The matrix

Algebra Level 5

{ a x + b y + c z = d e x + f y + g z = h j x + k y + l z = m \left\{\begin{matrix} ax+by+cz=d\\ ex+fy+gz=h\\ jx+ky+lz=m \end{matrix}\right.

Let's say there is a system of three equations with variables x , y , z x, y, z as above.

What is the minimum number of coefficients (denoted by letters a , b , c , e , f , g , j , k , l a,b,c,e,f,g,j,k,l ) that must be zero such that the system cannot be solved using Cramer's rule?


The answer is 0.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

Chew-Seong Cheong
Jun 12, 2015

The system of equations cannot be solved using Cramer's rule, when the determinant of coefficients is 0 0 . The determinant of coefficients equals 0 0 when any two rows or columns of coefficients are dependent. For example, when a = n e a = ne , b = n f b=nf and c = n g c=ng or a = n b a=nb , e = n f e=nf , j = n k j=nk for any value of n n . Therefore, the required answer is 0 \boxed{0} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

And of course, it can't be solved (for specific values) by any other method either.

Nicely explained!

Hobart Pao - 6 years ago

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