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I need to calculate the determinant of the following $4 \times 4$ matrix:

\begin{bmatrix}x&1&1&1\\1&x&1&1\\1&1&x&1\\1&1&1&x\end{bmatrix}

I heard there is a way by separating the matrix into blocks, but I couldn't succeed in doing that.

Rodrigo de Azevedo
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ליאור כהן
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3 Answers3

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Useful trick - if the sum of each row of the matrix is the same (in this case - $x+3$), then you can simplify the determinant via the following elementary operations:

  1. Add all columns to the first column.

  2. Subtract the first row from all rows.

In your case:

$$\begin{vmatrix}x&1&1&1\\1&x&1&1\\1&1&x&1\\1&1&1&x\end{vmatrix} =\begin{vmatrix}x+3&1&1&1\\x+3&x&1&1\\x+3&1&x&1\\x+3&1&1&x\end{vmatrix} =\begin{vmatrix}x+3&1&1&1\\0&x-1&0&0\\0&0&x-1&0\\0&0&0&x-1\end{vmatrix}$$

Your matrix is now triangular, and the determinant is the product of diagonal elements.

GSofer
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  • Thank you very much – ליאור כהן Jul 21 '22 at 11:20
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    Does this work in all cases where the sum of the rows is the same? It seems to be a bit special to the case where the off diagonal entries are all the same. – Michael Burr Jul 21 '22 at 13:27
  • @MichaelBurr It does work, although not as nicely. In this special case we ended up with a triangular matrix, whose determinant we can compute immediately. In the general case, we end up with a matrix whose first column is almost all zeros, which reduced you to computing the determinant of a smaller matrix (which is much easier than what you started with). – GSofer Jul 21 '22 at 13:57
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Alternative:

$A=\begin{bmatrix}x&1&1&1\\1&x&1&1\\1&1&x&1\\1&1&1&x\end{bmatrix}$

Since each row sum is $x+3$ , it's an eigenvalue of $A$ $($ see here $) $

It is clear that if $x=1$ , then $A$ will be a rank $1$ matrix.

Hence $x-1$ is an eigenvalue as $\det(A-(x-1) I) =0$

Infact Eigen space corresponding to $x-1$ has dimension $3$ as

$A-(x-1)I=\begin{bmatrix}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{bmatrix}$

has $3$ dimensional null space.

Hence Algebraic multiplicity of $(x-1) $ is $3$ $($ as it is bigger than geometric multiplicity and no. of distinct eigen values can't exceed $4$ $) $

Now $\det(A) =\Pi_{\lambda\in \operatorname{spec}{A}} \space {\lambda}=(x-1)^3(x+3)$

(determinant is the product of all eigenvalues )

Sourav Ghosh
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In the special case $$ \left( {\begin{array}{*{20}c} A & B \\ B & A \\ \end{array}} \right) $$ where $A,B$ are $2×2$ blocks you can use the formula $$ \det \left( {\begin{array}{*{20}c} A & B \\ B & A \\ \end{array}} \right) = \det \left( {A - B} \right)\det \left( {A + B} \right) $$ In our case you have $$ \det \left( {A - B} \right) = \det \left( {\begin{array}{*{20}c} {x - 1} & 0 \\ 0 & {x - 1} \\ \end{array}} \right) = (x - 1)^2 $$ while $$ \det \left( {A + B} \right) = \det \left( {\begin{array}{*{20}c} {x + 1} & 2 \\ 2 & {x + 1} \\ \end{array}} \right) = x^2 + 2x - 3 = \left( {x - 1} \right)\left( {x + 3} \right) $$ whence $$ \det \left( {\begin{array}{*{20}c} A & B \\ B & A \\ \end{array}} \right) = \left( {x - 1} \right)^3 \left( {x + 3} \right) $$

Sourav Ghosh
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Luca Goldoni Ph.D.
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