## 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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