Matrices - Linear Algebra

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Question

$A = \begin{bmatrix} \frac{1}{4} & \frac{\sqrt{3}}{4} \ \ \frac{\sqrt{3}}{4} & \frac{3}{4} \end{bmatrix}$

True or false:

is an example of an idempotent matrix.

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Answer

is an idempotent matrix, by definition, if $A^{2} = A$ . Multiply by itself by multiplying rows by columns - multiplying elements in corresponding positions and adding the products:

$A ^{2}= \begin{bmatrix} \frac{1}{4} & \frac{\sqrt{3}}{4} \ \ \frac{\sqrt{3}}{4} & \frac{3}{4} \end{bmatrix} \begin{bmatrix} \frac{1}{4} & \frac{\sqrt{3}}{4} \ \ \frac{\sqrt{3}}{4} & \frac{3}{4} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{4} \cdot \frac{1}{4} + \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{4} & \frac{1}{4} \cdot \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \cdot \frac{3}{4} \ \\ \ \frac{\sqrt{3}}{4} \cdot \frac{1}{4} + \frac{3}{4} \cdot \frac{\sqrt{3}}{4} & \frac{\sqrt{3}}{4} \cdot \frac{\sqrt{3}}{4} + \frac{3}{4} \cdot \frac{3}{4} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{16} + \frac{ 3}{16} & \frac{\sqrt{3}}{16} + \frac{3\sqrt{3}}{16} \ \\ \ \frac{\sqrt{3}}{16} +\frac{3\sqrt{3}}{16} & \frac{3}{16} + \frac{9}{16} \end{bmatrix}$

$= \begin{bmatrix} \frac{1}{4} & \frac{\sqrt{3}}{4} \ \ \frac{\sqrt{3}}{4} & \frac{3}{4} \end{bmatrix} =A$ .

$A^{2} = A$ , making idempotent.

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Question

True or false: A matrix whose determinant is neither 0 nor 1 cannot be an idempotent matrix.

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Answer

is an idempotent matrix, by definition, if $A^{2} = A$ . Since the determinant of the product of two matrices is equal to the product of their determinants, it follows that

$\det A^{2 }= (\det A )^{2}$

and, since $A^{2} = A$ ,

$\det A^{2 }= \det A$ .

By transitivity,

$(\det A )^{2} = \det A$ .

The only two numbers equal to their own squares are 0 and 1, so

$\det A = 0$ or $\det A = 1$ .

This makes the statement true.

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Question

$A = \begin{bmatrix} 2+ i & 3 - i & i \ -4 & 1+ i & 3 \end{bmatrix}$

Calculate $AA^{T}$ .

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Answer

$A^{T}$ , the transpose of , is the result of switching the rows of with the columns.

$A = \begin{bmatrix} 2+ i & 3 - i & i \ -4 & 1+ i & 3 \end{bmatrix}$ ,

so

$A^{T} = \begin{bmatrix} 2+ i & -4 \ 3- i & 1+ i \i & 3 \end{bmatrix}$

Find the product $AA^{T}$ by multiplying the rows of by the columns of $A^{T}$ ; that is, add the product of the terms in corresponding positions:

$AA^{T}$

$= \begin{bmatrix} 2+ i & 3 - i & i \ -4 & 1+ i & 3 \end{bmatrix} \begin{bmatrix} 2+ i & -4 \ 3- i & 1+ i \i & 3 \end{bmatrix}$

$= \begin{bmatrix}( 2+ i )( 2+ i ) + (3 - i) (3 - i)+ i(i) & ( 2+ i )( -4 ) + (3 - i) (1+i)+ i(3) \ ( -4 )( 2+ i ) + (1+i)(3 - i)+ 3(i) & -4(-4) +(1+ i)(i+1) + 3(3) \end{bmatrix}$

$= \begin{bmatrix}10-2i & -4+i \ -4+i & 25+2i \end{bmatrix}$

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Question

Compute $A\cdot B$ , where

$A=\begin{bmatrix}8&4 \ 2&4 \ 6&1 \end{bmatrix}$ $b=\begin{bmatrix}4&2 \ 4&9 \ 2&1 \end{bmatrix}$

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Answer

In order to be able to multiply matrices, the number of columns of the 1st matrix must equal the number of rows in the second matrix. Here, the first matrix has dimensions of (3x2). This means it has three rows and two columns. The second matrix has dimensions of (3x2), also three rows and two columns. Since $\tiny \tiny \small 2\neq3$ , we cannot multiply these two matrices together

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Question

Compute $A\cdot B$ , where

$A = \begin{bmatrix} 3&4 \ 2&7 \ 6&5 \end{bmatrix}$

$B = \begin{bmatrix} 2&2 \ 5&7 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (3x2). The product matrix equals, $\begin{bmatrix} (3\cdot 2+4\cdot 5)&(3\cdot 2 +4\cdot 7) \ (2\cdot 2 +7\cdot 5)&(2\cdot 2 +7\cdot 7) \ (6\cdot 2 +5\cdot 5)& (6\cdot 2 +5\cdot 7) \end{bmatrix}$

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Question

Compute $A\cdot B$ , where

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

$B=\begin{bmatrix} 8&4 \ 3&5 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (2x2). Next, we notice that matrix A is the identity matrix. Any matrix multiplied by the identity matrix remains unchanged.

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Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 3&4 &0 \ 2&7 &1 \ 6&5 &7 \end{bmatrix}$

$B =\begin{bmatrix} 2&2 &8 \ 5&7 &0 \ 6&4 &3 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (3x3). The product matrix equals, $\begin{bmatrix} (3\cdot 2+4\cdot 5+0\cdot 6)&(3\cdot 2 +4\cdot 7+0\cdot 4)&(3\cdot 8 +4\cdot 0+0\cdot 3) \ (2\cdot 2 +7\cdot 5+1\cdot 6)&(2\cdot 2 +7\cdot 7+1\cdot 4)&(2\cdot 8 +7\cdot 0+1\cdot 3) \ (6\cdot 2 +5\cdot 5+7\cdot 6)& (6\cdot 2 +5\cdot 7+7\cdot 4)&(7\cdot 8 +5\cdot 0+7\cdot 3) \end{bmatrix}$

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Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 4&5 & 6 \end{bmatrix}$

$B=\begin{bmatrix} 0&1 & 4 \end{bmatrix}$

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Answer

In order to be able to multiply matrices, the number of columns of the 1st matrix must equal the number of rows in the second matrix. Here, the first matrix has dimensions of (1x3). This means it has one row and three columns. The second matrix has dimensions of (1x3), also one row and three columns. Since $\tiny \tiny \small 3\neq1$ , we cannot multiply these two matrices together

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Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 4&5 &2 \end{bmatrix}$

$B=\begin{bmatrix} 0\ 1 \ 3 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (1x1). $\begin{bmatrix} (4\cdot 0+5\cdot 1+3\cdot 2) \end{bmatrix}$

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Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 4& 5 \end{bmatrix}$

$B=\begin{bmatrix} 0& 4&2 \ 7&0 &1 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (1x3). The product matrix equals, $\begin{bmatrix} (4\cdot 0+5\cdot 7)&(4\cdot 4 +5\cdot 0)&(4\cdot 2 +5\cdot 1) \end{bmatrix}$

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Question

Compute $A\cdot B$ where,

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

$B=\begin{bmatrix} 7\ 5 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (2x1). The product matrix equals, $\begin{bmatrix} (0\cdot 7+1\cdot 5)\(1\cdot 7+0\cdot 5)\end{bmatrix}$

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Question

Compute $A\cdot B$ where,

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

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

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (2x2). The product matrix equals, $\begin{bmatrix} (0\cdot 1+0\cdot 0)&(0\cdot 1 +0\cdot 0)\(1\cdot 1 +1\cdot 0)&(1\cdot 1 +1\cdot 0) \end{bmatrix}$

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Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 1\ 2\ 5 \end{bmatrix}$

$B=\begin{bmatrix} 1\ 5 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix does not equal the number of rows in the second matrix, you cannot multiply these two matrices.

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Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 12 \end{bmatrix}$

$B=\begin{bmatrix} 4&5 &1 &0 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (1x4). The product matrix equals, $\begin{bmatrix} 12\cdot 4&12\cdot 5&12\cdot 1 &12\cdot 0 \end{bmatrix}$

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Question

Compute $A\cdot B$ where,

$A=\begin{bmatrix} 0&1 &4 \ 4&0 &7 \ 2& 5& 0 \end{bmatrix}$

$B=\begin{bmatrix} 1\ 2\ 0 \end{bmatrix}$

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Answer

Since the number of columns in the first matrix equals the number of rows in the second matrix, we know that these matrices can be multiplied together. To determine the dimensions of the product matrix, we take the number of rows in the first matrix and the number of columns in the second matrix. For this example, our product matrix will have dimensions of (3x1). The product matrix equals, $\begin{bmatrix} (0\cdot 1+1\cdot 2+4\cdot 0)\(4\cdot 1+0\cdot 2+7\cdot 0)\(2\cdot 1+5\cdot 2+0\cdot 0)\end{bmatrix}$

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Question

$C = \begin{bmatrix} 4& 0.7\ 7& -0.4 \end{bmatrix}$ .

Evaluate $C^{T}C$ .

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Answer

The transpose of a matrix switches the rows and the columns. Therefore, the first column of $C^{T}$ has the same entries, in order, as the first row of , and so forth. Since

$C = \begin{bmatrix} 4& 0.7\ 7& -0.4 \end{bmatrix}$ ,

$C^{T} = \begin{bmatrix} 4& 7\ 0.7& -0.4 \end{bmatrix}$ .

The entry in column , row of the product $C^{T}C$ is the product of row of $C^{T}$ and column of - the sum of the products of the numbers that appear in the corresponding positions of the row and the column. $C^{T}C$ can consequently be calculated as follows:

$C^{T}C$

$= \begin{bmatrix} 4& 7\ 0.7& -0.4 \end{bmatrix} \begin{bmatrix} 4& 0.7\ 7& -0.4 \end{bmatrix}$

$= \begin{bmatrix} 4 (4) + 7 (7) & 4 (0.7) + 7 (-0.4)\ 0.7 (4) + (-0.4) (7) & 0.7 (0.7) + (-0.4) (-0.4) \end{bmatrix}$

$= \begin{bmatrix}16+49 & 2.8 + (-2.8) \ 2.8 + (-2.8) & 0.49 + 0.16 \end{bmatrix}$

$= \begin{bmatrix}65 & 0 \ 0 & 0.65 \end{bmatrix}$ ,

the correct choice.

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Question

Your friend Hector wants to multiply two matrices and as follows: $A\cdot B$ . Unfortunately, Hector knows nothing about matrix dimensions. Which of the following statements will help Hector figure out whether it is possible for him to multiply $A\cdot B$ ?

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Answer

Whenever we multiple two matrices together we must always check first that the number of columns in the first matrix is equal to the number of rows in the second matrix. For example, consider these two matrices

$\begin{pmatrix} 1 & 2 & 3\ 4&5&6 \end{pmatrix} \cdot \begin{pmatrix} 10 & 9 & 8 & 7\ 6&5&4&3\ 2&1&0&-1 \end{pmatrix}$

The first matrix has 3 columns, and the second matrix has 3 rows. We can multiple these two matrices together in this order. However, if we switch the order around, we will not be able to multiply these two matrices.

$\begin{pmatrix} 10 & 9 & 8 & 7\ 6&5&4&3\ 2&1&0&-1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3\ 4&5&6 \end{pmatrix} = ?$

Now the first matrix has 4 columns and the second matrix has 2 rows. We cannot multiply these two matrices in this order.

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Question

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

Evaluate $C^{T}C$ .

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Answer

The transpose of a matrix switches the rows and the columns. Therefore, the first column of $C^{T}$ has the same entries, in order, as the first row of , and so forth. Since

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

it follows that

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

The entry in column , row of the product $C^{T}C$ is the product of row of $C^{T}$ and column of - the sum of the products of the numbers that appear in the corresponding positions of the row and the column. Therefore,

$C^{T}C$

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

$= \begin{bmatrix} 3(3)+ (-7)(-7)&3(0)+ (-7)(1)& 3(6)+ (-7)(2) \ 0(3)+1(-7)&0(0)+ 1(1) & 0(6)+1(2) \ 6(3)+2(-7)&6(0)+ 2 (1) & 6(6)+ 2(2) \end{bmatrix}$

$= \begin{bmatrix} 9+49 &0 + (-7)& 18+ (-14 ) \ 0 + (-7)&0+1 & 0+2 \ 18 + (-14)&0+2& 36+4\end{bmatrix}$

$= \begin{bmatrix} 58 &-7 & 4 \ -7& 1 & 2 \ 4 & 2& 40\end{bmatrix}$

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Question

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

Evaluate $C^{2}$ .

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Answer

Two matrices can be multiplied if and only if the number of columns in the first matrix is equal to the number of rows in the second matrix. Therefore, for the square of a matrix to be defined, the number of rows and columns in that matrix must be the same; that is, it must be a square matrix. , having two rows and three columns, is not square, so $C^{2}$ cannot exist.

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Question

Find the product of these two matrices, if it exists.

$\begin{pmatrix} 2 & 0 & 1 & 6\ 1 & 9 & 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2\ 5 & 0 \ -3& 2\ 1& 3\end{pmatrix}$

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Answer

First we check that the dimensions match. The first matrix has 4 columns, and the second matrix has 4 rows. So the matrix product does exist. We find the product by taking the dot product of rows and columns.

$\begin{pmatrix} {\color{Red} 2} & {\color{blue} 0} & {\color{DarkGreen} 1} & {\color{Purple} 6}\ 1 & 9 & 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} {\color{Red} 1} & 2\ {\color{blue} 5} & 0 \ {\color{DarkGreen} -3}& 2\ {\color{Purple} 1}& 3\end{pmatrix} = \begin{pmatrix} {\color{Red} 2\cdot 1} + {\color{Blue} 0\cdot 5} + {\color{DarkGreen} 1\cdot -3} + {\color{Purple} 6\cdot 1} & ?\ ?&? \end{pmatrix} =\begin{pmatrix} 5 & ?\ ? & ? \end{pmatrix}$

$\begin{pmatrix} {\color{Red} 2} & {\color{blue} 0} & {\color{DarkGreen} 1} & {\color{Purple} 6}\ 1 & 9 & 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} 1 & {\color{Red} 2} \ 5 & {\color{blue} 0} \ -3 & {\color{DarkGreen} 2}\ 1 & {\color{Purple} 3}\end{pmatrix} = \begin{pmatrix} 5 & {\color{Red} 2\cdot 2} + {\color{Blue} 0\cdot 0} + {\color{DarkGreen} 1\cdot 2} + {\color{Purple} 6\cdot 3} \ ?&? \end{pmatrix} =\begin{pmatrix} 5 & 24 \ ? & ? \end{pmatrix}$

We fill in the rest of the entries in the product matrix in the same way.

$\begin{pmatrix} 2 & 0 & 1 & 6\ 1 & 9 & 9 & 6 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2\ 5 & 0 \ -3& 2\ 1& 3\end{pmatrix} = \begin{pmatrix} 5 & 24\ 25 & 38 \end{pmatrix}$

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