Question:In Exercises 27 and 28, A and B are \(n \times n\) matrices. Mark each statement True or False. Justify each answer.

28. a. If three row interchanges are made in succession, then the new determinant equals the old determinant.

b. The determinant of A is the product of the diagonal entries in A.

c. If det A is zero, then two rows or two columns are the same, or a row or a column is zero.

d. \(det{\rm{ }}{A^{ - 1}} = \left( { - 1} \right)det{\rm{ }}A\).

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

According to the statement, three rows are interchanged in succession.

For one interchange, it gives a negative sign with the samedeterminant.For two interchanges, it gives a positive sign with the same determinant.

And forthree interchanges,the sign of the determinant becomes negative again.

This means the new determinant cannot be equal to the old determinant.

Therefore, statement (a) is false.

If asquare matrixA is row reduced to echelon form, such that it becomes atriangular matrix, then the determinant of matrix A is the product of the diagonal entries.

If the reduced matrix is not triangular, then the determinant cannot be the product of diagonal entries.

So, ifA is a triangular matrix, then the determinant is the product of diagonal entries.

The given statement is not always true.

Therefore, statement (b) is false.

Consider a matrix in which two rows or columns are not the same, or a row or column is not zero.

Let the matrix be\(A = \left[ {\begin{array}{*{20}{c}}1&3&2\\0&0&5\\0&0&{ - 3}\end{array}} \right]\).

Obtain the determinant of the matrix\(A = \left[ {\begin{array}{*{20}{c}}1&3&2\\0&0&5\\0&0&{ - 3}\end{array}} \right]\).

\(\begin{aligned}{}\det A &= \left| {\begin{aligned}{{}}1&3&2\\0&0&5\\0&0&{ - 3}\end{aligned}} \right|\\ &= 1\left( {0\left( { - 3} \right) - 0\left( 5 \right)} \right) - 3\left( {0\left( { - 3} \right) - 5\left( 0 \right)} \right) + 2\left( {0 - 0} \right)\\ &= 0 - 0 + 0\\ &= 0\end{aligned}\)

So, the determinant of A is 0.

It means the given statement is not always true.

Therefore, statement (c) is false.

Check the equality\[\det {\rm{ }}{A^{ - 1}} = \left( { - 1} \right)\det {\rm{ }}A\].

Let matrix A be \(A = \left[ {\begin{array}{*{20}{c}}1&2\\0&1\end{array}} \right]\).

Obtain the determinant of matrix A.

\(\begin{aligned}{}\det \left( A \right) &= \left| {\begin{aligned}{{}}1&2\\0&1\end{aligned}} \right|\\ &= 1 - 0\\ &= 1\end{aligned}\)

The inverse of matrix A is shown below:

\(\begin{aligned}{}{A^{ - 1}} &= \frac{1}{{1\left( 1 \right) - 2\left( 0 \right)}}\left( {\begin{aligned}{{}}1&{ - 2}\\0&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}}1&{ - 2}\\0&1\end{aligned}} \right)\end{aligned}\)

Obtain the determinant of the inverse of the matrix.

\(\begin{aligned}{}\det \left( {{A^{ - 1}}} \right) &= \left| {\begin{aligned}{{}}1&{ - 2}\\0&1\end{aligned}} \right|\\ &= 1 + 2\left( 0 \right)\\ &= 1\end{aligned}\)

Here, the sign is the same, that is,\(\det {\rm{ }}{A^{ - 1}} = \left( { - 1} \right)\det {\rm{ }}A\).

It means the given statement is not always true.

Therefore, statement (d) is false.