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Linear Algebra : Linear Algebra

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Example Questions

Example Question #191 : Linear Algebra

$\begin{align*}&\text{Find }Tr(A)\text{ for }A=\begin{bmatrix}1&6&-4&13\\9&19&1&-7\\-16&5&11&-3\\-17&-10&-14&-9\end{bmatrix}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Tr, that is to say the trace, of a square nxn matrix is the its diagonal elements:}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{For the matrix }A=\begin{bmatrix}1&6&-4&13\\9&19&1&-7\\-16&5&11&-3\\-17&-10&-14&-9\end{bmatrix}\\&\text{We find: }\\&Tr(A)=(1)+(19)+(11)+(-9)=22\end{align*}$

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Example Question #14 : The Trace

$\begin{align*}&\text{Find }Tr(A)\text{ for }A=\begin{bmatrix}6&19&-11&7&-9&7&8&-18\\-10&-11&7&14&-6&12&7&-20\\4&-5&17&-20&-2&-3&-2&11\\-7&12&-1&-19&-13&9&-1&-14\\-7&4&-13&10&-11&17&-9&11\\-13&-9&-17&3&8&2&-3&6\\6&7&6&18&-12&9&-11&-16\\4&-2&-2&7&11&-6&7&-3\end{bmatrix}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Tr, that is to say the trace, of a square nxn matrix is the its diagonal elements:}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{For the matrix }A=\begin{bmatrix}6&19&-11&7&-9&7&8&-18\\-10&-11&7&14&-6&12&7&-20\\4&-5&17&-20&-2&-3&-2&11\\-7&12&-1&-19&-13&9&-1&-14\\-7&4&-13&10&-11&17&-9&11\\-13&-9&-17&3&8&2&-3&6\\6&7&6&18&-12&9&-11&-16\\4&-2&-2&7&11&-6&7&-3\end{bmatrix}\\&\text{We find: }\\&Tr(A)=(6)+(-11)+(17)+(-19)+(-11)+(2)+(-11)+(-3)=-30\end{align*}$

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Example Question #114 : Operations And Properties

$\begin{align*}&\text{Calculate the trace of the matrix}\\&\begin{bmatrix}15&-10&-7&-16&18\\6&-1&6&2&6\\2&9&1&20&-12\\-16&-16&-18&-4&-2\\-6&11&5&11&18\end{bmatrix}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The trace of a square nxn matrix is the sum of its diagonal elements:}\\&\sum_{i=1}^{n}A_{i,i}\\&\text{For the matrix }\begin{bmatrix}15&-10&-7&-16&18\\6&-1&6&2&6\\2&9&1&20&-12\\-16&-16&-18&-4&-2\\-6&11&5&11&18\end{bmatrix}\\&\text{This becomes: }\\&(15)+(-1)+(1)+(-4)+(18)=29\end{align*}$

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Example Question #14 : The Trace

$\begin{align*}&\text{Compute }Tr\begin{bmatrix}1&15&-1&-4&7\\10&1&-6&-14&4\\-10&-19&10&-11&-2\\8&-6&10&-4&8\\8&-2&-20&-7&-3\end{bmatrix}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Tr in the context of linear algebra represents the trace.}\\&\text{We can find this quantity for a square nxn matrix by summing the elements of its diagonal:}\\&\sum_{i=1}^{n}A_{i,i}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{Doing so we find for}\begin{bmatrix}1&15&-1&-4&7\\10&1&-6&-14&4\\-10&-19&10&-11&-2\\8&-6&10&-4&8\\8&-2&-20&-7&-3\end{bmatrix}\\&\text{We find:}\\&(1)+(1)+(10)+(-4)+(-3)=5\end{align*}$

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Example Question #192 : Linear Algebra

$\begin{align*}&\text{Find }Tr(A)\text{ for }A=\begin{bmatrix}13&10&-5&-12\\12&18&-7&7\\-3&14&11&-14\\15&20&1&16\end{bmatrix}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Tr, that is to say the trace, of a square nxn matrix is the its diagonal elements:}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{For the matrix }A=\begin{bmatrix}13&10&-5&-12\\12&18&-7&7\\-3&14&11&-14\\15&20&1&16\end{bmatrix}\\&\text{We find: }\\&Tr(A)=(13)+(18)+(11)+(16)=58\end{align*}$

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Example Question #15 : The Trace

$\begin{align*}&\text{Find }Tr(A)\text{ for }A=\begin{bmatrix}12&-7&1&-17&-16&-15&7\\0&-13&0&-14&-18&14&2\\18&8&3&13&16&20&-20\\15&5&20&1&-1&12&-11\\0&16&3&14&10&4&-10\\7&-17&5&7&9&16&20\\11&3&18&3&-20&-16&15\end{bmatrix}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{Tr, that is to say the trace, of a square nxn matrix is the its diagonal elements:}\\&\text{i.e. }Tr\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}=a+e+i\\&\text{For the matrix }A=\begin{bmatrix}12&-7&1&-17&-16&-15&7\\0&-13&0&-14&-18&14&2\\18&8&3&13&16&20&-20\\15&5&20&1&-1&12&-11\\0&16&3&14&10&4&-10\\7&-17&5&7&9&16&20\\11&3&18&3&-20&-16&15\end{bmatrix}\\&\text{We find: }\\&Tr(A)=(12)+(-13)+(3)+(1)+(10)+(16)+(15)=44\end{align*}$

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Example Question #193 : Linear Algebra

$A = \begin{bmatrix} 3&0 & 0\\ 0& 6& 0\\ 0&0 &x \end{bmatrix}$ ,

where is a complex number. The trace of $A^{2}$ is 50.

Evaluate .

Possible Answers:

$\pm i\sqrt{5}$

$\pm 5 i$

$\pm \sqrt{5}$

$\pm 5$

Correct answer:

$\pm \sqrt{5}$

Explanation:

is a diagonal matrix - its only nonzero elements are on the main diagonal, from upper left to lower right - so its square can be taken by simply squaring the diagonal elements. Since

$A = \begin{bmatrix} 3&0 & 0\\ 0& 6& 0\\ 0&0 &x \end{bmatrix}$

it follows that

$A ^{2}= \begin{bmatrix} 3^{2}&0 & 0\\ 0& 6^{2}& 0\\ 0&0 &x^{2} \end{bmatrix} = \begin{bmatrix} 9&0 & 0\\ 0& 36 & 0\\ 0&0 &x^{2} \end{bmatrix}$

The trace of a matrix is equal to the sum of the elements in its main diagonal, so the trace of $A^{2}$ is

$Tr(A^{2}) = 9 + 36 + x^{2}$

$Tr(A^{2}) = x^{2} + 45$

Since the trace is given to be 50, set this equal to 50 and solve for :

$x^{2} + 45= 50$

$x^{2} =5$

$x= \pm \sqrt{5}$

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Example Question #15 : The Trace

$A = \begin{bmatrix} 3& -1& 2& a\\ 2& a& 4&-3 \\ 1& 4& a& 2\\ a& -4& 6& 3 \end{bmatrix}$

Evaluate so that the trace of is equal to 10.

Possible Answers:

The trace of cannot be 10 regardless of the value of

a = 3

a = 2

a = 1

a= 4

Correct answer:

a = 2

Explanation:

The trace of a matrix is equal to the sum of the elements in its main diagonal - the elements going from upper left to lower right. Therefore,

$\textrm{Tr} (A) = 3+a+a+3$

$\textrm{Tr} (A) = 2a+6$

Since the trace of is equal to 10, set 2a+6 equal to this and solve for :

2a+6 = 10

2a=4

a = 2

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Example Question #21 : The Trace

$C = \begin{bmatrix} a& 0& -7& 3\\ 0& a& 2& -1\\ 0& 0&2 & 6\\ 0& 0& 0& 5 \end{bmatrix}$

where is a real number.

Which expression is equal to the trace of ?

Possible Answers:

2a+8

2a+7

$10a^{2}$

Correct answer:

2a+7

Explanation:

The trace of a matrix is equal to the sum of the elements in its main diagonal - the elements going from upper left to lower right. Therefore,

$\textrm{Tr}(A) = a+a+2+5 = 2a + 7$

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Example Question #22 : The Trace

$A = \begin{bmatrix} 2&0 &0 &0 \\ 0& x& 0& 0\\ 0& 0&6 &0 \\ 0&0 &0 & -4 \end{bmatrix}$ , $B= \begin{bmatrix} -4&0 &0 &0 \\ 0& 2& 0& 0\\ 0& 0&x &0 \\ 0&0 &0 & -3 \end{bmatrix}$

Give the trace of .

Possible Answers:

2x-1

-24x

8x+4

$x^{2}-x-20$

$-1,152x^{2}$

Correct answer:

8x+4

Explanation:

and are both diagonal matrices - their only nonzero elements are on their main diagonals, from upper left to lower right - so their product can be found by multiplying their corresponding diagonal elements:

$AB= \begin{bmatrix} 2&0 &0 &0 \\ 0& x& 0& 0\\ 0& 0&6 &0 \\ 0&0 &0 & -4 \end{bmatrix} \begin{bmatrix} -4&0 &0 &0 \\ 0& 2& 0& 0\\ 0& 0&x &0 \\ 0&0 &0 & -3 \end{bmatrix}$

$= \begin{bmatrix} 2(-4)&0 &0 &0 \\ 0& x(2)& 0& 0\\ 0& 0&6(x) &0 \\ 0&0 &0 & -4(-3) \end{bmatrix}$

$= \begin{bmatrix} -8&0 &0 &0 \\ 0&2x & 0& 0\\ 0& 0&6x&0 \\ 0&0 &0 & 12\end{bmatrix}$

The trace of a matrix is equal to the sum of the elements in its main diagonal, so the trace of is

$\textrm{Tr}(AB) = -8 + 2x + 6x + 12 = 8x+4$

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Linear Algebra : Linear Algebra

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #191 : Linear Algebra

Example Question #14 : The Trace

Example Question #114 : Operations And Properties

Example Question #14 : The Trace

Example Question #192 : Linear Algebra

Example Question #15 : The Trace

Example Question #193 : Linear Algebra

Example Question #15 : The Trace

Example Question #21 : The Trace

Example Question #22 : The Trace

All Linear Algebra Resources

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