My Math Forum Simple Matrix(Matrices)

 Linear Algebra Linear Algebra Math Forum

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.
Attached Images
 Matrix 3x3 to 2x2..JPG (17.6 KB, 10 views)

Last edited by skipjack; August 8th, 2017 at 09:33 AM.

 August 8th, 2017, 03:43 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,502 Thanks: 1739 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: 3,261 Thanks: 894 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 $\displaystyle x^2-20= 0$. Last edited by skipjack; August 8th, 2017 at 09:32 AM.

 Tags matrices, matrix, matrixmatrices, simple

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post coolkiller Linear Algebra 2 January 22nd, 2014 01:33 AM matrixmath Linear Algebra 0 December 14th, 2012 04:36 AM maximus101 Algebra 0 March 14th, 2011 09:41 AM evant8950 Linear Algebra 2 February 1st, 2011 10:14 AM josh3light New Users 0 May 30th, 2009 09:49 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top