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.

1.\(\left( {\begin{aligned}{*{20}{c}}5&7\\{ - 3}&{ - 6}\end{aligned}} \right)\)

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

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

1. Ais an invertible matrix.
2. Ais row equivalent to the identity matrix of \(n \times n\) 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.

Theorem 4 states that\(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\). If\(ad - bc \ne 0\), then A is invertible.

Since the columns of the matrix \(\left( {\begin{aligned}{*{20}{c}}5&7\\{ - 3}&{ - 6}\end{aligned}} \right)\) are not multiples, they are linearly independent. The matrix is invertible according to part (e) of the invertible matrix theorem.

Moreover, the matrix is invertible according to theorem 4 since its determinant is non-zero.

Thus, the matrix \(\left( {\begin{aligned}{*{20}{c}}5&7\\{ - 3}&{ - 6}\end{aligned}} \right)\) is invertible.