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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

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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.

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Most popular questions from this chapter

In Exercises 27 and 28, view vectors in \({\mathbb{R}^n}\) as \(n \times 1\) matrices. For \({\mathop{\rm u}\nolimits} \) and \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^n}\), the matrix product \({{\mathop{\rm u}\nolimits} ^T}v\) is a \(1 \times 1\) matrix, called the scalar product, or inner product, of u and v. It is usually written as a single real number without brackets. The matrix product \({{\mathop{\rm uv}\nolimits} ^T}\) is an \(n \times n\) matrix, called the outer product of u and v. The products \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \) and \({{\mathop{\rm uv}\nolimits} ^T}\) will appear later in the text.

27. Let \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\3\\{ - 4}\end{aligned}} \right)\) and \({\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}a\\b\\c\end{aligned}} \right)\). Compute \({{\mathop{\rm u}\nolimits} ^T}{\mathop{\rm v}\nolimits} \), \({{\mathop{\rm v}\nolimits} ^T}{\mathop{\rm u}\nolimits} \),\({{\mathop{\rm uv}\nolimits} ^T}\), and \({\mathop{\rm v}\nolimits} {{\mathop{\rm u}\nolimits} ^T}\).

A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.

37. Construct a random \({\bf{4}} \times {\bf{4}}\) matrix Aand test whether \(\left( {A + I} \right)\left( {A - I} \right) = {A^2} - I\). The best way to do this is to compute \(\left( {A + I} \right)\left( {A - I} \right) - \left( {{A^2} - I} \right)\) and verify that this difference is the zero matrix. Do this for three random matrices. Then test \(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^{\bf{2}}}\) the same way for three pairs of random \({\bf{4}} \times {\bf{4}}\) matrices. Report your conclusions.

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 \({A^{ - 1}}B\).

1. Find the inverse of the matrix \(\left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{6}}\\{\bf{5}}&{\bf{4}}\end{aligned}} \right)\).

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.

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