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Linear Algebra : Operations and Properties

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Example Questions

Example Question #23 : Orthogonal Matrices

$A = \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$

True or false: is an example of a unitary matrix.

Possible Answers:

True

False

Correct answer:

True

Explanation:

is unitary if $AA^{*} = I$ , where $A^{*}$ is its conjugate transpose.

$A = \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$ ,

so transpose rows and columns to get

$A^{T} = \begin{bmatrix} 0 & 0 &- i \\ 0 &1 & 0 \\ i & 0 & 0 \end{bmatrix}$

Now change each entry to its complex conjugate:

$A^{*} = \begin{bmatrix} 0 & 0 &- i \\ 0 &1 & 0 \\ i & 0 & 0 \end{bmatrix}$

Find $AA^{*}$ by multiplying rows of by columns of $A^{*}$ - adding the products of corresponding entries, as follows:

$AA^{*} = \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & i \\ 0 &1 & 0 \\- i & 0 & 0 \end{bmatrix}$

$= \begin{bmatrix} 0(0)+0(0) + i(-i) & 0(0)+0(1)+ i(0) & 0(i)+0(0)+ i(0) \\ 0(0)+1(0) +0(-i) & 0(0)+1(1)+0(0) & 0(i)+1(0)+0i(0) \\-i(0)+0(0) + 0(-i) & -i(0)+0(1)+0(0) & -i(i)+0(0)+ 0(0) \end{bmatrix}$

$= \begin{bmatrix} 1 & 0 & 0 \\ 0& 1 & 0 \\0& 0 & 1\end{bmatrix}$

$AA^{*} = I$ , so is unitary.

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Example Question #21 : Orthogonal Matrices

Is the following matrix M orthogonal?

$\textbf{M}=\begin{bmatrix} \cos(\theta)& -\sin(\theta)\\ \sin(\theta)& \cos(\theta) \end{bmatrix}$

Possible Answers:

Yes, M is orthogonal

No, M is not orthogonal

It is impossible to determine from the information given.

Correct answer:

Yes, M is orthogonal

Explanation:

An orthogonal matrix is a real, square matrix that satisfies the following condition:

$\textbf{M}^T\textbf{M}=\textbf{M}\textbf{M}^T=\textbf{I}$

where I is the identity matrix.

We can calculate the product of M and its transpose:

$\textbf{M}\textbf{M}^T=\begin{bmatrix} \cos(\theta)& -\sin(\theta)\\ \sin(\theta)& \cos(\theta) \end{bmatrix} \begin{bmatrix} \cos(\theta)& \sin(\theta)\\ -\sin(\theta)& \cos(\theta) \end{bmatrix}$

$=\begin{bmatrix} \cos(\theta)\cos(\theta)+(-\sin(\theta))(-\sin(\theta))& \cos(\theta)\sin(\theta)+(-\sin(\theta))\cos(\theta)\\ \sin(\theta)\cos(\theta)+\cos(\theta)(-\sin(\theta))& \sin(\theta)\sin(\theta)+\cos(\theta)\cos(\theta) \end{bmatrix}$

$=\begin{bmatrix} \cos^2(\theta)+\sin^2(\theta)& \cos(\theta)\sin(\theta)-\sin(\theta)\cos(\theta)\\ \sin(\theta)\cos(\theta)-\sin(\theta)\cos(\theta)& \sin^2(\theta)+\cos^2(\theta) \end{bmatrix}$

$\begin{bmatrix} 1& 0\\ 0& 1 \end{bmatrix}=\textbf{I}$

Therefore, M is indeed an orthogonal matrix.

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Example Question #1 : Range And Null Space Of A Matrix

Calculate the Null Space of the following Matrix.

$A=\left[\begin{matrix} 1 & 1& 1& -1 \\ 2 & 4&5 &6 \\ 3& 9& 5& 4 \end{matrix}\right]$

Possible Answers:

$N(A)=Span\begin{bmatrix} 0\\ \\ -\frac{5}{14}\\ \\ -\frac{17}{7} \\ 0 \end{bmatrix}$

$N(A)=Span\begin{bmatrix} 0\\ \\ 0\\ \\ 0 \\ 1 \end{bmatrix}$

There is no Null Space

$N(A)=Span\begin{bmatrix} \frac{53}{14}\\ \\ -\frac{5}{14}\\ \\ -\frac{17}{7} \\ 1 \end{bmatrix}$

$N(A)=Span\begin{bmatrix} \frac{53}{14}\\ \\ 0\\ \\ -\frac{17}{7} \\ 1 \end{bmatrix}$

Correct answer:

$N(A)=Span\begin{bmatrix} \frac{53}{14}\\ \\ -\frac{5}{14}\\ \\ -\frac{17}{7} \\ 1 \end{bmatrix}$

Explanation:

The first step is to create an augmented matrix having a column of zeros.

$\left[\begin{matrix} 1 & 1& 1& -1 \\ 2 & 4&5 &6 \\ 3& 9& 5& 4 \end{matrix}\right|\left.\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right]$

The next step is to get this into RREF.

$2R_1-R_2\rightarrow R_2$

$\left[\begin{matrix} 1 & 1& 1& -1 \\ 0 & -2&-3 &-8 \\ 3& 9& 5& 4 \end{matrix}\right|\left.\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right]$

$3R_1-R_3\rightarrow R_3$

$\left[\begin{matrix} 1 & 1& 1& -1 \\ 0 & -2&-3 &-8 \\ 0& -6& -2& -7 \end{matrix}\right|\left.\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right]$

$-3R_2+R_3\rightarrow R_3$

$\left[\begin{matrix} 1 & 1& 1& -1 \\ 0 & -2&-3 &-8 \\ 0& 0& 7& 17 \end{matrix}\right|\left.\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right]$

$\frac{1}{2}R_2+R_1\rightarrow R_1$

$\left[\begin{matrix} 1 & 0& -\frac{1}{2}& -5 \\ 0 & -2&-3 &-8 \\ 0& 0& 7& 17 \end{matrix}\right|\left.\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right]$

We can simplify to

$\left[\begin{matrix} 1 & 0& -\frac{1}{2}& -5 \\ 0 & 1&\frac{3}{2} &4 \\ 0& 0& 1& \frac{17}{7} \end{matrix}\right|\left.\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right]$

$-\frac{3}{2}R_3+R_2\rightarrow R_2$

$\left[\begin{matrix} 1 & 0& -\frac{1}{2}& -5 \\ 0 & 1&0 & \ \frac{5}{14} \\ 0& 0& 1& \frac{17}{7} \end{matrix}\right|\left.\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right]$

$\frac{1}{2}R_3+R_1\rightarrow R_1$

$\left[\begin{matrix} 1 & 0& 0& -\frac{53}{14} \\ 0 & 1&0 & \ \frac{5}{14} \\ 0& 0& 1& \frac{17}{7} \end{matrix}\right|\left.\begin{matrix} 0\\ 0\\ 0 \end{matrix}\right]$

This tells us the following.

$\\x_1-\frac{53}{14}x_4=0 \\ \\ x_2+\frac{5}{14}x_4=0 \\ \\ x_3+\frac{17}{7}x_4=0$

$\\x_1=\frac{53}{14}x_4 \\ \\ x_2=-\frac{5}{14}x_4 \\ \\ x_3=-\frac{17}{7}x_4$

Now we need to write this as a linear combination.

$\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4 \end{bmatrix} =x_4\begin{bmatrix} \frac{53}{14}\\ \\ -\frac{5}{14}\\ \\ -\frac{17}{7} \\ 1 \end{bmatrix}$

The null space is then

$N(A)=Span\begin{bmatrix} \frac{53}{14}\\ \\ -\frac{5}{14}\\ \\ -\frac{17}{7} \\ 1 \end{bmatrix}$

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Example Question #1 : Range And Null Space Of A Matrix

Find a basis for the range space of the transformation given by the matrix $A=\begin{bmatrix} 1& 2& 3&4 \\ -1&-3 &-2 &-5 \\ 0& 1& 1& 3 \end{bmatrix}$ .

Possible Answers:

None of the other answers.

$\begin{Bmatrix} \begin{bmatrix} 0\\ 0\\ 0\\ \end{bmatrix} \end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix} _,\begin{bmatrix} 0\\ 0\\ 1\\ \end{bmatrix} _, \begin{bmatrix} -3\\ 2\\ 1\\ \end{bmatrix} \end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix} \end{Bmatrix}$

Correct answer:

None of the other answers.

Explanation:

We can find a basis for 's range space first by finding a basis for the column space of its reduced row echelon form.

Using a calculator or row reduction, we obtain

$\begin{bmatrix} 1& 0& 0&-3 \\ 0& 1&0 & 2\\ 0&0 &1 & 1 \end{bmatrix}$

for the reduced row echelon form.

The fourth column in this matrix can be seen by inspection to be a linear combination of the other three columns, so it is not included in our basis. Hence the first three columns form a basis for the column space of the reduced row echelon form of , and therefore the first three columns of form a basis for its range space.

$\begin{Bmatrix} \begin{bmatrix} 1\\ -1\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 2\\ -3\\ 1\\ \end{bmatrix} _,\begin{bmatrix} 3\\ -2\\ 1\\ \end{bmatrix} \end{Bmatrix}$ .

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Example Question #1 : Range And Null Space Of A Matrix

Find a basis for the range space of the transformation given by the matrix $A=\begin{bmatrix} 1& 2& 3 \\ 0&0&2\\ 0& 0& -1\end{bmatrix}$ .

Possible Answers:

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 3\\ 2\\ -1\\ \end{bmatrix}\end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 2\\ 0\\ 0\\ \end{bmatrix}_,\begin{bmatrix} 3\\ 2\\ -1\\ \end{bmatrix}\end{Bmatrix}$

None of the other answers

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 2\\ 0\\ 0\\ \end{bmatrix}_,\begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix}\end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 0\\ 1\\ 0\\ \end{bmatrix}\end{Bmatrix}$

Correct answer:

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 3\\ 2\\ -1\\ \end{bmatrix}\end{Bmatrix}$

Explanation:

We can find a basis for 's range space first by finding a basis for the column space of its reduced row echelon form.

Using a calculator or row reduction, we obtain

$R=\begin{bmatrix} 1& 2& 0 \\ 0& 0&1\\ 0&0&0 \end{bmatrix}$

for the reduced row echelon form.

The second column in this matrix can be seen by inspection to be a linear combination of the first column, so it is not included in our basis for . Hence the first and the third columns form a basis for the column space of , and therefore the first and the third columns of form a basis for the range space of .

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ \end{bmatrix} _, \begin{bmatrix} 3\\ 2\\ -1\\ \end{bmatrix}\end{Bmatrix}$

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Example Question #1 : Range And Null Space Of A Matrix

Find a basis for the range space of the transformation given by the matrix $A=\begin{bmatrix} 0& 1& 1&0\\ 0&0&1&1\\ 1&0&0&1\\1&1&0&0\end{bmatrix}$ .

Possible Answers:

$\begin{Bmatrix} \begin{bmatrix} 0\\ 0\\ 1\\ 1\\ \end{bmatrix}_,\begin{bmatrix} 1\\ 0\\ 0\\ 1\\ \end{bmatrix}_, \begin{bmatrix} 1\\ 1\\ 0\\ 0\\ \end{bmatrix} \end{Bmatrix}$

None of the other answers

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ 0\\ \end{bmatrix}_,\begin{bmatrix} 0\\ 1\\ 0\\ 0\\ \end{bmatrix}_, \begin{bmatrix} 0\\ 0\\ 1\\ 0\\ \end{bmatrix}_, \begin{bmatrix} 1\\ -1\\ 1\\ 0\\ \end{bmatrix} \end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ 0\\ 0\\ \end{bmatrix}_,\begin{bmatrix} 0\\ 1\\ 0\\ 0\\ \end{bmatrix}_, \begin{bmatrix} 0\\ 0\\ 1\\ 0\\ \end{bmatrix} \end{Bmatrix}$

Correct answer:

$\begin{Bmatrix} \begin{bmatrix} 0\\ 0\\ 1\\ 1\\ \end{bmatrix}_,\begin{bmatrix} 1\\ 0\\ 0\\ 1\\ \end{bmatrix}_, \begin{bmatrix} 1\\ 1\\ 0\\ 0\\ \end{bmatrix} \end{Bmatrix}$

Explanation:

We can find a basis for 's range space first by finding a basis for the column space of its reduced row echelon form.

Using a calculator or row reduction, we obtain

$R=\begin{bmatrix} 1& 0& 0&1\\ 0& 1&0&-1\\ 0&0&1&1\\0&0&0&0 \end{bmatrix}$

for the reduced row echelon form.

The fourth column in this matrix can be seen by inspection to be a linear combination of the first three columns, so it is not included in our basis for . Hence the first three columns form a basis for the column space of , and therefore the first three columns of form a basis for the range space of .

$\begin{Bmatrix} \begin{bmatrix} 0\\ 0\\ 1\\ 1\\ \end{bmatrix}_,\begin{bmatrix} 1\\ 0\\ 0\\ 1\\ \end{bmatrix}_, \begin{bmatrix} 1\\ 1\\ 0\\ 0\\ \end{bmatrix} \end{Bmatrix}$

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Example Question #3 : Range And Null Space Of A Matrix

Find a basis for the null space of the matrix $A= \begin{bmatrix} 5& 2\\ 1& 0 \end{bmatrix}$ .

Possible Answers:

$\begin{Bmatrix} \begin{bmatrix} 0\\ 1\\ \end{bmatrix} \end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ \end{bmatrix} \end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\ \end{bmatrix}_, \begin{bmatrix} 0\\ 1\\ \end{bmatrix} \end{Bmatrix}$

None of the other answers

$\begin{Bmatrix} \begin{bmatrix} 0\\ 0\\ \end{bmatrix} \end{Bmatrix}$

Correct answer:

$\begin{Bmatrix} \begin{bmatrix} 0\\ 0\\ \end{bmatrix} \end{Bmatrix}$

Explanation:

The null space of the matrix is the set of solutions to the equation

$\begin{bmatrix} 5 &2\\ 1& 0 \end{bmatrix} \begin{bmatrix} x_1\\x_2\end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix}$ .

We can solve the above system by row reducing using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0\\0 & 1 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \end{bmatrix}= \begin{bmatrix} 0\\0 \end{bmatrix} \Longrightarrow \begin{align*} x_1 &=0 \\x_2 &=0 \end{align*} \Longrightarrow$

$X= \begin{bmatrix} 0\\0 \end{bmatrix}$

Hence a basis for the null space is just the zero vector;

$\begin{Bmatrix} \begin{bmatrix} 0\\ 0\\ \end{bmatrix} \end{Bmatrix}$ .

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Example Question #1 : Range And Null Space Of A Matrix

Find the null space of the matrix operator $T_M = \begin{bmatrix} 1 &1 & 0\\ 0& -1& 1\\ 9& 0& 1\end{bmatrix}$ .

Possible Answers:

None of the other answers

$Span\begin{Bmatrix} \begin{bmatrix} 1\\0 \\0 \end{bmatrix} \end{Bmatrix}$

$Span\begin{Bmatrix} \begin{bmatrix} 1\\0 \\2 \end{bmatrix} \end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \end{Bmatrix}$

$Span\begin{Bmatrix} \begin{bmatrix} 1\\1 \\0 \end{bmatrix} \end{Bmatrix}$

Correct answer:

$\begin{Bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \end{Bmatrix}$

Explanation:

The null space of the operator T_M is the set of solutions to the equation

$\begin{bmatrix} 1 &1 & 0\\ 0& -1& 1\\ 9& 0& 1\end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix} 0\\0 \\0 \end{bmatrix}$ .

We can solve the above system by row reducing our $3 \times 3$ matrix using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0 & 0\\0 & 1& 0\\ 0& 0&1 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \\x_3 \end{bmatrix}= \begin{bmatrix} 0\\0 \\0 \end{bmatrix} \Longrightarrow \begin{align*} x_1 &=0 \\x_2 &=0 \\ x_3&=0 \end{align*}$

Hence the null space consists of only the zero vector. $\begin{Bmatrix} \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \end{Bmatrix}$

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Example Question #1 : Range And Null Space Of A Matrix

Find the null space of the matrix $A = \begin{bmatrix} 2 & 0& 0\\1 &0 &0 \\9 &8 &2 \end{bmatrix}$ .

Possible Answers:

$Span\begin{Bmatrix} \begin{bmatrix} 0\\ -1\\4 \\ \end{bmatrix} \end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 0\\ 0\\0 \\ \end{bmatrix} \end{Bmatrix}$

$Span\begin{Bmatrix} \begin{bmatrix} 1\\ -2\\0 \\ \end{bmatrix} \end{Bmatrix}$

$Span\begin{Bmatrix} \begin{bmatrix} 1\\ 0\\1 \\ \end{bmatrix} \end{Bmatrix}$

None of the other answers

Correct answer:

$Span\begin{Bmatrix} \begin{bmatrix} 0\\ -1\\4 \\ \end{bmatrix} \end{Bmatrix}$

Explanation:

The null space of the matrix is the set of solutions to the equation

$\begin{bmatrix} 2 &0 & 0\\ 1& 0& 0\\ 9& 8& 2\end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3 \end{bmatrix} = \begin{bmatrix} 0\\0 \\0 \end{bmatrix}$ .

We can solve the above system by row reducing using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0 & 0\\0 & 1& 1/4\\ 0& 0&0 \end{bmatrix}\begin{bmatrix} x_1\\x_2 \\x_3 \end{bmatrix}= \begin{bmatrix} 0\\0 \\0 \end{bmatrix} \Longrightarrow \begin{align*} x_1 &=0 \\x_2 &=-(1/4) x_3 \\ x_3&=x_3 \end{align*} \Longrightarrow$

$X= \begin{bmatrix} 0\\-1/4 \\1 \end{bmatrix}t$

Multiplying this vector by gets rid of the fraction, and does not affect our answer, since there is an arbitrary constant behind it.

Hence the null space consists of all vectors spanned by $\begin{bmatrix} 0\\-1 \\4 \end{bmatrix}$ ;

$Span\begin{Bmatrix} \begin{bmatrix} 0\\ -1\\4 \end{bmatrix} \end{Bmatrix}$ .

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Example Question #2 : Range And Null Space Of A Matrix

Find the null space of the matrix $A = \begin{bmatrix} 2 & 0\\2 &0 \end{bmatrix}$ .

Possible Answers:

$Span\begin{Bmatrix} \begin{bmatrix} 0\\1 \end{bmatrix} \end{Bmatrix}$

$\begin{Bmatrix} \begin{bmatrix} 0\\0 \end{bmatrix} \end{Bmatrix}$

$Span\begin{Bmatrix} \begin{bmatrix} 1\\0 \end{bmatrix} \end{Bmatrix}$

$\mathbb{R}^2$

$Span\begin{Bmatrix} \begin{bmatrix} 1\\1 \end{bmatrix} \end{Bmatrix}$

Correct answer:

$Span\begin{Bmatrix} \begin{bmatrix} 0\\1 \end{bmatrix} \end{Bmatrix}$

Explanation:

The null space of the matrix is the set of solutions to the equation

$\begin{bmatrix} 2 &0\\ 2& 0\\ \end{bmatrix} \begin{bmatrix} x_1\\x_2 \end{bmatrix} = \begin{bmatrix} 0\\0 \end{bmatrix}$ .

We can solve the above system by row reducing using either row reduction, or a calculator to find its reduced row echelon form. After that, our system becomes

$\begin{bmatrix} 1 &0 \\0 &0\end{bmatrix}\begin{bmatrix} x_1\\x_2 \end{bmatrix}= \begin{bmatrix} 0\\0 \end{bmatrix} \Longrightarrow \begin{align*} x_1 &=0 \\x_2 &=x_2 \end{align*} \Longrightarrow$

$X= \begin{bmatrix} 0\\1 \end{bmatrix}t$

Hence the null space consists of all vectors spanned by $\begin{bmatrix} 0\\1 \end{bmatrix}$ ;

$Span\begin{Bmatrix} \begin{bmatrix} 0\\ 1\\\end{bmatrix} \end{Bmatrix}$ .

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Linear Algebra : Operations and Properties

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #23 : Orthogonal Matrices

Example Question #21 : Orthogonal Matrices

Example Question #1 : Range And Null Space Of A Matrix

Example Question #1 : Range And Null Space Of A Matrix

Example Question #1 : Range And Null Space Of A Matrix

Example Question #1 : Range And Null Space Of A Matrix

Example Question #3 : Range And Null Space Of A Matrix

Example Question #1 : Range And Null Space Of A Matrix

Example Question #1 : Range And Null Space Of A Matrix

Example Question #2 : Range And Null Space Of A Matrix

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