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Find the inverses of the matrices in Exercises 29–32, if they exist. Use the algorithm introduced in this section.

30. \(\left( {\begin{aligned}{*{20}{c}}5&{10}\\4&7\end{aligned}} \right)\)

Short Answer

Expert verified

The inverse of the matrix is \(\left( {\begin{aligned}{*{20}{c}}{ - 7/5}&2\\{4/5}&{ - 1}\end{aligned}} \right)\).

Step by step solution

01

Write the algorithm for obtaining \({A^{ - 1}}\)

The inverse of an\(m \times m\)matrix A can be computed using theaugmented matrix \(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\), where\(I\)is theidentity matrix. Matrix Ahas an inverse if\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)isrow equivalent to \(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\).

02

Obtain the inverse of matrix A

Consider the matrix\(A = \left( {\begin{aligned}{*{20}{c}}5&{10}\\4&7\end{aligned}} \right)\).

Write the augmented matrix\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right)\)as shown below:

\(\left( {\begin{aligned}{*{20}{c}}A&I\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}5&{10}&1&0\\4&7&0&1\end{aligned}} \right)\)

Row reduce the augmented matrix as shown below:

Multiply row one by\(\frac{1}{5}\).

\(\left( {\begin{aligned}{*{20}{c}}5&{10}&1&0\\4&7&0&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&2&{1/5}&0\\4&7&0&1\end{aligned}} \right)\)

Use the\({x_1}\)term in the first equation to eliminate the\(4{x_1}\)term from the second equation. Add\( - 4\)times row one to row two.

\(\left( {\begin{aligned}{*{20}{c}}1&2&{1/5}&0\\4&7&0&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&2&{1/5}&0\\0&{ - 1}&{ - 4/5}&1\end{aligned}} \right)\)

Multiply row two by\( - 1\).

\(\left( {\begin{aligned}{*{20}{c}}1&2&{1/5}&0\\0&{ - 1}&{ - 4/5}&1\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&2&{1/5}&0\\0&1&{4/5}&{ - 1}\end{aligned}} \right)\)

Use the\({x_2}\)term in the second equation to eliminate the\(2{x_2}\)term from the first equation. Add\( - 2\)times row two to row one.

\(\left( {\begin{aligned}{*{20}{c}}1&2&{1/5}&0\\0&1&2&{ - 1}\end{aligned}} \right) \sim \left( {\begin{aligned}{*{20}{c}}1&0&{ - 7/5}&2\\0&1&{4/5}&{ - 1}\end{aligned}} \right)\)

By comparing with\(\left( {\begin{aligned}{*{20}{c}}I&{{A^{ - 1}}}\end{aligned}} \right)\), you get\({A^{ - 1}} = \left( {\begin{aligned}{*{20}{c}}{ - 7/5}&2\\{4/5}&{ - 1}\end{aligned}} \right)\).

Thus, theinverse of the matrix is \(\left( {\begin{aligned}{*{20}{c}}{ - 7/5}&2\\{4/5}&{ - 1}\end{aligned}} \right)\).

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Most popular questions from this chapter

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, 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. \[\left[ {\begin{array}{*{20}{c}}X&{\bf{0}}\\Y&Z\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&{\bf{0}}\\B&C\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\{\bf{0}}&I\end{array}} \right]\]

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.

In the rest of this exercise set and in those to follow, you should assume that each matrix expression is defined. That is, the sizes of the matrices (and vectors) involved match appropriately.

Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{1}}}\\{\bf{5}}&{ - {\bf{2}}}\end{aligned}} \right)\). Compute \({\bf{3}}{I_{\bf{2}}} - A\) and \(\left( {{\bf{3}}{I_{\bf{2}}}} \right)A\).

Suppose Ais an \(m \times n\) matrix and there exist \(n \times m\) matrices C and D such that \(CA = {I_n}\) and \(AD = {I_m}\). Prove that \(m = n\) and \(C = D\). (Hint: Think about the product CAD.)

In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. In Exercises 5–8, 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.]

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

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