Chapter 3

Chapter 3



For $n \ge 2$, the determinant of an $n \times n$ matrix $A = [a_{ij}]$ is the sum of $n$ terms of the form $\pm a_{1j} \ \mathrm{det} \ A_{1j}$, with plus and minus signs alternating, where the entries $a_{11}, a_{12}, ... , a_{1n}$ are from the first row of $A$. Symbolically,

\begin{align} \mathrm{det} \ A &= a_{11} \ \mathrm{det} \ A_{11} - a_{12} \ \mathrm{det} \ A_{12} + \cdot \cdot \cdot + (-1)^{1+n} \ a_{1n} \ \mathrm{det} \ A_{1n} \newline \newline &= \sum_{j=1}^{n} (-1)^{1+j} \ a_{1j} \ \mathrm{det} \ A_{1j} \end{align}

If A is an nxn matrix, the following is true:
1) Let $det(A) = x$. If any row or column in matrix A is multiplied by the scalar k, the new matrix B will have the determinant $det(B) = kx$.
2) A row swap of $r_i$ and $r_j$ will give the new matrix B and the determinant of A will be equal to the negative determinant of B. In other words, $det(A) = -det(B)$.
3) The row operation $r_i = r_i + kr_j$ on matrix A will give matrix B and the determinants of A and B will be equal. In other words, $det(A) = det(B)$.


A cofactor is the quantity obtained from a determinant or a square matrix by removal of the row and column containing a specified element. The (i, j)-cofactor of an nxn matrix A is $C_{ij}=(-1)^{i+j}det(A_{ij})$. Then the cofactor expansion across the first row of A can be given by Theorem 1.

Theorem 1

The determinant of an $n \times n$ matrix $A$ can be computed by a cofactor expansion across any row or down any column. The expansion across the $i$th row using the cofactors in (4) is

\begin{align} \mathrm{det} \ A &= a_{i1}C_{i1} + a_{i2}C_{i2}+ \cdot \cdot \cdot + a_{in}C_{in} \end{align}

The cofactor expansion down the $j$th column is

\begin{align} \mathrm{det} \ A &= a_{1j}C_{1j} + a_{2j}C_{2j}+ \cdot \cdot \cdot + a_{nj}C_{nj} \end{align}


The sign of the $(i, j)$-cofactor depends on the position of $a_{ij}$ as follows:

\begin{bmatrix} + & - & + & \dotsb\\ - & + & - & \\ + & - & + & \\ \vdots & & & \ddots \end{bmatrix}

Theorem 2

If $A$ is a triangular matrix, then $\mathrm{det} (A)$ is the product of the entries on the main diagonal of $A$.

Theorem 3

Let $A$ be a square matrix.

a. If a multiple of one row of $A$ is added to another row to produce a matrix $B$ , then $\mathrm{det} (B) = \mathrm{det} (A)$.
b. If two rows of $A$ are interchanged to produce $B$, then $\mathrm{det} (B) = -(\mathrm{det} (A))$.
c. If one row of $A$ is multiplied by $k$ to produce $B$, then $\mathrm{det} (B) = k (\mathrm{det} (A))$

Theorem 4

A square matrix A is invertible if and only if $\mathrm {det} (A)$ $\neq$ 0

Theorem 5

If $A$ is an n x n matirx, then $\mathrm {det} (A)^T$ = $\mathrm {det} (A)$

Theorem 6

Multiplicative Property

If A and B are nxn matrices, then $\mathrm {det} (AB)$ = $\mathrm {det} (A)$ $\mathrm {det} (B)$

Homework Problems

19. Consider the matrix $A=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$. What row operation transforms this to $B=\begin{bmatrix} c & d\\ a & b \end{bmatrix}$ and how does the determinant change?
Notice a row swap was performed on rows 1 and 2. Now: $det(A) = ad - bc$ and $det(B) = bc - ad$. Thus, swapping two rows negates the determinant of a matrix.

25-30 find the determinant:


\begin{align} det \begin{bmatrix}1&0&0\\0&1&0\\0&k&1\end{bmatrix}=1\cdot det\begin{bmatrix}1&0\\0&1\end{bmatrix}=1(1-0)=1 \end{align}


\begin{align} det \begin{bmatrix}1&0&0\\0&1&0\\k&0&1\end{bmatrix}=1\cdot det\begin{bmatrix}1&0\\0&1\end{bmatrix}=1(1-0)=1 \end{align}


\begin{align} det \begin{bmatrix}k&0&0\\0&1&0\\0&0&1\end{bmatrix}=k\cdot det\begin{bmatrix}1&0\\0&1\end{bmatrix}=k(1-0)=k \end{align}


\begin{align} det \begin{bmatrix}1&0&0\\0&k&0\\0&0&1\end{bmatrix}=1\cdot det\begin{bmatrix}k&0\\0&1\end{bmatrix}=1(k-0)=k \end{align}


\begin{align} det \begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}=-1\cdot det\begin{bmatrix}1&0\\0&1\end{bmatrix}=-1(1-0)=-1 \end{align}


\begin{align} det \begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}=1\cdot det\begin{bmatrix}0&1\\1&0\end{bmatrix}=1(0-1)=-1 \end{align}

31. Prove $det(A^{-1})=\frac{1}{det(A)}$ if $A$ is invertible.
By definition, we know $A A^{-1} = I$. Then
$det(A A^{-1}) = det(I)$
$det(A) det(A^{-1}) = 1$
$det(A^{-1}) = \frac{1}{det(A)}$, as desired.

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