Math 204 - Midterm - March 2018_Solutions.pdf - MATH 204...

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MATH 204Winter 2018 MidtermSolutionQ1 [5 marks](a). Find the inverse ofA=3-191-142-210(b). Solve the following equation for matrixX:100011012X2134=251001Solution(a).3-191001-140102-210001r1r2----→1-140103-191002-210001R2=r2-3r1-------→R3=r3-2r11-1401002-31-300020-21R2=r2/2------→1-1401001-3/21/2-3/200020-21R1=r1+r2-------→105/21/2-1/2001-3/21/2-3/200020-21R3=r3/2------→105/21/2-1/2001-3/21/2-3/200010-11/2R1=r1-(5/2)r3----------→R2=r2+(3/2)r31001/22-5/40101/2-33/40010-11/2A-1=1/22-5/41/2-33/40-11/2(b). Is the matrix100011012invertible?100100011010012001R3=r3-r2-------→1001000110100010-11R2=r2-r3-------→10010001002-10010-11Yes,100011012is invertible.2·4-1·3 = 8-3 = 56= 0:2134is invertible too. And its inverse is:2134-1=154-1-32Using those two results (invertibility of the matrices), we can solve:1 of 5
MATH 204Winter 2018 MidtermSolution100011012X2134=251001100011012-1100011012|{z}I3X21342134-1|{z}I2=100011012-12510012134-1X=100011012-12510012

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