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Chapter 2: Q2.3-30Q (page 93)

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.

Short Answer

Expert verified

The transformation is invertible.

Step by step solution

01

State the invertible matrix theorem

The statements are identical according to the invertible matrix theorem, as shown below:

  1. The matrix is invertible.
  2. For some b, the matrix equation\[A{\bf{x}} = {\bf{b}}\]has no unique solution (more than one solution).
  3. The linear transformation\(x| \to Ax\)is one-to-one.
  4. The mapping of \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\) is equivalent to the linear transformation \(x| \to Ax\).
02

Define the transformation

According to the invertible matrix theorem, if the matrix equation\[A{\bf{x}} = {\bf{b}}\]has more than one solution, the transformation is one-to-one.

From the given statement, the linear transformation\(x| \to Ax\)is one-to-one. So, the matrix equation must have more than one solution for some b, and matrix A should be invertible.

Therefore, both the matrix and the transformation are invertible.

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Most popular questions from this chapter

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 \(E = \left( {\begin{aligned}{*{20}{c}}{ - 5}\\3\end{aligned}} \right)\)

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

In Exercise 10 mark each statement True or False. Justify each answer.

10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.

b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.

c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.

d. If A can be row reduced to the identity matrix, then A must be invertible.

e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).

If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.

In Exercises 33 and 34, Tis a linear transformation from \({\mathbb{R}^2}\) into \({\mathbb{R}^2}\). Show that T is invertible and find a formula for \({T^{ - 1}}\).

33. \(T\left( {{x_1},{x_2}} \right) = \left( { - 5{x_1} + 9{x_2},4{x_1} - 7{x_2}} \right)\)

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

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