Linear Algebra and Its Applications, Review Exercise 1.6

Review exercise 1.6. (a) For each of the each of the 2 by 2 matrices containing only 0 or 1 as entries, determine whether the matrix is invertible or not. (b) Of the 10 by 10 matrices containing only 0 or 1 as entries, is a particular matrix chosen at random more likely to be invertible or not?

Answer: (a) A 2 by 2 matrix has four entries. If each entry can be either 0 or 1 then there are 16 possible matrices of this type:

A_0 = \begin{bmatrix} 0&0 \\ 0&0 \end{bmatrix} \quad A_1 = \begin{bmatrix} 0&0 \\ 0&1 \end{bmatrix} \quad A_2 = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix} \quad A_3 = \begin{bmatrix} 0&1 \\ 0&1 \end{bmatrix}

A_4 = \begin{bmatrix} 0&0 \\ 1&0 \end{bmatrix} \quad A_5 = \begin{bmatrix} 0&0 \\ 1&1 \end{bmatrix} \quad A_6 = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} \quad A_7 = \begin{bmatrix} 0&1 \\ 1&1 \end{bmatrix}

A_8 = \begin{bmatrix} 1&0 \\ 0&0 \end{bmatrix} \quad A_9 = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix} \quad A_{10} = \begin{bmatrix} 1&1 \\ 0&0 \end{bmatrix} \quad A_{11} = \begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix}

A_{12} = \begin{bmatrix} 1&0 \\ 1&0 \end{bmatrix} \quad A_{13} = \begin{bmatrix} 1&0 \\ 1&1 \end{bmatrix} \quad A_{14} = \begin{bmatrix} 1&1 \\ 1&0 \end{bmatrix} \quad A_{15} = \begin{bmatrix} 1&1 \\ 1&1 \end{bmatrix}

Of these matrices, the following ten matrices are singular:

A_0, A_1, A_2, A_3, A_4, A_5, A_8, A_{10}, A_{12}, A_{15}

This can be most easily shown by computing the value ad - bc, which is zero for all these matrices.

The remaining six matrices are nonsingular and have inverses as follows:

A_{6}^{-1} = \frac{1}{0 \cdot 0 - 1 \cdot 1} \begin{bmatrix} 0&-1 \\ -1&0 \end{bmatrix} = \frac{1}{-1} \begin{bmatrix} 0&-1 \\ -1&0 \end{bmatrix} = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix} = A_{6}

A_{7}^{-1} = \frac{1}{0 \cdot 1 - 1 \cdot 1} \begin{bmatrix} 1&-1 \\ -1&0 \end{bmatrix} = \frac{1}{-1} \begin{bmatrix} 1&-1 \\ -1&0 \end{bmatrix} = \begin{bmatrix} -1&1 \\ 1&0 \end{bmatrix}

A_{9}^{-1} = I^{-1} = I = A_{9}

A_{11}^{-1} = \frac{1}{1 \cdot 1 - 1 \cdot 0} \begin{bmatrix} 1&-1 \\ 0&1 \end{bmatrix} = \frac{1}{1} \begin{bmatrix} 1&-1 \\ 0&1 \end{bmatrix} = \begin{bmatrix} 1&-1 \\ 0&1 \end{bmatrix}

A_{13}^{-1} = \frac{1}{1 \cdot 1 - 0 \cdot 1} \begin{bmatrix} 1&0 \\ -1&1 \end{bmatrix} = \frac{1}{1} \begin{bmatrix} 1&0 \\ -1&1 \end{bmatrix} = \begin{bmatrix} 1&0 \\ -1&1 \end{bmatrix}

A_{14}^{-1} = \frac{1}{1 \cdot 0 - 1 \cdot 1} \begin{bmatrix} 0&-1 \\ -1&1 \end{bmatrix} = \frac{1}{-1} \begin{bmatrix} 0&-1 \\ -1&1 \end{bmatrix} = \begin{bmatrix} 0&1 \\ 1&-1 \end{bmatrix}

(b) The 10 by 10 matrices have 100 entries, and if each can be 0 or 1 that means there are 2^{100} possible matrices of this type. At present I don’t know of a good method to determine whether the majority of those matrices are invertible or not. I’ll update this post later if and when I have time to work on this some more.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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