## Linear Algebra and Its Applications, Exercise 1.6.8

Exercise 1.6.8. The matrix $A = \begin{bmatrix} 1&1 \\ 3&3 \end{bmatrix}$

has no inverse. Demonstrate this by trying to solve the following: $\begin{bmatrix} 1&1 \\ 3&3 \end{bmatrix} \begin{bmatrix} a&b \\ c&d \end{bmatrix} = \begin{bmatrix} 1&0 \\ 0&1 \end{bmatrix}$

Answer: Multiplying the first row of the first marix by the first column of the second matrix gives $a + c = 1$

However multiplying the second row of the first matrix by the first column of the second matrix gives us $3a + 3c = 0 \rightarrow a + c = 0$

We thus have a contradiction, and conclude that the matrix is not invertible.

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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