## 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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