Science:Math Exam Resources/Courses/MATH221/April 2009/Question 02

MATH221 April 2009

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Question 02
Compute the determinant of the matrix $A={\begin{bmatrix}1&1&x&1\\x&1&1&1\\1&1&1&x\\1&x&1&1\end{bmatrix}}$

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Can we use row operations to make the matrix easier to handle? How do different row operations affect the determinant?

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Before computing the determinant of $A$, we could first use row operations to put it into an easier form. Recall that the following relationships hold between row operations and the determinant: If we add or subtract multiples of one row to another, the determinant is unchanged If we swap rows, the determinant changes by (-1) If we multiply a row by a constant, the determinant also multiplies by that constant The first rule is the most powerful since we can manipulate the matrix without changing the determinant. If we want to use co-factor expansion to compute the determinant then it will be useful if we can place a lot of zeros in either a row or a column. If we subtract the first row from the third and fourth columns this will put zeros in the first column of each row. If we subtract $x$ multiples of row 1 from row 2 then this will place a zero in column 1 or row 2. Let's perform these operations, ${\begin{bmatrix}1&1&x&1\\x&1&1&1\\1&1&1&x\\1&x&1&1\end{bmatrix}}{\underset {R3-R1,R4-R1}{\rightarrow }}{\begin{bmatrix}1&1&x&1\\0&1-x&1-x^{2}&1-x\\0&0&1-x&x-1\\0&x-1&1-x&0\end{bmatrix}}.$ Since we used rule 1 above of row operations, we didn't change the determinant at all! If we were to co-factor expand now, the first column is a good choice because only the entry in the first row will contribute anything. Therefore, $\det(A)=(1)\det \left({\begin{bmatrix}1-x&1-x^{2}&1-x\\0&1-x&x-1\\x-1&1-x&0\end{bmatrix}}\right).$ We could co-factor expand this final matrix as is but we can make it even easier if we add the first row to the third row since the the first column of the third row will be zero and the co-factor expansion will be easier to do, ${\begin{bmatrix}1-x&1-x^{2}&1-x\\0&1-x&x-1\\x-1&1-x&0\end{bmatrix}}{\underset {R3+R1}{\rightarrow }}{\begin{bmatrix}1-x&1-x^{2}&1-x\\0&1-x&x-1\\0&2-x-x^{2}&1-x\end{bmatrix}}$ and, again, if we co-factor expand along the first column, only the entry in the first row contributes. Therefore, $\det(A)=(1)(1-x)\det \left({\begin{bmatrix}1-x&-(1-x)\\(x+2)(1-x)&1-x\end{bmatrix}}\right)$ where we have factored the quadratic term to easily see the factor of $1-x$. Since there is a factor of $1-x$ in each row we can remove that from the matrix and by rule 3 above, this will change the determinant by $(1-x)^{2}$. Therefore, $\det(A)=(1)(1-x)^{3}\det \left({\begin{bmatrix}1&-1\\(x+2)&1\end{bmatrix}}\right)=(1-x)^{3}(x+3).$ If we expand this out we get $(1-x)^{3}(x+3)=3-8x+6x^{2}-x^{4}$ but this simplification is an optional step.

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. As a brute-force alternative we can compute the determinant directly, using co-factor expansion. Expanding on the first row gives ${\begin{aligned}\det(A)&=\displaystyle (1){\text{det}}\left({\begin{bmatrix}1&1&1\\1&1&x\\x&1&1\end{bmatrix}}\right)-(1){\text{det}}\left({\begin{bmatrix}x&1&1\\1&1&x\\1&1&1\end{bmatrix}}\right)\\&\quad +(x){\text{det}}\left({\begin{bmatrix}x&1&1\\1&1&x\\1&x&1\end{bmatrix}}\right)-(1){\text{det}}\left({\begin{bmatrix}x&1&1\\1&1&1\\1&x&1\end{bmatrix}}\right)\end{aligned}}$ Performing one row swap in the first, third and fourth terms (which changes the sign of the determinant) yields ${\begin{aligned}\det(A)&=\displaystyle -(1){\text{det}}\left({\begin{bmatrix}x&1&1\\1&1&x\\1&1&1\end{bmatrix}}\right)-(1){\text{det}}\left({\begin{bmatrix}x&1&1\\1&1&x\\1&1&1\end{bmatrix}}\right)\\&\quad -(x){\text{det}}\left({\begin{bmatrix}x&1&1\\1&x&1\\1&1&x\end{bmatrix}}\right)+(1){\text{det}}\left({\begin{bmatrix}x&1&1\\1&x&1\\1&1&1\end{bmatrix}}\right)\end{aligned}}$ Now the first two terms are the same so we combine them to get ${\begin{aligned}\det(A)&=\displaystyle -(2){\text{det}}\left({\begin{bmatrix}x&1&1\\1&1&x\\1&1&1\end{bmatrix}}\right)-(x){\text{det}}\left({\begin{bmatrix}x&1&1\\1&x&1\\1&1&x\end{bmatrix}}\right)\\&\quad +(1){\text{det}}\left({\begin{bmatrix}x&1&1\\1&x&1\\1&1&1\end{bmatrix}}\right)\end{aligned}}$ To compute the remaining $3\times {}3$ determinants we can use a variety of techniques. One such technique is to co-factor expand again. For example for the first matrix, if we expand along the first row, $\det \left({\begin{bmatrix}x&1&1\\1&1&x\\1&1&1\end{bmatrix}}\right)=(x)(1-x)-(1)(1-x)+(1)(0)=-x^{2}+2x-1.$ Similarly for the other two matrices we have ${\text{det}}\left({\begin{bmatrix}x&1&1\\1&x&1\\1&1&x\end{bmatrix}}\right)=(x)(x^{2}-1)-(1)(x-1)+(1)(1-x)=x^{3}-3x+2$ and ${\text{det}}\left({\begin{bmatrix}x&1&1\\1&x&1\\1&1&1\end{bmatrix}}\right)=x(x-1)-(1)(0)+(1)(1-x)=x^{2}-2x+1$ Combining everything, we have ${\begin{aligned}\det(A)&=-2(-x^{2}+2x-1)-(x)(x^{3}-3x+2)+(x^{2}-2x+1)\\&=-x^{4}+6x^{2}-8x+3\end{aligned}}$

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