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Chapter 3: Q42E (page 165)

Question 42: Let \(A = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\)and \(B = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\). Show that \(\det \left( {A + B} \right) = \det A + \det B\) if and only if \(a + d = 0\).

Short Answer

Expert verified

It is proved that \(\det \left( {A + B} \right) = \det A + \det B\) if and only if \(a + d = 0\).

Step by step solution

01

Determine the matrix \(A + B\)

Compute the matrix \(A + B\) as shown below:

\(\begin{array}{c}A + B = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + a}&{0 + b}\\{0 + c}&{1 + d}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{1 + a}&b\\c&{1 + d}\end{array}} \right]\end{array}\)

02

Show that \(\det \left( {A + B} \right) = \det A + \det B\) if and only if \(a + d = 0\)

The determinant of the matrix \(A + B\)is shown below:

\(\begin{array}{c}\det \left( {A + B} \right) = \left| {\begin{array}{*{20}{c}}{1 + a}&b\\c&{1 + d}\end{array}} \right|\\ = \left( {1 + a} \right)\left( {1 + d} \right) - cb\\ = 1 + a + d + ad - cb\\ = \det A + a + d + \det B\end{array}\)

Therefore, \(\det \left( {A + B} \right) = \det A + \det B\)if \(a + d = 0\).

Thus, it is proved that \(\det \left( {A + B} \right) = \det A + \det B\) if and only if \(a + d = 0\).

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Most popular questions from this chapter

In Exercises 24鈥�26, use determinants to decide if the set of vectors is linearly independent.

24. \(\left( {\begin{aligned}{*{20}{c}}4\\6\\2\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{ - 7}\\0\\7\end{aligned}} \right)\), \(\left( {\begin{aligned}{*{20}{c}}{ - 3}\\{ - 5}\\{ - 2}\end{aligned}} \right)\)

Question: Use Cramer鈥檚 rule to compute the solutions of the systems in Exercises1-6.

3. \(\begin{array}{c}3{x_1} - 2{x_2} = 3\\ - 4{x_1} + 6{x_2} = - 5\end{array}\)

Let \(A = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\) and let \(k\) be a scalar. Find a formula that relates \(\det kA\) to \(k\) and \(\det A\).

Find the determinants in Exercises 5-10 by row reduction to echelon form.

\(\left| {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{3}}&{ - {\bf{1}}}&{\bf{0}}&{ - {\bf{2}}}\\{\bf{0}}&{\bf{2}}&{ - {\bf{4}}}&{ - {\bf{2}}}&{ - {\bf{6}}}\\{ - {\bf{2}}}&{ - {\bf{6}}}&{\bf{2}}&{\bf{3}}&{{\bf{10}}}\\{\bf{1}}&{\bf{5}}&{ - {\bf{6}}}&{\bf{2}}&{ - {\bf{3}}}\\{\bf{0}}&{\bf{2}}&{ - {\bf{4}}}&{\bf{5}}&{\bf{9}}\end{array}} \right|\)

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.

atr

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.

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