Saturday , December 4 2021

# linear algebra – Equation with an unknown variable how to solve easily The elements of the first column are all multiples of \$ x \$; the items in the third column are all multiples of \$ x-2 \$. The determinant can be calculated in this way:

\$\$ 0 = left | begin {array} {ccc}
phantom {-} x & -2 and 3x-6 \
2x and ghost {-} 0 and 2-x \
-x and phantom {-} 5 and x-2
End {array} right | =
x (x-2) left | begin {array} {ccc}
phantom {-} 1 & -2 and phantom {-} 3 \
phantom {-} 2 and phantom {-} 0 & -1 \
-1 & phantom {-} 5 and phantom {-} 1
End {array} right | =

The rules of the determinants allow to eliminate the common multipliers from a single column (or row):

\$\$ left | begin {array} {ccc}
pa & b & c \
pd & e & f \
ph & i & j
End {array} right | ; = ; p left | begin {array} {ccc}
a & b & c \
d & e & f \
h & i & j
End {array} right | \$\$

The justification is immediately clear if one is familiar with the relationship between a determinant and the volume of the parallelepiped determined by the column vectors. But it also makes sense if you pay attention to how the determining factors are expanded. Here is a quick example with \$ 2 times 2 \$:

\$\$ left | begin {array} {cc} pa & b \ pc & d end {array} right | = pa cdot d -pc cdot b = p (ad-cb) = p left | begin {array} {cc} a & b \ c & d end {array} right | \$\$
Each term in the expansion includes the multiplier, which can be decomposed.

However … For the problem in question, I calculated \$ x \$ and \$ x-2 \$ from the first and third columns. Then I calculated the value of the remaining determinant (just to make sure it was not zero, in which case \$ x \$ could be nothing).