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typo corrected
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Scott Carter
Scott Carter
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My answer is not very highfaluting, but it is what I use to remember. Switch the diagonals, change the signs of the off-diagonals and divide by the determinant. Since the inverse of a diagonal matrix is easy, the switch should be easy to remember. On the other hand such mnemonics are dangerous. The critical points of a cubic $Ax^3+Bx^2+Cx+D$ are at $\frac{-B\pm\sqrt{B^2-3AC}}{6A}$$\frac{-B\pm\sqrt{B^2-3AC}}{3A}$, or so I remember.

My answer is not very highfaluting, but it is what I use to remember. Switch the diagonals, change the signs of the off-diagonals and divide by the determinant. Since the inverse of a diagonal matrix is easy, the switch should be easy to remember. On the other hand such mnemonics are dangerous. The critical points of a cubic $Ax^3+Bx^2+Cx+D$ are at $\frac{-B\pm\sqrt{B^2-3AC}}{6A}$, or so I remember.

My answer is not very highfaluting, but it is what I use to remember. Switch the diagonals, change the signs of the off-diagonals and divide by the determinant. Since the inverse of a diagonal matrix is easy, the switch should be easy to remember. On the other hand such mnemonics are dangerous. The critical points of a cubic $Ax^3+Bx^2+Cx+D$ are at $\frac{-B\pm\sqrt{B^2-3AC}}{3A}$, or so I remember.

Source Link
Scott Carter
Scott Carter
  • 5.3k
  • 2
  • 28
  • 35

My answer is not very highfaluting, but it is what I use to remember. Switch the diagonals, change the signs of the off-diagonals and divide by the determinant. Since the inverse of a diagonal matrix is easy, the switch should be easy to remember. On the other hand such mnemonics are dangerous. The critical points of a cubic $Ax^3+Bx^2+Cx+D$ are at $\frac{-B\pm\sqrt{B^2-3AC}}{6A}$, or so I remember.