## Linear Algebra and Its Applications, Review Exercise 1.7

Review exercise 1.7. For each of the each of the 2 by 2 matrices containing only -1 or 1 as entries, determine whether the matrix is invertible or not.

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

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

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

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

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

Of these matrices, the following eight matrices are singular:

$A_0, A_3, A_5, A_6, A_{9}, A_{10}, A_{12}, A_{15}$

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

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

$A_{1}^{-1} = \frac{1}{(-1) \cdot 1 - (-1) \cdot (-1)} \begin{bmatrix} -1&-1 \\ -(-1)&-1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -1&-1 \\ 1&-1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} = \frac{1}{2} A_{1}$

$A_{2}^{-1} = \frac{1}{(-1) \cdot (-1) - 1 \cdot (-1)} \begin{bmatrix} -1&-1 \\ -(-1)&-1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -1&-1 \\ 1&-1 \end{bmatrix} = \frac{1}{2} A_{4}$

$A_{4}^{-1} = \frac{1}{(-1) \cdot (-1) - (-1) \cdot 1} \begin{bmatrix} -1&-(-1) \\ -1&-1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -1&1 \\ -1&-1 \end{bmatrix} = \frac{1}{2} A_{2}$

$A_{7}^{-1} = \frac{1}{-1 \cdot 1 - 1 \cdot 1} \begin{bmatrix} 1&-1 \\ -1&-1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} 1&-1 \\ -1&-1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} -1&1 \\ 1&1 \end{bmatrix} = \frac{1}{2} A_{7}$

$A_{8}^{-1} = \frac{1}{1 \cdot (-1) - (-1) \cdot (-1)} \begin{bmatrix} -1&-(-1) \\ -(-1)&1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -1&1 \\ 1&1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1&-1 \\ -1&-1 \end{bmatrix} = \frac{1}{2} A_{8}$

$A_{11}^{-1} = \frac{1}{1 \cdot 1 - 1 \cdot (-1)} \begin{bmatrix} 1&-1 \\ -(-1)&1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1&-1 \\ 1&1 \end{bmatrix} = \frac{1}{2} A_{13}$

$A_{13}^{-1} = \frac{1}{1 \cdot 1 - (-1) \cdot 1} \begin{bmatrix} 1&-(-1) \\ -1&1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1&1 \\ -1&1 \end{bmatrix} = \frac{1}{2} A_{11}$

$A_{14}^{-1} = \frac{1}{1 \cdot (-1) - 1 \cdot 1} \begin{bmatrix} -1&-1 \\ -1&1 \end{bmatrix} = -\frac{1}{2} \begin{bmatrix} -1&-1 \\ -1&1 \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1&1 \\ 1&-1 \end{bmatrix} = \frac{1}{2} A_{14}$

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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