 0 & 0 \\ e & f & g & h \\ x & y & z & w \end{bmatrix}.$ Prove that $\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}.$
Solution 1. $\quad$ We can use the row operations of the Gauss-Jordan elimination process to turn $A$ into a lower-triangular matrix, the determinant of which is the product of its diagonal entries.
$\quad$ First, we note that if $b$ and $d$ are both zero, or $h$ and $w$ are both zero, then the identity is satisfied trivially since $A$ is a lower-triangular matrix with diagonal entries equal to zero. Thus, for each pair of entries $(b, d)$ and $(h, w),$ we will assume that at least one entry is nonzero. This gives us three main cases: (1) $b = 0$ (2) $d = 0,$ and (3) $b, d$ both nonzero, each containing three subcases: (1) $h = 0,$ (2) $w = 0,$ and (3) $h, w$ both nonzero.
(1) $\quad$ $b = 0. \quad$ In this case, the top left-hand block is lower-triangular, and we can move onto the lower right-hand block, giving us three subcases:
$\qquad$ (1.1) $\quad h = 0. \quad$ In this case, $A$ can be rewritten as a lower-triangular matrix,
\begin{align*}
A &= \begin{bmatrix} a & 0 & 0 & 0 \\ c & d & 0 & 0 \\ e & f & g & 0 \\ x & y & z & w \end{bmatrix}
\end{align*}
whose determinant is given by the product $\det A = (ad)(gw).$ But because $b$ and $h$ are both zero, this satisfies the identity $$\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}.$$
$\qquad$ (1.2) $\quad w = 0. \quad$ In this case, we can turn $A$ into a lower triangular matrix by interchanging the bottom two rows, giving us
\begin{align*}
\det A &= - \det \begin{bmatrix} a & b & 0 & 0 \\ c & d & 0 & 0 \\ x & y & z & w \\ e & f & g & h \\ \end{bmatrix} \\
&= - \det \begin{bmatrix} a & 0 & 0 & 0 \\ c & d & 0 & 0 \\ x & y & z & 0 \\ e & f & g & h \\ \end{bmatrix} \\
&= -(ad)(zh) \\
&= (ad)(-zh).
\end{align*}
But as we can see, because $b = h = 0,$ this also satisfies the identity $$\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}.$$
$\qquad$ (1.3) $\quad h, w$ both nonzero. $\quad$ In this case, we can multiply the fourth row by $-h$ and add it to $w$ times the third row to give
\begin{align*}
\det A &= -\frac{1}{hw}\det \begin{bmatrix} a & 0 & 0 & 0 \\ c & d & 0 & 0 \\ ew - hx & fw - hy & gw - hz & 0 \\ -hx & -hy & -hz & -hw \end{bmatrix} \\
&= (ad)(gw - hz) \\
\end{align*}
But because $b = 0$ in this case, $\det A$ satisfies the identity
$$\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}.$$
(2) $\quad d = 0. \quad$ In this case, we first interchange the top two rows of $A$ to give us
\begin{align*}
\det A &= -\begin{bmatrix} c & 0 & 0 & 0 \\ a & b & 0 & 0 \\ e & f & g & h \\ x & y & z & w \end{bmatrix}.
\end{align*}
Here, we find three subcases.
$\qquad$ (2.1) $\quad h = 0. \quad $ In this case, the resulting matrix is lower-triangular and we have
\begin{align*}
\det A &= -\begin{bmatrix} c & 0 & 0 & 0 \\ a & b & 0 & 0 \\ e & f & g & 0 \\ x & y & z & w \end{bmatrix} \\
&= -(bc)(gw)
\end{align*}
But because $d = 0$ and $h = 0,$ this satisfies the identity
$$\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}.$$
$\qquad$ (2.2) $\quad w = 0.$ In this case, interchanging the third and fourth rows gives us a lower-triangular matrix and thus
\begin{align*}
\det A &= \begin{bmatrix} c & 0 & 0 & 0 \\ a & b & 0 & 0 \\ x & y & z & 0 \\ e & f & g & h \\ \end{bmatrix} \\
&= (bc)(hz) \\
&= (-bc)(-hz)
\end{align*}
But because $d = 0$ and $w = 0,$ this satisfies the identity
$$\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}.$$
$\qquad$ (2.3) $\quad$ $h, w$ both nonzero. $\quad$ In this case, we apply the same elimination steps as in case (1.3) to give us
\begin{align*}
\det A &= \frac{1}{hw}\det \begin{bmatrix} c & 0 & 0 & 0 \\ a & b & 0 & 0 \\ ew - hx & fw - hy & gw - hz & 0 \\ -hx & -hy & -hz & -hw \end{bmatrix} \\
&= (-bc)(gw - hz) \\
\end{align*}
And because $d = 0,$ this satisfies the identity
$$\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}.$$
(3) $\quad b, d$ both nonzero. $\quad$ In this case, we first turn the upper left-hand block of the matrix into a lower-triangular matrix by multiplying the first row by $-d$ and the second row by $b,$ and adding the second row to the first to give \begin{align*} \det A &= -\frac{1}{bd} \det \begin{bmatrix} bc -ad & 0 & 0 & 0 \\ bc & bd & 0 & 0 \\ e & f & g & h \\ x & y & z & w \end{bmatrix} \end{align*} But because the three subcases follow the same logic as when $b = 0$ and $d = 0,$ we can apply the same process as in cases (1) and (2) to find that in every case, the matrix $A$ can be transformed into a lower triangular matrix whose determinant satisfies the identity $$\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}. \quad \blacksquare $$
Solution 2. $\quad$ We may also use the row operations of the Gauss-Jordan elimination process to turn $A$ into an equivalent block-diagonal matrix. To do so, we first divide the top row by $-a$ and the second row by $-d,$ giving us \begin{align*} \det A &= (ad) \det \begin{bmatrix} -1 & -\frac{b}{a} & 0 & 0 \\ -\frac{c}{d} & -1 & 0 & 0 \\ e & f & g & h \\ x & y & z & w \end{bmatrix}. \end{align*} Then, we add the first row to the third row $e$ times and to the fourth row $x$ times (leaving the determinant unchanged each time). This gives us \begin{align*} \det A &= (ad) \det \begin{bmatrix} -1 & -\frac{b}{a} & 0 & 0 \\ -\frac{c}{d} & -1 & 0 & 0 \\ 0 & f - \frac{be}{a} & g & h \\ 0 & y - \frac{bx}{a} & z & w \end{bmatrix}. \end{align*} Then, adding the second row to the third and fourth rows $(f - be/a)$ times and $(y - bx/a)$ times, respectively (this once again leaves the determinant unchanged), we have \begin{align*} \det A &= (ad) \det \begin{bmatrix} -1 & -\frac{b}{a} & 0 & 0 \\ -\frac{c}{d} & -1 & 0 & 0 \\ 0 & 0 & g & h \\ 0 & 0 & z & w \end{bmatrix}. \end{align*} Then, multiplying the first row by $-a$ and the second row by $-d,$ we have \begin{align*} \det A &= \det \begin{bmatrix} a & b & 0 & 0 \\ c & d & 0 & 0 \\ 0 & 0 & g & h \\ 0 & 0 & z & w \end{bmatrix}. \end{align*} As we can see, $A$ can be rewritten as the equivalent block-diagonal matrix $A = \begin{bmatrix}B & O \\ O & C \end{bmatrix}$ with blocks $B = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ and $C = \begin{bmatrix} g & h \\ z & w \end{bmatrix}.$ But as shown in Theorem 3.7, $\det A = (\det B)(\det C),$ or in other words: $$\det A = \det \begin{bmatrix} a & b \\ c & d \end{bmatrix} \det \begin{bmatrix} g & h \\ z & w \end{bmatrix}. \quad \blacksquare $$