Question

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.

3.\(\left( {\begin{aligned}{*{20}{c}}5&0&0\\{ - 3}&{ - 7}&0\\8&5&{ - 1}\end{aligned}} \right)\)

Short answer

The matrix \(\left( {\begin{aligned}{*{20}{c}}5&0&0\\{ - 3}&{ - 7}&0\\8&5&{ - 1}\end{aligned}} \right)\) is invertible.

Step-by-step solution

State the invertible matrix theorem

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 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.

Apply the row operation

At row one, multiply row one by \(\frac{1}{5}\).

\(\left( {\begin{aligned}{*{20}{c}}1&0&0\\{ - 3}&{ - 7}&0\\8&5&{ - 1}\end{aligned}} \right)\)

At row two, multiply row one by 3 and add it to row two.

\(\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{ - 7}&0\\8&5&{ - 1}\end{aligned}} \right)\)

At row three, multiply row one by 8 and subtract it from row three.

\(\left( {\begin{aligned}{*{20}{c}}1&0&0\\0&{ - 7}&0\\0&5&{ - 1}\end{aligned}} \right)\)

Determine whether the matrix is invertible

A \(3 \times 3\) matrix has three pivot positions. It is invertible according to part (c) of the invertible matrix theorem.

Thus, the matrix \(\left( {\begin{aligned}{*{20}{c}}5&0&0\\{ - 3}&{ - 7}&0\\8&5&{ - 1}\end{aligned}} \right)\) is invertible.