MHB Calculating Determinants of Matrices: A How-To Guide

[mathmari]

Hey!

I want to calculate the determinant of the following matrices:

1. $$A=\begin{pmatrix}-3 & -11 & -11 & 45 \\ 1 & 11 & 10 & -83 \\ 1 & -6 & -5 & 81 \\ 0 & -3 & -3 & 42\end{pmatrix}$$
2. $$B=\begin{pmatrix}1+a_1 & a_2 & \ldots & a_n \\ a_1 & 1+a_2 & \ldots & a_n \\ \ldots & \ldots & \ldots & \ldots \\ a_1 & a_2 & \ldots & 1+a_n\end{pmatrix}$$
3. $$C=(c_{ij})\in \mathbb{Q}^{n\times n} \text{ with } c_{ij}=\left\{\begin{matrix}
   0 & i=j\\
   1 & i\neq j
   \end{matrix}\right.$$

I have done the following:

1. $$|A|=\begin{vmatrix}-3 & -11 & -11 & 45 \\ 1 & 11 & 10 & -83 \\ 1 & -6 & -5 & 81 \\ 0 & -3 & -3 & 42\end{vmatrix}=-3\begin{vmatrix} 11 & 10 & -83 \\ -6 & -5 & 81 \\ -3 & -3 & 42\end{vmatrix}-\begin{vmatrix} -11 & -11 & 45 \\ -6 & -5 & 81 \\ -3 & -3 & 42\end{vmatrix}+\begin{vmatrix} -11 & -11 & 45 \\ 11 & 10 & -83 \\ -3 & -3 & 42\end{vmatrix}-0\cdot \begin{vmatrix} -11 & -11 & 45 \\ 11 & 10 & -83 \\ -6 & -5 & 81 \end{vmatrix} =\ldots =42$$

Could you give me a hint how we could calculate the determinant of $B$ and $C$ ? (Wondering)

 

[Evgeny.Makarov]

You can use the third elementary row operation (it does not change the determinant) to make as many elements zero as possible in a certain column. For example, in problem 2 you can subtract the last row from all other rows. Then the matrix will become the identity matrix except for the last row and column. Then subtract from the last row the first row multiplied by $a_1$, the second row multiplied by $a_2$ and so on to eliminate all elements of the last row except the bottom right element.

 

[mathmari]

Ah ok! So, we have the following:

2. $$|B|=\begin{vmatrix}1+a_1 & a_2 & \ldots & a_n \\ a_1 & 1+a_2 & \ldots & a_n \\ \ldots & \ldots & \ldots & \ldots \\ a_1 & a_2 & \ldots & 1+a_n\end{vmatrix}$$

By the row operation $R_i=R_i-R_n$ for each $i\in \{1, \ldots , n-1\}$ we get
$$|B|=\begin{vmatrix}1 & 0 & \ldots & -1 \\ 0 & 1 & \ldots & -1 \\ \ldots & \ldots & \ldots & \ldots \\ a_1 & a_2 & \ldots & 1+a_n\end{vmatrix}$$

By the row operation $R_n=R_n-a_1\cdot R_1-a_2\cdot R_2-\ldots -a_{n-1}\cdot R_{n-1}$ we get
$$|B|=\begin{vmatrix}1 & 0 & \ldots & -1 \\ 0 & 1 & \ldots & -1 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & 1+\sum_{i=0}^na_i\end{vmatrix}=\begin{vmatrix} 1 & \ldots & -1 \\ \ \ldots & \ldots & \ldots \\ 0 & \ldots & 1+\sum_{i=0}^na_i\end{vmatrix}= \ldots =\begin{vmatrix}1 & -1 \\ 0 &1+\sum_{i=0}^na_i\end{vmatrix}=1+\sum_{i=0}^na_i$$

Is this correct? (Wondering) 3. $$|C|=\begin{vmatrix}0 & 1 & \ldots & 1 \\ 1 & 0 & \ldots & 1 \\ \ldots & \ldots & \ldots & \ldots \\ 1 & 1 & \ldots & 0\end{vmatrix}$$

By the row operation $R_1=R_1+R_2+\ldots +R_n$ we get
$$|C|=\begin{vmatrix}n-1 & n-1 & \ldots & n-1 \\ 1 & 0 & \ldots & 1 \\ \ldots & \ldots & \ldots & \ldots \\ 1 & 1 & \ldots & 0\end{vmatrix}=(n-1)\begin{vmatrix}1 & 1 & \ldots & 1 \\ 1 & 0 & \ldots & 1 \\ \ldots & \ldots & \ldots & \ldots \\ 1 & 1 & \ldots & 0\end{vmatrix}$$

By the row operation $R_i=R_i-R_1$ for $i\in \{2, \ldots , n\}$ we get
$$|C|=(n-1)\begin{vmatrix}1 & 1 & \ldots & 1 \\ 0 & -1 & \ldots & 0 \\ \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & \ldots & -1\end{vmatrix}=(n-1)\begin{vmatrix} -1 & \ldots & 0 \\ \ldots & \ldots & \ldots \\ 0 & \ldots & -1\end{vmatrix}=(n-1)(-1)^{n-1}\begin{vmatrix} 1 & \ldots & 0 \\ \ldots & \ldots & \ldots \\ 0 & \ldots & 1\end{vmatrix}=(n-1)(-1)^{n-1}|I_{n-1}|=(n-1)(-1)^{n-1}$$

Is this correct? (Wondering)

 

[mathmari]

Suppose we have the matrix $A_n=(A_{ij})\in \mathbb{C}^{n\times n}$ with $a_{ij}=\left\{\begin{matrix}
1 , & i=j\\
-1 , & i=j-1\\
j^2, & i=j+1\\
0 , & \text{ otherwise}
\end{matrix}\right.$ for $1\leq i,j\leq n$.

I want to find the determinant using induction.

I have done the following:

To see the patern I have made some examples:

For $n=2$ we have the matrix $A_2=\begin{pmatrix}1& -1 \\ 1& 1\end{pmatrix}$. The determinant is $|A_2|=\begin{vmatrix}1& -1 \\ 1& 1\end{vmatrix}=2=n(n-1)$.

For $n=3$ we have the matrix $A_3=\begin{pmatrix}1& -1& 0 \\ 1 & 1 & -1 \\ 0 & 4 & 1\end{pmatrix}$. The determinant is $|A_3|=\begin{vmatrix}1& -1& 0 \\ 1 & 1 & -1 \\ 0 & 4 & 1\end{vmatrix} \ \ \overset{R_2=R_2-R_1}{ = } \ \ \begin{vmatrix}1& -1& 0 \\ 0 & 2 & -1 \\ 0 & 4 & 1\end{vmatrix}=\begin{vmatrix} 2 & -1 \\ 4 & 1\end{vmatrix}=2-(-4)=6=n(n-1)$. So, we want to prove that the determinant is $|A_n|=n(n-1)$, for $n\geq 2$, right? (Wondering) Induction on $n$.

Base case: For $n=2$, as shown above, it holds.

Inductive hypothesis: We suppose that it holds for $n=k$, i.e., $|A_k|=k(k-1)$.

Inductive step: We want to prove it for $n=k+1$, i.e., that it holds that $|A_{k+1}|=(k+1)k$. When we consider the matrix without the last row and without the last column, it is the matrix $A_k$.

But what is the relation between the determinant of $A_k$ and that of $A_{k+1}$ ? (Wondering)

 

[Evgeny.Makarov]

I agree for problems 2 and 3 from post #1. For the problem about induction, you may want to create a new thread.

 

[mathmari]

I agree for problems 2 and 3 from post #1. For the problem about induction, you may want to create a new thread.

Ok, I will do that.

Thank you very much! (Smile)