## Linear Algebra and Its Applications, Exercise 1.6.9

Exercise 1.6.9. Given the singular matrix

$A = \begin{bmatrix} 2&1&4&6 \\ 0&3&8&5 \\ 0&0&0&7 \\ 0&0&0&9 \end{bmatrix}$

show that A has no inverse. If it did have an inverse then multiplying the third row of $A^{-1}$ by the columns of A should give the third row of I. Explain why this is not possible.

Answer: For the 4 by 4 case the third row of I is

$\begin{bmatrix} 0&0&1&0 \end{bmatrix}$

A zero value for the first entry in this row would be produced by multiplying the third row of $A^{-1}$ by the first column of A. The last three entries in the first column of A are zero and thus would produce zero terms in the product. Since the first entry of the first column of A is nonzero, in order for the first entry of the product to be zero the first entry of the third row of $A^{-1}$ must be zero.

A zero value for the second entry in the third row of I would be produced by multiplying the third row of $A^{-1}$ by the second column of A. The last two entries in the first column of A are zero, and thus would produce zero terms in the product. Similarly the first entry of the third row of $A^{-1}$ is zero (from the previous paragraph) and would also produce a zero term in the product when multiplying the first entry of the second column of A. Since the first term and the last two terms are zero, and the second entry in the second column of A is nonzero, in order for the second term of the product to be zero (and thus the overall product to be zero) the second entry of the third row of $A^{-1}$ must also be zero.

A value of 1 for the third entry in the third row of I would be produced by multiplying the third row of $A^{-1}$ by the third column of A. The last two entries in the first column of A are zero, and thus would produce zero terms in the product. Similarly the first and second entries of the third row of $A^{-1}$ are zero (from the previous two paragraphs) and would also produce zero terms in the product when multiplying the first and second entries of the second column of A.

But this means that the product of the third row of $A^{-1}$ and the third column of A must be zero, which means that the product of $A^{-1}$ and A cannot be I. We conclude therefore that $A^{-1}$ does not exist and that A 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.

This entry was posted in linear algebra. Bookmark the permalink.