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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 $$\begin{bmatrix}x^2&3x \\ 1 & 2\end{bmatrix} \cdot \begin{bmatrix}1 & x \\ 3 & x\end{bmatrix} = \begin{bmatrix}x^2+9x & x^33x^2 \\ 7 & x\end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}$$ so either I am mistaken or there is a problem somewhere... 
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: $\displaystyle \left  \begin{array} {c c} a & b \\ c & d \end{array} \right  =adbc$ So we get: $\displaystyle \begin{align}(2x^23x)(x3x)&=20 \\ (2x^23x)(4x)&=2 \\ (2x^23x)(2x)&=1 \end{align}$ Which I believe that you correctly derived. WA 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. 
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 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,888 Thanks: 767 Math Focus: Wibbly wobbly timeywimey stuff.  
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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determinants, inverses, matrices, square 
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