# Determinant of Matrix whose Diagonal Entries are 6 and 2 Elsewhere

## Problem 380

Find the determinant of the following matrix

\[A=\begin{bmatrix}

6 & 2 & 2 & 2 &2 \\

2 & 6 & 2 & 2 & 2 \\

2 & 2 & 6 & 2 & 2 \\

2 & 2 & 2 & 6 & 2 \\

2 & 2 & 2 & 2 & 6

\end{bmatrix}.\]

(*Harvard University, Linear Algebra Exam Problem*)

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## Hint.

Computing the determinant directly by hand is tedious.

So use the fact that the determinant of a matrix $A$ is the product of all eigenvalues of $A$.

However, finding the eigenvalue of $A$ itself is as complicated as computing the determinant of $A$.

Instead, first determine the eigenvalues of $B=A-4I$.

Then use the fact that if $\lambda$ is an eigenvalue of $B$, then $\lambda+4$ is an eigenvalue of $A$.

For a proof of this fact, see the post “Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$“.

## Solution.

Let $B=A-4I$, where $I$ is the $5 \times 5$ identity matrix.

Then every entry of $B$ is $2$.

By elementary row operations, we can reduced the matrix $B$ into

\[B=\begin{bmatrix}

2 & 2 & 2 & 2 &2 \\

2 & 2 & 2 & 2 & 2 \\

2 & 2 & 2 & 2 & 2 \\

2 & 2 & 2 & 2 & 2 \\

2 & 2 & 2 & 2 & 2

\end{bmatrix}\to

\begin{bmatrix}

1 & 1 & 1 & 1 &1 \\

0 & 0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0 & 0 \\

0 & 0 & 0 & 0 & 0

\end{bmatrix}.\]
Thus, the rank of $B$ is $1$, hence the nullity is $4$ by the rank-nullity theorem.

It follows that $0$ is an eigenvalue of $B$ and its geometric multiplicity is $4$.

Since all entries of $B$ are equal, we compute

\[B\begin{bmatrix}

1 \\

1 \\

1 \\

1 \\

1

\end{bmatrix}=10\begin{bmatrix}

1 \\

1 \\

1 \\

1 \\

1

\end{bmatrix}.\]
This yields that $10$ is an eigenvalue of $B$ and the vector $[1\, 1\, 1\, 1\, 1]^{\trans}$ is an eigenvector corresponding to $10$.

Combining these observations, we see that the matrix $B$ has eigenvalues $0$ and $10$ with (algebraic) multiplicities $4$ and $0$, respectively.

Since $A=B+4I$, the eigenvalues of $A$ are $4$ and $14$ with algebraic multiplicities $4$ and $0$, respectively.

(See Problem Eigenvalues and algebraic/geometric multiplicities of matrix $A+cI$.)

The determinant of $A$ is the product of all the eigenvalues of $A$ (counting multiplicities). Thus we have

\[\det(A)=4^4\cdot 14=3584.\]

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## 1 Response

[…] is also a problem of a Linear Algebra final exam at Harvard University. See the post “Determinant of matrix whose diagonal entries are 6 and 2 elsewhere” for a […]