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Chapter 5: Q5.2-26E (page 267)

Question: Let \(A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\). Use formula (1) for a determinant (given before Example 2) to show that \(\det A = ad - bc\). Consider two cases: \(a \ne 0\) and \(a = 0\).

Short Answer

Expert verified

It is proved that \({\rm{det}}A = ad - cb\) when \(a \ne 0\) and \(a = 0\).

Step by step solution

01

Consider \(a \ne 0\) and determine the determinant

Consider \(a \ne 0\), apply the row operations on the matrix \(A\).

\[\begin{array}A = \left( {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}a&b\\0&{d - c{a^{ - 1}}b}\end{array}} \right)\;\;\;\;\;\;\;\;\;\;{R_2} \to {R_2} - \frac{1}{a}{R_1}\\ = U\end{array}\]

Find the determinant of \(A\).

\(\begin{array}{\rm{det}}A = a\left( {d - c{a^{ - 1}}b} \right) - b\left( 0 \right)\\ = ad - cb\end{array}\)

Thus, it is proved that \({\rm{det}}A = ad - cb\) when \(a \ne 0\).

02

Consider \(a = 0\) and determine the determinant

Consider \(a = 0\), apply the row operations on the matrix \(A\).

\[\begin{array}A = \left( {\begin{array}{*{20}{c}}0&b\\c&d\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}c&d\\0&b\end{array}} \right)\;\;\;\;\;\;\;\;\;\;{R_2} \leftrightarrow {R_1}\\ = U\end{array}\]

Find the determinant of \(A\).

\[\begin{array}{\rm{det}}A = {\left( { - 1} \right)^1}\left( {cb} \right)\\ = 0 - bc\\ = ad - bc\end{array}\]

Thus, it is proved that \({\rm{det}}A = ad - cb\) when \(a = 0\).

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Most popular questions from this chapter

Exercises 19–23 concern the polynomial \(p\left( t \right) = {a_{\bf{0}}} + {a_{\bf{1}}}t + ... + {a_{n - {\bf{1}}}}{t^{n - {\bf{1}}}} + {t^n}\) and \(n \times n\) matrix \({C_p}\) called the companion matrix of \(p\): \({C_p} = \left( {\begin{aligned}{*{20}{c}}{\bf{0}}&{\bf{1}}&{\bf{0}}&{...}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}&{}&{\bf{0}}\\:&{}&{}&{}&:\\{\bf{0}}&{\bf{0}}&{\bf{0}}&{}&{\bf{1}}\\{ - {a_{\bf{0}}}}&{ - {a_{\bf{1}}}}&{ - {a_{\bf{2}}}}&{...}&{ - {a_{n - {\bf{1}}}}}\end{aligned}} \right)\).

20. Let \(p\left( t \right){\bf{ = }}\left( {t{\bf{ - 2}}} \right)\left( {t{\bf{ - 3}}} \right)\left( {t{\bf{ - 4}}} \right){\bf{ = - 24 + 26}}t{\bf{ - 9}}{t^{\bf{2}}}{\bf{ + }}{t^{\bf{3}}}\). Write the companion matrix for \(p\left( t \right)\), and use techniques from chapter \({\bf{3}}\) to find the characteristic polynomial.

Define\(T:{{\rm P}_3} \to {\mathbb{R}^4}\)by\(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 3} \right)}\\{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 1 \right)}\\{{\bf{p}}\left( 3 \right)}\end{aligned}} \right)\).

  1. Show that \(T\) is a linear transformation.
  2. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2},{t^3}} \right\}\)for \({{\rm P}_3}\)and the standard basis for \({\mathbb{R}^4}\).

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

12. \(\left[ {\begin{array}{*{20}{c}}- 1&0&1\\- 3&4&1\\0&0&2\end{array}} \right]\)

(M)Use a matrix program to diagonalize

\(A = \left( {\begin{aligned}{*{20}{c}}{ - 3}&{ - 2}&0\\{14}&7&{ - 1}\\{ - 6}&{ - 3}&1\end{aligned}} \right)\)

If possible. Use the eigenvalue command to create the diagonal matrix \(D\). If the program has a command that produces eigenvectors, use it to create an invertible matrix \(P\). Then compute \(AP - PD\) and \(PD{P^{{\bf{ - 1}}}}\). Discuss your results.

Assume the mapping\(T:{{\rm P}_2} \to {{\rm P}_{\bf{2}}}\)defined by \(T\left( {{a_0} + {a_1}t + {a_2}{t^2}} \right) = 3{a_0} + \left( {5{a_0} - 2{a_1}} \right)t + \left( {4{a_1} + {a_2}} \right){t^2}\) is linear. Find the matrix representation of\(T\) relative to the bases \(B = \left\{ {1,t,{t^2}} \right\}\).
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