Cramer is on Brilliant?

Algebra Level 1

x = ∣ 1 − 4 1 5 1 2 11 − 1 − 3 ∣ ∣ 3 − 4 1 5 1 2 1 − 1 − 3 ∣ , y = ∣ 3 1 1 5 5 2 1 11 − 3 ∣ ∣ 3 − 4 1 5 1 2 1 − 1 − 3 ∣ , z = ∣ 3 − 4 1 5 1 5 1 − 1 11 ∣ ∣ 3 − 4 1 5 1 2 1 − 1 − 3 ∣ x = \frac{\left| \begin{array}{ccc} 1 & -4 & 1\\ 5 & 1 & 2 \\ 11 & -1 & -3 \end{array} \right|}{\left| \begin{array}{ccc} 3 & -4 & 1\\ 5 & 1 & 2 \\ 1 & -1 & -3 \end{array} \right|}, \quad \quad y = \frac{\left| \begin{array}{ccc} 3 & 1 & 1\\ 5 & 5 & 2 \\ 1 & 11 & -3 \end{array} \right|}{\left| \begin{array}{ccc} 3 & -4 & 1\\ 5 & 1 & 2 \\ 1 & -1 & -3 \end{array} \right|}, \quad \quad z = \frac{\left| \begin{array}{ccc} 3 & -4 & 1\\ 5 & 1 & 5 \\ 1 & -1 & 11 \end{array} \right|}{\left| \begin{array}{ccc} 3 & -4 & 1\\ 5 & 1 & 2 \\ 1 & -1 & -3 \end{array} \right|}

Using Cramer's rule, which equation is not solved by the solutions above?

2 x + 3 y − z = 5 2x + 3y - z = 5 5 x + y + 2 z = 5 5x + y + 2z = 5 3 x − 4 y + z = 1 3x - 4y + z = 1 x − y − 3 z = 11 x - y - 3z = 11

Andrew Ellinor by Andrew Ellinor , 33, USA

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1 solution

Jason Carrier
Sep 21, 2018

Cramer’s Rule involves division by the determinant of the coefficients matrix. So, we match the rows of the denominators to answers, and see which isn’t present.

The row 3 -4 1 corresponds to 3 x − 4 y + z = 1 3x-4y+z=1

The row 5 1 2 corresponds to 5 x + y + 2 z = 5 5x+y+2z=5

The row 1 -1 -3 corresponds to x − y − 3 z = 11 x-y-3z=11

The only choice not represented is 2 x + 3 y − z = 5 \boxed{2x+3y-z=5}

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