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Let \(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{2}}}&{ - {\bf{7}}}&{ - {\bf{9}}}\\{\bf{2}}&{\bf{5}}&{\bf{6}}\\{\bf{1}}&{\bf{3}}&{\bf{4}}\end{aligned}} \right)\). Find the third column of \({A^{ - {\bf{1}}}}\) without computing the other columns.

Short Answer

Expert verified

\(\left( {\begin{aligned}{*{20}{c}}3\\{ - 6}\\4\end{aligned}} \right)\)

Step by step solution

01

Find the expression \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{3}}}}\end{aligned}} \right)\)

The expression \(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right)\) can be calculated as shown below:

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 2}&{ - 7}&{ - 9}\\2&5&6\\1&3&4\end{aligned}\,\,\begin{aligned}{*{20}{c}}0\\0\\1\end{aligned}} \right)\)

02

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{3}}}}\end{aligned}} \right)\)

Interchange rows one and three, i.e., \({R_1} \leftrightarrow {R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\2&5&6\\{ - 2}&{ - 7}&{ - 9}\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\0\\0\end{aligned}} \right)\)

For row three, multiply row one by 2 and add it to row three, i.e., \({R_3} \to {R_3} + 2{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\2&5&6\\0&{ - 1}&{ - 1}\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\0\\2\end{aligned}} \right)\)

03

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{3}}}}\end{aligned}} \right)\)

For row two, multiply row one by 2 and subtract it from row two, i.e., \({R_2} \to {R_2} - 2{R_1}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\0&{ - 1}&{ - 2}\\0&{ - 1}&{ - 1}\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\{ - 2}\\2\end{aligned}} \right)\)

04

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{3}}}}\end{aligned}} \right)\)

At row three, subtract row two from row three, i.e., \({R_3} \to {R_3} - {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\0&{ - 1}&{ - 2}\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\{ - 2}\\4\end{aligned}} \right)\)

05

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{3}}}}\end{aligned}} \right)\)

At row two, multiply row two by 2 and add it to row two, i.e., \({R_2} \to {R_2} + 2{R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&4\\0&{ - 1}&0\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}1\\6\\4\end{aligned}} \right)\)

06

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{3}}}}\end{aligned}} \right)\)

At row one, multiply row three by 4 and subtract it from row one, i.e., \({R_1} \to {R_1} - 4{R_3}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&3&0\\0&{ - 1}&0\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}{ - 15}\\6\\4\end{aligned}} \right)\)

At row one, multiply row two by 3 and add it to row one, i.e., \({R_1} \to {R_1} + 3{R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{ - 1}&0\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}3\\6\\4\end{aligned}} \right)\)

07

Apply row operation to the matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{3}}}}\end{aligned}} \right)\)

At row two, multiply row two by \( - 1\), i.e., \({R_2} \to - {R_2}\).

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{aligned}\,\,\begin{aligned}{*{20}{c}}3\\{ - 6}\\4\end{aligned}} \right)\)

So, the third column of the inverse matrix is \(\left( {\begin{aligned}{*{20}{c}}3\\{ - 6}\\4\end{aligned}} \right)\).

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Most popular questions from this chapter

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

36. Write the command(s) that will create a \(6 \times 4\) matrix with random entries. In what range of numbers do the entries lie? Tell how to create a \(3 \times 3\) matrix with random integer entries between \( - {\bf{9}}\) and 9. (Hint:If xis a random number such that 0 < x < 1, then \( - 9.5 < 19\left( {x - .5} \right) < 9.5\).

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).

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