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

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.

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.

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

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.