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Found in: Page 165

### Linear Algebra and its Applications

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

1. The given statement is true.
2. The given statement is false.
3. The given statement is true.
4. The given statement is false.
See the step by step solution

## (a) Step 1: Mark the first statement true or 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.

## (b) Step 2: Mark the second statement true or false

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.

## (c) Step 3: Mark the third statement true or 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.

## (d) Step 4: Mark the fourth statement true or false

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}

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