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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 1โ€“9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5โ€“8, find formulas for X, Y, and Zin terms of A, B, and C, and justify your calculations. In some cases, you may need to make assumptions about the size of a matrix in order to produce a formula. [Hint:Compute the product on the left, and set it equal to the right side.]

6. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

In Exercises 1โ€“9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1โ€“4.

2. \[\left[ {\begin{array}{*{20}{c}}E&{\bf{0}}\\{\bf{0}}&F\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\]

Let T be a linear transformation that maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\). Is \({T^{ - 1}}\) also one-to-one?

Suppose block matrix \(A\) on the left side of (7) is invertible and \({A_{{\bf{11}}}}\) is invertible. Show that the Schur component \(S\) of \({A_{{\bf{11}}}}\) is invertible. [Hint: The outside factors on the right side of (7) are always invertible. Verify this.] When \(A\) and \({A_{{\bf{11}}}}\) are invertible, (7) leads to a formula for \({A^{ - {\bf{1}}}}\), using \({S^{ - {\bf{1}}}}\) \(A_{{\bf{11}}}^{ - {\bf{1}}}\), and the other entries in \(A\).

In Exercises 1 and 2, compute each matrix sum or product if it is defined. If an expression is undefined, explain why. Let

\(A = \left( {\begin{aligned}{*{20}{c}}2&0&{ - 1}\\4&{ - 5}&2\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}7&{ - 5}&1\\1&{ - 4}&{ - 3}\end{aligned}} \right)\), \(C = \left( {\begin{aligned}{*{20}{c}}1&2\\{ - 2}&1\end{aligned}} \right)\), \(D = \left( {\begin{aligned}{*{20}{c}}3&5\\{ - 1}&4\end{aligned}} \right)\) and \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

\(A + 2B\), \(3C - E\), \(CB\), \(EB\).

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