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7 0

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Determine if the specified linear transformation is (a) one-to-one and (b) onto. Justify your answer. T: , T()(,),

xz_007 [3.2K]

Answer:

C. T is not one-to-one because the standard matrix A has a free variable.

Step-by-step explanation:

Given

$T(x_1,x_2,x_3) = (x_1-5x_2+4x_3,x_2 - 6x_3)$

Required

Determine if it is linear or onto

Represent the above as a matrix.

$T(x_1,x_2,x_3) = \left[\begin{array}{ccc}1&-5&4\\0&1&-6\\0&0&0\end{array}\right] \left[\begin{array}{c}x_1&x_2&x_3\end{array}\right]$

From the above matrix, we observe that the matrix does not have a pivot in every column.

This means that the column are not linearly independent, & it has a free variable and as such T is not one-on-one

8 0

2 years ago

Solve each equation? (n+2) ( n+1)= 0

Delicious77 [7]

Answer:

$n=-2,\:n=-1$

Step-by-step explanation:

$\left(n+2\right)\left(n+1\right)=0\\\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)\\\mathrm{Solve\:}\:n+2=0:\quad n=-2\\n+2=0\\\mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}\\n+2-2=0-2\\\mathrm{Simplify}\\n=-2\\\mathrm{Solve\:}\:n+1=0:\quad n=-1\\n+1=0\\\mathrm{Subtract\:}1\mathrm{\:from\:both\:sides}\\n+1-1=0-1\\Simplify\\n=-1\\$ $\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\n=-2,\:n=-1$