# A and B are n \times nn\times n matrices. Mark each statement True or Fals

A and B are $$\displaystyle{n}\times\cap\times{n}$$ matrices. Mark each statement True or False. Justify your answer.
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 $$\displaystyle{\left(-{1}\right)}^{{{r}}}$$ , 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 $$\displaystyle{A}={0}.{d}.{\det{{\left({A}+{B}\right)}}}={\det{{A}}}+{\det{{B}}}$$.

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Theirl1972
(a)Thestatementistrue. Adding a multiple of one row to another row does not change the value of the determinant according to Theorem 3 (a) (a)True.
(b)Thestatementisfalse. This statement is only true for invertible matrices since their determinants are not equal to zero. Since pivots are defined as the first nonzero entries in each of the rows of the row echelon form of a matrix, it follows that their product is also a nonzero number. Thus, determinant of a singular matrix can never be the product of its pivots multiplied by $$\displaystyle{\left(-{1}\right)}^{{{r}}}$$, where r is the number of row interchanges made during its transformation to its row echelon form., (b) False
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Edward Belanger
(c)Thestatementistrue. There are multiple ways to prove this. Since the columns of A are linearly dependent, it follows that the rows of $$\displaystyle{A}^{{{T}}}$$ are linearly dependent as well. Therefore, its row echelon form contains a zero row. Keep in mind that the row echelon form of a matrix is always an upper triangular matrix. Since the determinant of the row echelon form of $$\displaystyle{A}^{{{T}}}$$ has the same determinant as the matrix $$\displaystyle{A}^{{{T}}}$$ , and the determinant of a triangular matrix is equal to the product of its diagonal entries, it follows that det $$\displaystyle{\left({4}\right)}={0}$$ (at least one of the rows of its row echelon form is a zero row). Now, acoording to Theorem 5, it follows that det $$\displaystyle{\left({A}\right)}={0}$$. (c) True
user_27qwe

(d) Thestatementis false. Consider the following example of $$2\times2$$ matrices.

$$det(A)=\left|\begin{matrix} 1 & 1 \\ 0 & 1\end{matrix}\right|=1,\ \det(B)=\left|\begin{matrix} 1 & 0 \\ 1 & 1\end{matrix}\right|=1,\ \det(A+B)=\left|\begin{matrix} 2 & 1 \\ 1 & 2\end{matrix}\right|=3$$

Therefore, it follows that generally $$det (A)+det(B)\neq det(A+B)$$.,

(d) False