Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 3 - Determinants - 3.1 The Determination of a Matrix - 3.1 Exercises - Page 111: 60


$ad-cb\neq 0$

Work Step by Step

The system can be written as follows $$ \left[\begin{array}{rrr}{a} & {b} \\ {c} & {d}\end{array}\right]\left[\begin{array}{rrr}{x} \\ {y} \end{array}\right]=\left[\begin{array}{rrr}{e} \\ {f} \end{array}\right]. $$ Now, the above system has solution if and only if the determinant of the coefficient matrix has inverse, that is, its determinant is non zero. Moreover, the solution is given by $$\left[\begin{array}{rrr}{x} \\ {y} \end{array}\right]=\frac{1}{ad-cb}\left[\begin{array}{rrr}{d} & {-b} \\ {-c} & {a}\end{array}\right]\left[\begin{array}{rrr}{e} \\ {f} \end{array}\right].$$ It is easy to see that the above solution is unique and exists if and only if $ad-cb\neq 0$ that is the determinant of the coefficient matrix is nonzero.
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