# Calculate Determinants of Matrices

## Problem 45

Calculate the determinants of the following $n\times n$ matrices.
$A=\begin{bmatrix} 1 & 0 & 0 & \dots & 0 & 0 &1 \\ 1 & 1 & 0 & \dots & 0 & 0 & 0 \\ 0 & 1 & 1 & \dots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\ 0 & 0 & 0 &\dots & 1 & 1 & 0\\ 0 & 0 & 0 &\dots & 0 & 1 & 1 \end{bmatrix}$

The entries of $A$ is $1$ at the diagonal entries, entries below the diagonal, and $(1, n)$-entry.
The other entries are zero.
$B=\begin{bmatrix} 1 & 0 & 0 & \dots & 0 & 0 & -1 \\ -1 & 1 & 0 & \dots & 0 & 0 & 0 \\ 0 & -1 & 1 & \dots & 0 & 0 & 0 \\ \vdots & \vdots & \vdots & \dots & \dots & \ddots & \vdots \\ 0 & 0 & 0 &\dots & -1 & 1 & 0\\ 0 & 0 & 0 &\dots & 0 & -1 & 1 \end{bmatrix}.$

The entries of $B$ is $1$ at the diagonal entries.
The entries below the diagonal and $(1,n)$-entry are $-1$.
The other entries are zero.

## Hint.

1. Calculate the first row cofactor expansion.
2. The determinant of a triangular matrix is the product of its diagonal entries.

## Solution.

Apply the cofactor expansion corresponding to the first row. We obtain
\begin{align*}
\det(A)&=
\begin{vmatrix}
1 & 0 & \dots & 0 & 0 & 0 \\
1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 &\dots & 1 & 1 & 0\\
0 & 0 &\dots & 0 & 1 & 1
\end{vmatrix}
+(-1)^{n+1}
\begin{vmatrix}
1 & 1 & 0 & \dots & 0 & 0 \\
0 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 &\dots & 1 & 1 \\
0 & 0 & 0 &\dots & 0 & 1
\end{vmatrix}
\end{align*}
The two smaller (minor) $n-1 \times n-1$ matrices are both triangular.
The determinant of a triangular matrix is the product of its diagonal entries.
Thus we see that
\begin{align*}
\det(A)&=1+(-1)^{n+1} \\
&= \begin{cases}
2 & \text{ if } n \text{ is odd}\\
0 & \text{ if } n \text{ is even}.
\end{cases}
\end{align*}
Next we calculate $\det(B)$. By the first row cofactor expansion , we obtain
\begin{align*}
\det(B)&=\\
&\begin{vmatrix}
1 & 0 & \dots & 0 & 0 & 0 \\
-1 & 1 & \dots & 0 & 0 & 0 \\
\vdots & \vdots & \dots & \ddots & \vdots & \vdots \\
0 & 0 &\dots & -1 & 1 & 0\\
0 & 0 &\dots & 0 & -1 & 1
\end{vmatrix}
+(-1)^{n+1}(-1)
\begin{vmatrix}
-1 & 1 & 0 & \dots & 0 & 0 \\
0 & -1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 &\dots & -1 & 1 \\
0 & 0 & 0 &\dots & 0 & -1
\end{vmatrix}.
\end{align*}
The two minor matrices are both triangular.
All the diagonal entries of the first minor matrix are $1$ and those of the second minor matrix are $-1$.
Thus we have
\begin{align*}
\det(B)&=1+(-1)^{n}(-1)^{n-1}=0.
\end{align*}

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