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Chapter 3: Q13SE (page 165)

Question: 13. Show that if A is invertible, then adj A is invertible, and \({\left( {adj\,A} \right)^{ - {\bf{1}}}} = \frac{{\bf{1}}}{{detA}}A\).

Short Answer

Expert verified

Hence, adj Ais invertible and \({\left( {{\rm{adj}}\,A} \right)^{ - 1}} = \frac{1}{{\det A}}A\).

Step by step solution

01

Use the definition of invertible

Given, A is invertible. Therefore,

\(A{A^{ - 1}} = {A^{ - 1}}A = I\).

Also, \({A^{ - 1}}\) is invertible and \({\left( {{A^{ - 1}}} \right)^{ - 1}} = A\).

02

Use the inverse formula

By inverse formula,

\({A^{ - 1}} = \frac{1}{{\det A}}{\rm{adj}}\,A\)

This implies \({\rm{adj}}\,A\) is also invertible.

03

 Perform the substitution

\(\begin{array}{c}{\left( {{A^{ - 1}}} \right)^{ - 1}} = A\\{\left( {\frac{1}{{\det A}}{\rm{adj}}\,A} \right)^{ - 1}} = A\\\det A{\left( {{\rm{adj}}\,A} \right)^{ - 1}} = A\\{\left( {{\rm{adj}}\,A} \right)^{ - 1}} = \frac{1}{{\det A}}A\end{array}\)

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Most popular questions from this chapter

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?

Question: 15. Let A, B, C, and D be \(n \times n\) matrices with A invertible.

  1. Find matrices X and Y to produce the block LU factorization \(\left( {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right) = \left( {\begin{array}{*{20}{c}}I&{\bf{0}}\\X&I\end{array}} \right)\left( {\begin{array}{*{20}{c}}A&B\\{\bf{0}}&Y\end{array}} \right)\)and then show that \({\bf{det}}\,\left( {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right) = \left( {{\bf{det}}\,A} \right) \cdot det\left( {D - C{A^{ - {\bf{1}}}}B} \right)\)
  1. Show that if \(AC = CA\), then \({\bf{det}}\,\left( {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right) = {\bf{det}}\,\left( {AD - CB} \right)\)

In Exercise 19-24, explore the effect of an elementary row operation on the determinant of a matrix. In each case, state the row operation and describe how it affects the determinant.

\[\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{0}}&{\bf{1}}\\{ - {\bf{3}}}&{\bf{4}}&{ - {\bf{4}}}\\{\bf{2}}&{ - {\bf{3}}}&{\bf{1}}\end{array}} \right],\left[ {\begin{array}{*{20}{c}}k&{\bf{0}}&k\\{ - {\bf{3}}}&{\bf{4}}&{ - {\bf{4}}}\\{\bf{2}}&{ - {\bf{3}}}&{\bf{1}}\end{array}} \right]\]

Find the determinants in Exercises 5-10 by row reduction to echelon form.

\(\left| {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{ - {\bf{4}}}\\{ - {\bf{1}}}&{ - {\bf{4}}}&{\bf{5}}\\{ - {\bf{2}}}&{ - {\bf{8}}}&{\bf{7}}\end{array}} \right|\)

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. \(det\left( {A + B} \right) = det{\rm{ }}A + det{\rm{ }}B\).
See all solutions

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