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Q27Q

Expert-verifiedFound in: Page 165

Book edition
5th

Author(s)
David C. Lay, Steven R. Lay and Judi J. McDonald

Pages
483 pages

ISBN
978-03219822384

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

- The given statement is true.
- The given statement is false.
- The given statement is true.
- The given statement is false.

Recall Theorem 3:

Consider a **square matrix** *A*. If any row of matrix *A* has a multiple that is added to another row to give a new matrix *B, *then the **determinant** of matrix *A* must be equal to the determinant of matrix *B*.

\(\det \left( A \right) = \det \left( B \right)\)

Therefore, statement (a) is true.

Recall that when *A* is **invertible,** then the determinant of matrix *A* can be written as shown below:

\(\det {\rm{ }}A = {\left( { - 1} \right)^r} \cdot \left( {{\rm{product of pivots in }}U} \right)\)

If it is not invertible, then the determinant of the matrix is 0.

Also, scaling a row by a constant (non-zero) during conversion to an echelon form may affect the value of the determinant.

Therefore, statement (b) is false.

The set of vectors \({{\bf{b}}_1},{{\bf{b}}_2},{{\bf{b}}_3},...,{{\bf{b}}_n}\) is said to be **linearly independent** if the determinant of the matrix \(\left[ {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}& \cdots &{{{\bf{b}}_n}}\end{aligned}} \right]\) is 0 (or \(\left| {\begin{aligned}{*{20}{c}}{{{\bf{b}}_1}}&{{{\bf{b}}_2}}& \cdots &{{{\bf{b}}_n}}\end{aligned}} \right| = 0\)).

So, the columns of matrix *A* are linearly independent, and the determinant of the matrix is zero.

Therefore, statement (c) is true.

Check the equality \(\det \left( {A + B} \right) = \det {\rm{ }}A + \det {\rm{ }}B\)**.**

** **

Let the matrices be \(A = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right]\) and \(B = \left[ {\begin{aligned}{*{20}{c}}u&v\\w&x\end{aligned}} \right]\).

Obtain the determinant of each matrix.

\(\begin{aligned}{c}\det {\rm{ }}A = \left| {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right|\\ = ad - bc\end{aligned}\)

And

\(\begin{aligned}{c}\det {\rm{ }}B = \left| {\begin{aligned}{*{20}{c}}u&v\\w&x\end{aligned}} \right|\\ = ux - vw\end{aligned}\)

Add both the determinants.

\(\begin{aligned}{c}\det {\rm{ }}A + \det {\rm{ }}B = \left( {ad - bc} \right) + \left( {ux - vw} \right)\\ = ad - bc + ux - vw\end{aligned}\)

Thus, \(\det {\rm{ }}A + \det {\rm{ }}B = ad - bc + ux - vw\).

Add both the matrices as shown below:

\(\begin{aligned}{c}A + B = \left[ {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right] + \left[ {\begin{aligned}{*{20}{c}}u&v\\w&x\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}{a + u}&{b + v}\\{c + w}&{d + x}\end{aligned}} \right]\end{aligned}\)

Obtain the determinant of \(A + B = \left[ {\begin{aligned}{*{20}{c}}{a + u}&{b + v}\\{c + w}&{d + x}\end{aligned}} \right]\).

\(\begin{aligned}{c}\det \left( {A + B} \right) = \left| {\begin{aligned}{*{20}{c}}{a + u}&{b + v}\\{c + w}&{d + x}\end{aligned}} \right|\\ = \left( {a + u} \right)\left( {d + x} \right) - \left( {c + w} \right)\left( {b + v} \right)\\ = ad + ax + ud + ux - cb - cv - bw - vw\end{aligned}\)

It is observed that \(\det \left( {A + B} \right) \ne \det {\rm{ }}A + \det {\rm{ }}B\).

Therefore, it is not true that \(\det \left( {A + B} \right) = \det {\rm{ }}A + \det {\rm{ }}B\).

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