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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+(√3)/2 OR 1-(√3)/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: 7,963 Thanks: 1148 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: 7,963 Thanks: 1148 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: 642 Thanks: 99 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: 7,963 Thanks: 1148 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 Tags determinants, inverses, matrices, square Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post starbug7 Linear Algebra 3 May 19th, 2015 07:05 AM norm abrams Algebra 10 January 13th, 2012 02:49 AM Pell's fish Number Theory 1 November 13th, 2011 05:51 PM prashantakerkar Linear Algebra 4 August 20th, 2011 10:18 AM matrixman42 Linear Algebra 2 November 4th, 2007 08:18 PM

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