Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Unless otherwise specified, assume that all matrices in these exercises are \(n \times n\). Determine which of the matrices in Exercises 1-10 are invertible. Use a few calculations as possible. Justify your answer.

7. \(\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 3}&0&1\\3&5&8&{ - 3}\\{ - 2}&{ - 6}&3&2\\0&{ - 1}&2&1\end{array}} \right]\)

Short Answer

Expert verified

The matrix \(\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 3}&0&1\\3&5&8&{ - 3}\\{ - 2}&{ - 6}&3&2\\0&{ - 1}&2&1\end{array}} \right]\) is invertible.

Step by step solution

01

State the invertible matrix theorem

Let Abe a square \(n \times n\) matrix. Then the following statements are equivalent.

For a given matrix A, the statements are either all true or all false.

  1. Ais an invertible matrix.
  2. Ais row equivalent to the identity matrix of the \(n \times n\) matrix.
  3. Ahas n pivot positions.
  4. The equation Ax = 0 has only a trivial solution.
  5. The columns of A form a linearly independent set.
  6. The linear transformation \(x \mapsto Ax\) is one-to-one.
  7. The equation \(Ax = b\) has at least one solution for each b in \({\mathbb{R}^n}\).
  8. The columns of Aspan \({\mathbb{R}^n}\).
  9. The linear transformation \(x \mapsto Ax\) maps \({\mathbb{R}^n}\) onto \({\mathbb{R}^n}\).
  10. There is an \(n \times n\) matrix Csuch that CA = I.
  11. There is an \(n \times n\) matrix Dsuch that DA = I.
  12. \({A^T}\) is an invertible matrix.
02

Apply the row operation

At row one, multiply row one by \( - 1\).

\(\left[ {\begin{array}{*{20}{c}}1&3&0&{ - 1}\\3&5&8&{ - 3}\\{ - 2}&{ - 6}&3&2\\0&{ - 1}&2&1\end{array}} \right]\)

At row two, multiply row one by 3 and subtract it from row two. And at row three, multiply row one by 2 and add it to row three.

\(\left[ {\begin{array}{*{20}{c}}1&3&0&{ - 1}\\0&{ - 4}&8&0\\0&0&3&0\\0&{ - 1}&2&1\end{array}} \right]\)

At row four, multiply row two by 1 and add it to row four.

\(\left[ {\begin{array}{*{20}{c}}1&3&0&{ - 1}\\0&{ - 4}&8&0\\0&0&3&0\\0&0&0&1\end{array}} \right]\)

03

Determine whether the matrix is invertible

The \(4 \times 4\) matrix \(\left[ {\begin{array}{*{20}{c}}1&3&0&{ - 1}\\0&{ - 4}&8&0\\0&0&3&0\\0&0&0&1\end{array}} \right]\) has four pivot positions. It is invertible according to part (c) of the invertible matrix theorem.

Thus, the matrix \(\left[ {\begin{array}{*{20}{c}}{ - 1}&{ - 3}&0&1\\3&5&8&{ - 3}\\{ - 2}&{ - 6}&3&2\\0&{ - 1}&2&1\end{array}} \right]\) is invertible.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose the transfer function W(s) in Exercise 19 is invertible for some s. It can be showed that the inverse transfer function \(W{\left( s \right)^{ - {\bf{1}}}}\), which transforms outputs into inputs, is the Schur complement of \(A - BC - s{I_n}\) for the matrix below. Find the Sachur complement. See Exercise 15.

\(\left[ {\begin{array}{*{20}{c}}{A - BC - s{I_n}}&B\\{ - C}&{{I_m}}\end{array}} \right]\)

Explain why the columns of an \(n \times n\) matrix Aare linearly independent when Ais invertible.

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

Let Abe an invertible \(n \times n\) matrix, and let \(B\) be an \(n \times p\) matrix. Explain why \({A^{ - 1}}B\) can be computed by row reduction: If\(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right) \sim ... \sim \left( {\begin{aligned}{*{20}{c}}I&X\end{aligned}} \right)\), then \(X = {A^{ - 1}}B\).

If Ais larger than \(2 \times 2\), then row reduction of \(\left( {\begin{aligned}{*{20}{c}}A&B\end{aligned}} \right)\) is much faster than computing both \({A^{ - 1}}\) and \({A^{ - 1}}B\).

Suppose Aand Bare \(n \times n\), Bis invertible, and ABis invertible. Show that Ais invertible. (Hint: Let C=AB, and solve this equation for A.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free