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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

Solve the equation \(AB = BC\) for A, assuming that A, B, and C are square and Bis invertible.

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 \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

Let \(A = \left[ {\begin{array}{*{20}{c}}B&{\bf{0}}\\{\bf{0}}&C\end{array}} \right]\), where \(B\) and \(C\) are square. Show that \(A\)is invertible if an only if both \(B\) and \(C\) are invertible.

If Ais an \(n \times n\) matrix and the transformation \({\bf{x}}| \to A{\bf{x}}\) is one-to-one, what else can you say about this transformation? Justify your answer.

Use matrix algebra to show that if A is invertible and D satisfies \(AD = I\) then \(D = {A^{ - {\bf{1}}}}\).

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