Science:Math Exam Resources/Courses/MATH152/April 2012/Question 05 (b)

MATH152 April 2012

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Question 05 (b)
The following matrix A depends on a parameter p. Find all the values of of the parameter p for which the corresponding matrix is not invertible. $A={\begin{bmatrix}1&2&3&4\\100&200&300&401\\0&p&p+1&p+2\\0&-p&-2p+1&-3p+2\end{bmatrix}}$

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
The determinant is your friend. A matrix is noninvertible if its determinant is zero. For which values of p is the determinant of this matrix zero?

Hint 2
To find the determinant simplify the matrix first by using row operations.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We begin by taking our matrix and applying row operations. First, we take 100 times the first row and subtract it from the second row to turn the matrix $A={\begin{bmatrix}1&2&3&4\\100&200&300&401\\0&p&p+1&p+2\\0&-p&-2p+1&-3p+2\end{bmatrix}}$ into ${\begin{bmatrix}1&2&3&4\\0&0&0&1\\0&p&p+1&p+2\\0&-p&-2p+1&-3p+2\end{bmatrix}}$ Next we subtract $p+2$ times the second row to the third row and also add $-3p+2$ times the second row to the fourth row giving ${\begin{bmatrix}1&2&3&4\\0&0&0&1\\0&p&p+1&0\\0&-p&-2p+1&0\end{bmatrix}}$ Next, we add the third row to the fourth row to get ${\begin{bmatrix}1&2&3&4\\0&0&0&1\\0&p&p+1&0\\0&0&-p+2&0\end{bmatrix}}$ Swap the second row with the third row and the resulting third row with the fourth row (so this introduces no change in the sign since we swap twice and so multiply the determinant $(-1)^{2}=1$) giving ${\begin{bmatrix}1&2&3&4\\0&p&p+1&0\\0&0&-p+2&0\\0&0&0&1\end{bmatrix}}$ Now we take determinants and notice that the determinant of A is the same as the determinant of the above matrix. Since the above matrix is upper triangular, the determinant can be taken by multiplying the entries on the diagonal to give $\det(A)=\det \left({\begin{bmatrix}1&2&3&4\\0&p&p+1&0\\0&0&-p+2&0\\0&0&0&1\end{bmatrix}}\right)=p(-p+2)$ This is zero when $\displaystyle p=0$ or when $\displaystyle p=2.$

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