In Exercises 31-36, mention an appropriate theorem in your explanation.

36. Find a formula for \(\det \left( {rA} \right)\) when \(A\) is an\(n \times n\) matrix.

Short Answer

Expert verified

The formula for \(\det \left( {rA} \right)\) is \({r^n}\det A\).

Step by step solution

01

Determine the formula for \(\det \left( {rA} \right)\)

Consider \(A\) as an\(n \times n\) matrix.

Each of the \(n\) rows can be factored by \(r\) as shown below:

\(\det \left( {rA} \right) = {r^n}\det A\)

02

CONCLUSION

Thus, the formula for \(\det \left( {rA} \right)\) is \({r^n}\det A\).

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Most popular questions from this chapter

Find the determinant in Exercise 17, where \[\left| {\begin{aligned}{*{20}{c}}{\bf{a}}&{\bf{b}}&{\bf{c}}\\{\bf{d}}&{\bf{e}}&{\bf{f}}\\{\bf{g}}&{\bf{h}}&{\bf{i}}\end{aligned}} \right| = {\bf{7}}\].

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

Question: In Exercise 16, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.

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

Compute the determinant in Exercise 5 using a cofactor expansion across the first row.

5. \(\left| {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{3}}&{ - {\bf{3}}}\\{\bf{4}}&{\bf{0}}&{\bf{3}}\\{\bf{6}}&{\bf{1}}&{\bf{5}}\end{aligned}} \right|\)

Question: In Exercise 13, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.

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

In Exercise 19-24, explore the effect of an elementary row operation on the determinant of a matrix. In each case, state the row operation and describe how it affects the determinant.

\(\left[ {\begin{array}{*{20}{c}}a&b&c\\{\bf{3}}&{\bf{2}}&{\bf{1}}\\{\bf{4}}&{\bf{5}}&{\bf{6}}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}{\bf{3}}&{\bf{2}}&{\bf{1}}\\a&b&c\\{\bf{4}}&{\bf{5}}&{\bf{6}}\end{array}} \right]\)

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