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 August 20th, 2018, 05:33 PM #1 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Determinants and Inverses of Square Matrices Solve |x^2 3X|* |1 x|= |2 0| |1 2 | |3 -x| |0 1| -8x^3 +12x^2 =2 8x^3 - 12x^2 = -2 4x^3 - 6x^2 = -1 2x (2x^2 - 3x) = -1 x = -1/2 OR -1 OR 2 This is how I calculated this question, but the model answers are 1/2 OR 1+(âˆš3)/2 OR 1-(âˆš3)/2.
 August 20th, 2018, 05:36 PM #2 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 The question should be a square matrice but I don't know how to type it in to the computer
 August 20th, 2018, 06:08 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,878 Thanks: 1087 Math Focus: Elementary mathematics and beyond Is this what you intended? $$\begin{bmatrix}x^2&3x \\ 1 & 2\end{bmatrix} \cdot \begin{bmatrix}1 & x \\ 3 & -x\end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}$$ To observe the code, right click on the math and select "Show Math As" > "TeX Commands" from the context menu or quote this post and examine the contents of the reply window.
 August 20th, 2018, 06:11 PM #4 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Yes, are there any steps wrong so I cannot calculate the correct answer?
 August 20th, 2018, 06:34 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,878 Thanks: 1087 Math Focus: Elementary mathematics and beyond Well, I get $$\begin{bmatrix}x^2&3x \\ 1 & 2\end{bmatrix} \cdot \begin{bmatrix}1 & x \\ 3 & -x\end{bmatrix} = \begin{bmatrix}x^2+9x & x^3-3x^2 \\ 7 & -x\end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}$$ so either I am mistaken or there is a problem somewhere... Thanks from hy2000
 August 20th, 2018, 08:07 PM #6 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 626 Thanks: 90 Math Focus: Electrical Engineering Applications Here is what I get: $\displaystyle \left | \begin{array} {c c} x^2 & 3x \\ 1 & 2 \end{array} \right | \cdot \left | \begin{array} {c c} 1 & x \\ 3 & -x \end{array} \right | = \left | \begin{array} {c c} 2 & 0 \\ 0 & 1 \end{array} \right |$ Per WP: $\displaystyle \left | \begin{array} {c c} a & b \\ c & d \end{array} \right | =ad-bc$ So we get: \displaystyle \begin{align}(2x^2-3x)(-x-3x)&=2-0 \\ (2x^2-3x)(-4x)&=2 \\ (2x^2-3x)(2x)&=-1 \end{align} Which I believe that you correctly derived. W|A gives the solutions as: $\displaystyle \frac{1}{2}, \quad \frac{1}{2} - \frac{\sqrt{3}}{2}, \quad \frac{1}{2} + \frac{\sqrt{3}}{2}$ so I think that the opening parenthesis were misplaced. Thanks from greg1313 and hy2000
 August 21st, 2018, 02:50 AM #7 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,878 Thanks: 1087 Math Focus: Elementary mathematics and beyond Ahh... determinants! My mistake. It's best to state the entire question in the post body to avoid such confusion (but then I could have been more careful).
August 21st, 2018, 05:48 AM   #8
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Quote:
 Originally Posted by hy2000 Solve |x^2 3X|* |1 x|= |2 0| |1 2 | |3 -x| |0 1|
A quick comment: Mathematics is "case sensitive." That means that X and x are not the same variable.

-Dan

 August 21st, 2018, 05:43 PM #9 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Thanks a lot, I have tried both ways; by determinants and just multiplying the matrices together, they have the same result

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