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Chapter 3: Q28Q (page 165)

Compute the determinants of the elementary matrices given in Exercise 25-30.

28. \(\left( {\begin{aligned}{*{20}{c}}k&0&0\\0&1&0\\0&0&1\end{aligned}} \right)\).

Short Answer

Expert verified

The determinant of the matrix is \(k\).

Step by step solution

01

Compute the determinant of the elementary matrix

If A is a triangular matrix, then according to theorem 2, det A is the product of the entries on its main diagonal.

The determinant of the matrix is the product of the diagonal entries because the matrix is triangular.

\(\begin{aligned}{c}\left| {\begin{aligned}{*{20}{c}}k&0&0\\0&1&0\\0&0&1\end{aligned}} \right| = \left( k \right)\left( 1 \right)\left( 1 \right)\\ = k\end{aligned}\)

Thus, the determinant of the matrix is \(k\).

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Most popular questions from this chapter

Each equation in Exercises 1-4 illustrates a property of determinants. State the property

\(\left| {\begin{array}{*{20}{c}}{\bf{0}}&{\bf{5}}&{ - {\bf{2}}}\\{\bf{1}}&{ - {\bf{3}}}&{\bf{6}}\\{\bf{4}}&{ - {\bf{1}}}&{\bf{8}}\end{array}} \right| = - \left| {\begin{array}{*{20}{c}}{\bf{1}}&{ - {\bf{3}}}&{\bf{6}}\\{\bf{0}}&{\bf{5}}&{ - {\bf{2}}}\\{\bf{4}}&{ - {\bf{1}}}&{\bf{8}}\end{array}} \right|\)

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

In Exercises 39 and 40, \(A\) is an \(n \times n\) matrix. Mark each statement True or False. Justify each answer.

39.

a. An \(n \times n\) determinant is defined by determinants of \(\left( {n - 1} \right) \times \left( {n - 1} \right)\) submatrices.

b. The \(\left( {i,j} \right)\)-cofactor of a matrix \(A\) is the matrix \({A_{ij}}\) obtained by deleting from A its \(i{\mathop{\rm th}\nolimits} \) row and \[j{\mathop{\rm th}\nolimits} \]column.

Question: In Exercises 31鈥�36, mention an appropriate theorem in your explanation.

34. Let A and P be square matrices, with P invertible. Show that \(det\left( {PA{P^{ - {\bf{1}}}}} \right) = det{\rm{ }}A\).

Question: 6. Use Cramer鈥檚 rule to compute the solution of the following system.

\(\begin{array}{c}{x_{\bf{1}}} + {\bf{3}}{x_{\bf{2}}} + \,{x_{\bf{3}}} = {\bf{4}}\\ - {x_{\bf{1}}} + \,\,\,\,\,\,\,\,\,\,{\bf{2}}{x_{\bf{3}}} = {\bf{2}}\\{\bf{3}}{x_{\bf{1}}} + \,{x_{\bf{2}}}\,\,\,\,\,\,\,\,\,\,\,\,\, = {\bf{2}}\end{array}\)
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