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Chapter 2: Q2.3-32Q (page 93)

Suppose Ais an \(n \times n\) matrix with the property that the equation \(Ax = 0\)has only the trivial solution. Without using the Invertible Matrix Theorem, explain directly why the equation \(Ax = b\) must have a solution for each b in \({\mathbb{R}^n}\).

Short Answer

Expert verified

The equation \(Ax = b\) has a solution for each b in \({\mathbb{R}^n}\).

Step by step solution

01

Show that the equation \[Ax = {\mathop{\rm b}\nolimits} \] has only a trivial solution

If \(Ax = 0\) has only a trivial solution, A must have a pivot in each of its \(n\) columns.

02

Explain that the equation Ax = b has exactly one solution

Theorem 4states that if \(A\) is an \({\rm{m}} \times n\) matrix, then the following statements are equivalent.

  1. For each \({\mathop{\rm b}\nolimits} \) in \({\mathbb{R}^m}\), the equation \(Ax = b\) has a solution.
  2. Each \({\mathop{\rm b}\nolimits} \) in \({\mathbb{R}^m}\) is a linear combination of the columns of A.
  3. The columns of \(A\) span .
  4. \(A\)has a pivot position in every row.

Since Ais a square matrix, itmust have a pivot in each row. According to theorem 4, the equation \(Ax = b\) has a solution for each b in \({\mathbb{R}^n}\).

Thus, the equation \(Ax = b\) has a solution for each b in \({\mathbb{R}^n}\).

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Most popular questions from this chapter

Suppose a linear transformation \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) has the property that \(T\left( {\mathop{\rm u}\nolimits} \right) = T\left( {\mathop{\rm v}\nolimits} \right)\) for some pair of distinct vectors u and v in \({\mathbb{R}^n}\). Can Tmap \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\)? Why or why not?

Suppose the third column of Bis the sum of the first two columns. What can you say about the third column of AB? Why?

Suppose \({A_{{\bf{11}}}}\) is an invertible matrix. Find matrices Xand Ysuch that the product below has the form indicated. Also,compute \({B_{{\bf{22}}}}\). [Hint:Compute the product on the left, and setit equal to the right side.]

\[\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}&{\bf{0}}\\X&I&{\bf{0}}\\Y&{\bf{0}}&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{A_{{\bf{1}}1}}}&{{A_{{\bf{1}}2}}}\\{{A_{{\bf{2}}1}}}&{{A_{{\bf{2}}2}}}\\{{A_{{\bf{3}}1}}}&{{A_{{\bf{3}}2}}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{{B_{11}}}&{{B_{12}}}\\{\bf{0}}&{{B_{22}}}\\{\bf{0}}&{{B_{32}}}\end{array}} \right]\]

a. Verify that \({A^2} = I\) when \(A = \left[ {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right]\).

b. Use partitioned matrices to show that \({M^2} = I\) when\(M = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\3&{ - 1}&0&0\\1&0&{ - 1}&0\\0&1&{ - 3}&1\end{array}} \right]\).

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 \({I_n}\).
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