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Chapter 2: Q30Q (page 93)

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

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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

Let \({{\bf{r}}_1} \ldots ,{{\bf{r}}_p}\) be vectors in \({\mathbb{R}^{\bf{n}}}\), and let Qbe an\(m \times n\)matrix. Write the matrix\(\left( {\begin{aligned}{*{20}{c}}{Q{{\bf{r}}_1}}& \cdots &{Q{{\bf{r}}_p}}\end{aligned}} \right)\)as a productof two matrices (neither of which is an identity matrix).

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 \(CA = {I_n}\)(the \(n \times n\) identity matrix). Show that the equation \(Ax = 0\) has only the trivial solution. Explain why Acannot have more columns than rows.

Use partitioned matrices to prove by induction that the product of two lower triangular matrices is also lower triangular. [Hint: \(A\left( {k + 1} \right) \times \left( {k + 1} \right)\) matrix \({A_1}\) can be written in the form below, where \[a\] is a scalar, v is in \({\mathbb{R}^k}\), and Ais a \(k \times k\) lower triangular matrix. See the study guide for help with induction.]

\({A_1} = \left[ {\begin{array}{*{20}{c}}a&{{0^T}}\\0&A\end{array}} \right]\).

[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.

  1. Display the submatrix of Afrom rows 15 to 20 and columns 5 to 10.
  2. Insert a \(5 \times 10\) matrix B into A, beginning at row 10 and column 20.
  3. Create a \(50 \times 50\) matrix of the form \(B = \left[ {\begin{array}{*{20}{c}}A&0\\0&{{A^T}}\end{array}} \right]\).

[Note: It may not be necessary to specify the zero blocks in B.]
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