Theorem
Let \(A \in M_{n \times n}\). If we perform the following row operations
- \(A \xrightarrow{R_i \leftrightarrow R_j} B \), then \(\det(B) = -\det(A)\)
- \(A \xrightarrow{R_i \rightarrow cR_i} B \), then \(\det(B) = c\det(A)\)
- \(A \xrightarrow{R_i \rightarrow R_i + cR_j} B \), then \(\det(B) = -\det(A)\)
We will only need to use Theorem 1 and Corollary 2 from last lecture to prove this result!
Proof
For (II) Let
$$
\begin{align*}
A &= \begin{pmatrix} a_1 \\ \vdots \\ a_i \\ \vdots \\ a_n \end{pmatrix},
B = \begin{pmatrix} a_1 \\ \vdots \\ ca_i \\ \vdots \\ a_n \end{pmatrix},
\end{align*}
$$
Using theorem 1 we know that \(\det(B) = c\det(A)\) as we wanted to show. For (III) Let
$$
\begin{align*}
A &= \begin{pmatrix} a_1 \\ \vdots \\ a_i \\ \vdots \\ a_j \\ \vdots \\ a_n \end{pmatrix},
B = \begin{pmatrix} a_1 \\ \vdots \\ a_i+ca_j \\ \vdots \\ a_j \\ \vdots \\ a_n \end{pmatrix},
C = \begin{pmatrix} a_1 \\ \vdots \\ a_j \\ \vdots \\ a_j \\ \vdots \\ a_n \end{pmatrix},
\end{align*}
$$
Using theorem 1 we know that \(\det(B) = \det(A) + c\det(C)\). But Corollary 2 implies that \(\det(C) = 0\) since \(C\) has two identical rows. Therefore, \(\det(B) = \det(A)\) as we wanted to show. For (I)
$$
\begin{align*}
A &= \begin{pmatrix} a_1 \\ \vdots \\ a_i \\ \vdots \\ a_j \\ \vdots \\ a_n \end{pmatrix},
B = \begin{pmatrix} a_1 \\ \vdots \\ a_j \\ \vdots \\ a_i \\ \vdots \\ a_n \end{pmatrix},
C = \begin{pmatrix} a_1 \\ \vdots \\ a_i+a_j \\ \vdots \\ a_i+a_j \\ \vdots \\ a_n \end{pmatrix},
\end{align*}
$$
One thing we know right away here is that \(\det(C)=0\) since \(C\) has two identical rows (by corollary 2) so
$$
\begin{align*}
0 = \det\begin{pmatrix} a_1 \\ \vdots \\ a_i+a_j \\ \vdots \\ a_i+a_j \\ \vdots \\ a_n \end{pmatrix} &= \det\begin{pmatrix} a_1 \\ \vdots \\ a_i \\ \vdots \\ a_i + a_j \\ \vdots \\ a_n \end{pmatrix} + \det\begin{pmatrix} a_1 \\ \vdots \\ a_j \\ \vdots \\ a_i + a_j \\ \vdots \\ a_n \end{pmatrix} \\
&= \det\begin{pmatrix} a_1 \\ \vdots \\ a_i \\ \vdots \\ a_i \\ \vdots \\ a_n \end{pmatrix} +
\det\begin{pmatrix} a_1 \\ \vdots \\ a_i \\ \vdots \\ a_j \\ \vdots \\ a_n \end{pmatrix} +
\det\begin{pmatrix} a_1 \\ \vdots \\ a_j \\ \vdots \\ a_i \\ \vdots \\ a_n \end{pmatrix} +
\det\begin{pmatrix} a_1 \\ \vdots \\ a_j \\ \vdots \\ a_j \\ \vdots \\ a_n \end{pmatrix} \\
&= \det\begin{pmatrix} a_1 \\ \vdots \\ a_i \\ \vdots \\ a_j \\ \vdots \\ a_n \end{pmatrix} +
\det\begin{pmatrix} a_1 \\ \vdots \\ a_j \\ \vdots \\ a_i \\ \vdots \\ a_n \end{pmatrix} \\
0 &= \det(A) + \det(B) \\
det(A) &= -\det(B)
\end{align*}
$$
Which is what we wanted to show. \(\blacksquare\)
Definition
\(A \in M_{n \times n }\) is upper(lower) triangular if all entries below (above) diagonal are zero.
For example any \(n \times n\) matrix in REF is upper triangular. One reason why the Upper/Lower triangular matrices are interesting is the following theorem.
Theorem
If \(A \in M_{2 \times 2}\) is upper or lower triangular then
$$
\begin{align*}
\det(A) = A_{11}A_{22}A_{nn}
\end{align*}
$$
Proof
For upper triangular matrices, by induction on \(n\).
Base Case: \(n = 2\)
$$
\begin{align*}
\det
\begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}
= A_{11}A_{22}.
\end{align*}
$$
Inductive Case: assume this is true for \(n-1 \geq 2\). Consider
$$
\begin{align*}
\begin{pmatrix}
A_{11} & A_{12} & \cdots & \cdots & A_{1n} \\
0 & A_{22} & \cdots & \cdots & \vdots \\
\vdots & 0 & \ddots & \cdots & \vdots \\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0 & \cdots & 0 & A_{nn} \\
\end{pmatrix}
\end{align*}
$$
We can compute the determinant by choosing to cofactor along the \(n\)th row and so,
$$
\begin{align*}
\det(A) = \sum^n_{j=1}(-1)^{n+j}A_{nj}\det(\tilde{A_{nj}}) \\
= (-1)^{n+n}A_{nn}\det(\tilde{A_{nn}}) \\
\end{align*}
$$
But \(\tilde{A_{nn}}\) is an \(n-1 \times n-1\) upper triangular matrix. So by the inductive hypothesis its determinant is the product of the diagonal entries and so
$$
\begin{align*}
\det(\tilde{A_{nn}} = A_{11}A_{22}...A_{n-1n-1}
\end{align*}
$$
And therefore,
$$
\begin{align*}
\det(A) = A_{11}A_{22}...A_{nn}
\end{align*}
$$
As we wanted to show. \(\blacksquare\).
Example
Compute the determinant for
\(A = \begin{pmatrix}
1 & 2 & 3 \\
1 & 0 & 2 \\
2 & 1 & 1
\end{pmatrix}\).
To use the theorem, we’ll put \(A\) in REF.
$$
\begin{align*}
\begin{pmatrix}
1 & 2 & 3 \\
1 & 0 & 2 \\
2 & 1 & 1
\end{pmatrix}
R_2 \rightarrow R_2 - R_1
R_3 \rightarrow R_3 - 2R_1
\begin{pmatrix}
1 & 2 & 3 \\
0 & -2 & -1 \\
0 & -3 & -5
\end{pmatrix}
\end{align*}
$$
Note here that these two operations will not change the value of the determinant per thoerem 1 from last lecture. Continuing with REF:
$$
\begin{align*}
\begin{pmatrix}
1 & 2 & 3 \\
0 & -2 & -1 \\
0 & -3 & -5
\end{pmatrix}
R_2 \rightarrow -\frac{1}{2}R_2
\begin{pmatrix}
1 & 2 & 3 \\
0 & 1 & \frac{1}{2} \\
0 & -3 & -5
\end{pmatrix}
\end{align*}
$$
Here, since we scaled by \(-\frac{1}{2}\) then we know that \(\det(A) = -2\det(RREF(A))\).
$$
\begin{align*}
\begin{pmatrix}
1 & 2 & 3 \\
0 & 1 & \frac{1}{2} \\
0 & -3 & -5
\end{pmatrix}
R_3 \rightarrow 3R_2 + R_3
\begin{pmatrix}
1 & 2 & 3 \\
0 & 1 & \frac{1}{2} \\
0 & 0 & -\frac{7}{2}
\end{pmatrix}
\end{align*}
$$
This operation will not further change the value of the determinant. So now we can apply the theorem to compute the determinant since \(A\) is an upper triangular matrix
$$
\begin{align*}
\det
\begin{pmatrix}
1 & 2 & 3 \\
1 & 0 & 2 \\
2 & 1 & 1
\end{pmatrix}
= -2\det
\begin{pmatrix}
1 & 2 & 3 \\
0 & 1 & \frac{1}{2} \\
0 & 0 & -\frac{7}{2}
\end{pmatrix}
= -2(1 * 1 * -\frac{7}{2}) = 7
\end{align*}
$$
Comparisons
Computing \(\det(A)\) for \(A \in M_{n \times n}\) using the inductive formula is roughly \(n!\) operations. Using row operations, it is roughly \(n^3\) operations.
References
- Video Lectures from Math416 by Ely Kerman.