Determinant of Matrix whose Diagonal Entries are 6 and 2 Elsewhere

Math exam problems and solutions at Harvard University

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. 04/16/2017

    […] 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 […]

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