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Use Theorem 3 (but not Theorem 4) to show that if two rows of a square matrix A are equal, then \(det A = 0\). The same is true for twocolumns. Why?

Short Answer

Expert verified

It is proved thatif two rows or columns of asquare matrixA are equal, then \(\det A = 0\).

Step by step solution

01

State the determinant of the matrix

According totheorem 3,if an interchange operation between any two rows in matrix A gives a new matrix B, \(\det {\rm{ }}B = - \det {\rm{ }}A\).

02

Step 2:Find the determinant of the matrix

An interchange between two rows or columnscan be written as

\(\det {\rm{ }}B = - \det {\rm{ }}A\).

If two rows or columns are the same, then theirinterchange does not change the matrix.Thus thedeterminant also does not change.

\(\det {\rm{ }}B = \det {\rm{ }}A\)

Both the cases,\(\det {\rm{ }}B = - \det {\rm{ }}A\)and\(\det {\rm{ }}B = \det {\rm{ }}A\), are satisfied only when\(\det {\rm{ }}A = 0\).

Hence, it is proved that if two rows or columns of a square matrix A are equal, then \(\det A = 0\).

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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 Exercise 14, compute the adjugate of the given matrix, and then use Theorem 8 to give the inverse of the matrix.

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

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.

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.

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

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

2. \(\begin{array}{l}4{x_1} + {x_2} = 6\\3{x_1} + 2{x_2} = 7\end{array}\)

Each equation in Exercises 1-4 illustrates a property of determinants. State the property

\(\left| {\begin{array}{*{20}{c}}{\bf{0}}&{\bf{5}}&{ - {\bf{2}}}\\{\bf{1}}&{ - {\bf{3}}}&{\bf{6}}\\{\bf{4}}&{ - {\bf{1}}}&{\bf{8}}\end{array}} \right| = - \left| {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{3}}}&{\bf{6}}\\{\bf{0}}&{\bf{5}}&{ - {\bf{2}}}\\{\bf{4}}&{ - {\bf{1}}}&{\bf{8}}\end{array}} \right|\)

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