Science:Math Exam Resources/Courses/MATH221/December 2009/Question 02 (a)

MATH221 December 2009

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Question 02 (a)
Find the determinant of the matrix $A={\begin{bmatrix}0&1&1&1&1\\1&0&1&1&1\\1&1&0&1&1\\1&1&1&0&1\\1&1&1&1&0\end{bmatrix}}$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
One can do this by just charging ahead and either using the definition or using row operations to get the matrix into an upper diagonal matrix (see solution 2) but there is a more elegant way to solve this (this will be solution 1).

Hint 2
Let $\displaystyle J={\begin{bmatrix}1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\end{bmatrix}}$ and let $\displaystyle I$ be the five by five identity matrix. Then $\displaystyle A=J-I$. In this form, eigenvectors and hence eigenvalues are very easy to compute. How can we relate eigenvalues and determinants?

Hint 3

The product of the eigenvalues $\displaystyle \lambda _{i}$ of a matrix $\displaystyle X$ is equal to the determinant. This follows since $\displaystyle \det(X-\lambda I)=(\lambda _{1}-\lambda )(\lambda _{2}-\lambda )...(\lambda _{n}-\lambda )$ Then plug in $\displaystyle \lambda =0$ into both sides of the equation. Thus compute the eigenvalues of $\displaystyle J-I$ and then multiply them together.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The following proof can be generalized very easily to an n by n matrix and so we present this solution first. Let $\displaystyle J={\begin{bmatrix}1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\\1&1&1&1&1\end{bmatrix}}$ and let $\displaystyle I$ be the five by five identity matrix. Then $\displaystyle A=J-I$. Next, let $\displaystyle e_{i}={\begin{bmatrix}0\\0\\\vdots \\0\\1\\0\\\vdots \\0\end{bmatrix}}$ be the standard basis vectors where the 1 above is in the ith position. Notice then that the vectors $\displaystyle e_{1,i}$ defined by $\displaystyle e_{1,i}=e_{1}-e_{i}$ for i from 2 to 5 (or in general, the size of the matrix J). A simple calculation shows us that $\displaystyle Je_{1,i}=0$ for each value of i from 2 to 5. Further, taking the vector $\displaystyle v=e_{1}+e_{2}+e_{3}+e_{4}+e_{5}$ (the all ones vector), we see that $\displaystyle Jv=5v$. As every vector is an eigenvector of the identity matrix, we have $\displaystyle Ae_{1,i}=(J-I)e_{1,i}=Je_{1,i}-e_{1,i}=-e_{1,i}$ for each i from 2 to 5 and $\displaystyle Av=(J-I)v=Jv-v=5v-v=(5-1)v=4v$ Hence, we get $\displaystyle (5-1)$ eigenvalues of $\displaystyle (-1)$ and one eigenvalue of $\displaystyle (5-1)$. Thus, we have $\displaystyle {\text{det}}(A)=(-1)^{5-1}(5-1)=4$ To generalize the above, just change every 5 above to an n.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Use the following determinant rules: The determinant of upper triangular matrix (i.e a matrix in row echelon form in this case) is equal to the product of its diagonal entries. If i$\neq$j , then subtracting some constant time row i from row j does not change the determinant. Although there are multiple ways of finding determinants, this solution will use rule 2 when performing row operations so that the determinant of the original matrix is the same as the matrix in row echelon form and then use rule 1 to find the determinant. Let $A={\begin{bmatrix}0&1&1&1&1\\1&0&1&1&1\\1&1&0&1&1\\1&1&1&0&1\\1&1&1&1&0\end{bmatrix}}$ Step 1: Do the following row operations on the matrix A in the following order: row 1 = row 1 + row 2 row 2 = row 2 - row 1 row 3 = row 3 - row 1 and we get ${\begin{bmatrix}1&1&2&2&2\\0&-1&-1&-1&-1\\0&0&-2&-1&-1\\0&0&-1&-2&-1\\0&0&-1&-1&-2\end{bmatrix}}$ The matrix is still not in upper triangular form. Then perform the following row operations: row 4 = row 4 - ${\frac {1}{2}}$row 3 row 5 = row 5 - ${\frac {1}{2}}$row 3 and we get : ${\begin{bmatrix}1&1&2&2&2\\0&-1&-1&-1&-1\\0&0&-2&-1&-1\\0&0&0&-3/2&-1/2\\0&0&0&-1/2&-3/2\end{bmatrix}}$ After doing one more row operation, row 5 = row 5 - ${\frac {1}{3}}$row 4 , we get the following upper triangular matrix: ${\begin{bmatrix}1&1&2&2&2\\0&-1&-1&-1&-1\\0&0&-2&-1&-1\\0&0&0&-3/2&-1/2\\0&0&0&0&-4/3\end{bmatrix}}$ Step 2: let $U={\begin{bmatrix}1&1&2&2&2\\0&-1&-1&-1&-1\\0&0&-2&-1&-1\\0&0&0&-3/2&-1/2\\0&0&0&0&-4/3\end{bmatrix}}$ Then ${\text{det}}(A)={\text{det}}(U)={\text{ product of diagonal entries}}=1\times -1\times -2\times -{\frac {3}{2}}\times -{\frac {4}{3}}=4$ Thus ${\text{det}}(A)=4$

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