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August 8th, 2017, 01:21 AM  #1 
Newbie Joined: Aug 2017 From: uk Posts: 1 Thanks: 0  Simple Matrix(Matrices)
Hey, Never looked at Matrices before, but am preparing for the SATs. Can someone help me with this question with a full explanation if possible? Question attached. Really appreciate it. Last edited by skipjack; August 8th, 2017 at 09:33 AM. 
August 8th, 2017, 03:43 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,063 Thanks: 1396 
Assuming that the determinants of the matrices have to be equal, calculate the determinant of the first matrix; the result is zero. For the determinant of the second matrix to be zero, $x$ must satisfy $x^2 = 5 \times 4 = 20$, so $x = \pm\sqrt{20} = \pm4.47$ (approximately). 
August 8th, 2017, 04:09 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,741 Thanks: 710 
The way this is stated, this is impossible! A 3 by 3 matrix is never equal to a 2 by 2 matrix. If, as skipjack suggests, these are determinants rather than matrices (in which case it should be "$\displaystyle \left\begin{array}{ccc}2 & 1 & 4 \\ 3 & 0 & 5 \\ 4 & 1 & 6 \end{array}\right= \left\begin{array}{cc}x & 4 \\ 5 & x \end{array}\right$", using vertical lines rather than parentheses) we can expand the determinant on the left by the middle column: 1(3*6 5*4)+ 1(2*5 4*3)= 1(2)+ 1(2)= 2 2= 0. The determinant on the left is $\displaystyle x^2 20$ so x must satisfy $\displaystyle x^220= 0$. Last edited by skipjack; August 8th, 2017 at 09:32 AM. 

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matrices, matrix, matrixmatrices, simple 
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