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\begin{array}{rrrrr|r} b & a & a & \cdot \cdot \cdot & a \\ a & b & a & \cdot \cdot \cdot & a \\ a & a & b & \cdot \cdot \cdot & a \\ \cdot & \cdot & \cdot & \space & \cdot\\ \cdot & \cdot & \cdot & \space & \cdot\\ a & a & a & \cdot \cdot \cdot & b \end{array}

I have the above matrix $A\in M_{n\times n}(F)$ where $F$ is a field and $n\geq1$, $a,b\in F$.

I'm trying to find out how to use row operations to make it into an upper triangular matrix in order to figure out the determinant. But I'm not sure how I would approach it.

Martin Sleziak
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JamieD
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2 Answers2

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To find the determinant, rather than transforming the matrix to an upper triangular form, you could express the matrix (of size $n\times n$) as $$M=(b-a)I_{n\times n}+a\mathbf{1}\cdot\mathbf{1}^T$$ where $I_{n\times n}$ is the identity matrix, and $\mathbf{1}$ is an $n\times 1$ vector all of whose elements are $1$ (note that matrix $a\mathbf{1}\cdot\mathbf{1}^T$ will have rank $1$).

Thus (using the Matrix determinant lemma) one of the eigenvalues of matrix $M$ will be $(b-a)+na$, and the rest (i.e. $n-1$) of the eigenvalues will be $b-a$. Thus the determinant will be $(b-a)^{n-1}(b+(n-1)a)$.

Alijah Ahmed
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This is a variant of https://math.stackexchange.com/a/1238109/62967

Consider the $n\times n$ matrix $A_a$, with all coefficients equal to $a$. Then $na$ is the only nonzero eigenvalue (if $a\ne0$) because the rank is $1$. So the characteristic polynomial of $A_a$ is $$ \det(A_a-XI_n)=(0-X)^{n-1}(na-X)=(-1)^{n-1}X^{n-1}(na-X) $$ Your matrix can be written as $A_a-(a-b)I_n$, so its determinant is the value of the characteristic polynomial for $X=a-b$: $$ (-1)^{n-1}(a-b)^{n-1}(na-a+b)=(b-a)^{n-1}((n-1)a+b) $$

Community
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egreg
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