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Chapter 3: Q8E (page 165)

Question: In Exercise 8, determine the values of the parameter s for which the system has a unique solution, and describe the solution.

8.

\(\begin{array}{c}{\bf{3}}s{x_{\bf{1}}} + {\bf{5}}{x_{\bf{2}}} = {\bf{3}}\\12{x_{\bf{1}}} + {\bf{5}}s{x_{\bf{2}}} = {\bf{2}}\end{array}\)

Short Answer

Expert verified

The solution of the given system is unique for \(s \ne \pm 2\). For such a system, the solution is \({x_1} = \frac{{3s - 2}}{{3\left( {{s^2} - 4} \right)}}\), and \({x_2} = \frac{{2s - 12}}{{5\left( {{s^2} - 4} \right)}}\).

Step by step solution

01

Write the matrix form

The given system is equivalent to \(Ax = b\).

Here, \(A = \left( {\begin{array}{*{20}{c}}{3s}&5\\{12}&{5s}\end{array}} \right)\), \(x = \left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right)\), and \(b = \left( {\begin{array}{*{20}{c}}3\\2\end{array}} \right)\).

Then, \({A_1}\left( b \right) = \left( {\begin{array}{*{20}{c}}3&5\\2&{5s}\end{array}} \right)\), and \({A_2}\left( b \right) = \left( {\begin{array}{*{20}{c}}{3s}&3\\{12}&2\end{array}} \right)\).

02

Determine the value of s

Note that thesolution is unique for\(\det A \ne 0\).

\(\begin{array}{c}\det A = \left| {\begin{array}{*{20}{c}}{3s}&5\\{12}&{5s}\end{array}} \right|\\ = 15{s^2} - 60\\\det A = 15\left( {{s^2} - 4} \right)\end{array}\)

When \(\det A = 0\), you get:

\(\begin{array}{c}15\left( {{s^2} - 4} \right) = 0\\{s^2} = 4\\s = \pm 2\end{array}\)

Hence, the solution of the given system is unique for \(s \ne \pm 2\).

03

Use Cramer’s rule

For such a system, the solution is obtained by using Cramer鈥檚 rule, that is,

\({x_i} = \frac{{\det {A_i}\left( b \right)}}{{\det A}}\), \(i = 1,2\).

Hence,

\(\begin{array}{c}{x_1} = \frac{{\det {A_1}\left( b \right)}}{{\det A}}\\ = \frac{{\left| {\begin{array}{*{20}{c}}3&5\\2&{5s}\end{array}} \right|}}{{15\left( {{s^2} - 4} \right)}}\\ = \frac{{15s - 10}}{{15\left( {{s^2} - 4} \right)}}\\{x_1} = \frac{{3s - 2}}{{3\left( {{s^2} - 4} \right)}}\end{array}\)

\(\begin{array}{c}{x_2} = \frac{{\det {A_2}\left( b \right)}}{{\det A}}\\ = \frac{{\left| {\begin{array}{*{20}{c}}{3s}&3\\{12}&2\end{array}} \right|}}{{15\left( {{s^2} - 4} \right)}}\\ = \frac{{6s - 36}}{{15\left( {{s^2} - 4} \right)}}\\{x_2} = \frac{{2s - 12}}{{5\left( {{s^2} - 4} \right)}}\end{array}\)

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Most popular questions from this chapter

Find the determinant in Exercise 16, where \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{\bf{d}}&{\bf{e}}&{\bf{f}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right| = {\bf{7}}\].

16. \[\left| {\begin{array}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{{\bf{5d}}}&{{\bf{5e}}}&{{\bf{5f}}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{array}} \right|\]

Question: In Exercises 31鈥�36, mention an appropriate theorem in your explanation.

31. Show that if A is invertible, then \(det{\rm{ }}{A^{ - 1}} = \frac{1}{{det{\rm{ }}A}}\).

In Exercises 27 and 28, A and B are \[n \times n\] matrices. Mark each statement True or False. Justify each answer.

27. a. A row replacement operation does not affect the determinant of a matrix.

b. The determinant of A is the product of the pivots in any echelon form U of A, multiplied by \({\left( { - {\bf{1}}} \right)^r}\), where r is the number of row interchanges made during row reduction from A to U.

c. If the columns of A are linearly dependent, then \(det\left( A \right) = 0\).

d. \(det\left( {A + B} \right) = det{\rm{ }}A + det{\rm{ }}B\).

Construct a random \({\bf{4}} \times {\bf{4}}\) matrix A with integer entries between \( - {\bf{9}}\) and 9, and compare det A with det\({A^T}\), \(det\left( { - A} \right)\), \(det\left( {{\bf{2}}A} \right)\), and \(det\left( {{\bf{10}}A} \right)\). Repeat with two other random \({\bf{4}} \times {\bf{4}}\) integer matrices, and make conjectures about how these determinants are related. (Refer to Exercise 36 in Section 2.1.) Then check your conjectures with several random \({\bf{5}} \times {\bf{5}}\) and \({\bf{6}} \times {\bf{6}}\) integer matrices. Modify your conjectures, if necessary, and report your results.

The expansion of a \({\bf{3}} \times {\bf{3}}\) determinant can be remembered by the following device. Write a second type of the first two columns to the right of the matrix, and compute the determinant by multiplying entries on six diagonals.

Add the downward diagonal products and subtract the upward products. Use this method to compute the determinants in Exercises 15-18. Warning: This trick does not generalize in any reasonable way to \({\bf{4}} \times {\bf{4}}\) or larger matrices.

15. \(\left| {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{4}}\\{\bf{2}}&{\bf{3}}&{\bf{2}}\\{\bf{0}}&{\bf{5}}&{ - {\bf{2}}}\end{array}} \right|\)
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