Linear Algebra - Matrix-Matrix Product

Let $A = \begin{bmatrix} \frac{2}{5} & \frac{1}{4}\end{bmatrix}$ and $B = \begin{bmatrix} \frac{3}{2} & \frac{1}{4} \ \ \frac{3}{4} &- \frac{1}{5}\end{bmatrix}$ .

Find .

$\begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

$\begin{bmatrix} \frac{53}{80} & \frac{1}{4} \end{bmatrix}$

$\begin{bmatrix} \frac{63}{80} \ \ \frac{1}{20} \end{bmatrix}$

$\begin{bmatrix} \frac{53}{80} \ \ \frac{1}{4} \end{bmatrix}$

is undefined.

Explanation

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$AB= \begin{bmatrix} \frac{2}{5} & \frac{1}{4}\end{bmatrix} \begin{bmatrix} \frac{3}{2} & \frac{1}{4} \ \ \frac{3}{4} &- \frac{1}{5}\end{bmatrix}$ .

$= \begin{bmatrix} \frac{2}{5}\left ( \frac{3}{2} \right ) + \frac{1}{4} \left ( \frac{3}{4} \right )& \frac{2}{5}\left ( \frac{1}{4} \right ) + \frac{1}{4} \left ( -\frac{1}{5} \right ) \end{bmatrix}$

$= \begin{bmatrix} \frac{3}{5} + \frac{3}{16} & \frac{1}{10} - \frac{1}{20} \end{bmatrix}$

$= \begin{bmatrix} \frac{63}{80} & \frac{1}{20} \end{bmatrix}$

Let $A = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \end{bmatrix}$ and $B = \begin{bmatrix} \sqrt{2} & \sqrt{6} \-2 \sqrt{6} & \sqrt{3} \end{bmatrix}$ .

Find .

$\begin{bmatrix} 4-6 \sqrt{2} & 3+ 4 \sqrt{3}\end{bmatrix}$

$\begin{bmatrix} 4&3 \ -6 \sqrt{2}& 4 \sqrt{3}\end{bmatrix}$

$\begin{bmatrix} 7 & 6 \sqrt{2}+ 4\sqrt{3} \end{bmatrix}$

$\begin{bmatrix} 4& -6 \sqrt{2} \3& 4 \sqrt{3}\end{bmatrix}$

is undefined.

Explanation

First, it must be established that is defined. This is the case if and only if has as many columns as has rows. Since has two columns and has two rows, is defined.

Matrices are multiplied by multiplying each row of the first matrix by each column of the second - that is, by adding the products of the entries in corresponding positions. Thus,

$A B = \begin{bmatrix} 2\sqrt{2} & \sqrt{3} \end{bmatrix} \begin{bmatrix} \sqrt{2} & \sqrt{6} \-2 \sqrt{6} & \sqrt{3} \end{bmatrix}$

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

$= \begin{bmatrix} 4-6 \sqrt{2} & 3+ 4 \sqrt{3}\end{bmatrix}$

What is dimension criteria to multiply two matrices $(A\times B)$ ?

$(A\times B)$ can only be multiplied if the number of columns in equals the number of rows in

$(A\times B)$ can only be multiplied if the number of rows in equals the number of rows in

$(A\times B)$ can only be multiplied if the number of columns in equals the number of columns in

$(A\times B)$ can only be multiplied if the number of rows in equals the number of columns in

Explanation

If is an $(m\times n)$ matrix and is an $(r\times s)$ matrix,

$(A\times B)$ can only be multiplied if , or the number of columns in equals the number of rows in . Otherwise there is a mismatch, and the two matrices can not be multiplied.

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

True

False

Explanation

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.

For any given value , how many $n \times n$ nonsingular idempotent matrices exist?

One

Zero

Two

Infinitely many

Explanation

is nonsingular, by definition, if it has an inverse - that is, if $A^{-1}$ exists. is an idempotent matrix, by definition, if

$A^{2} = A$

Premultiplying both sides of the equation by $A^{-1}$ , we get

$A^{-1}(A^{2}) = A^{-1}A$

$A^{-1}(AA )= I$ ,

where is the $n \times n$ identity matrix.

Matrix multiplication is associative, so

$(A^{-1}A)A = I$

Therefore, the only nonsingular idempotent matrix of a given dimension $n \times n$ is the identity.

Compute $A\cdot B$ , where

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

Not Possible

$\begin{bmatrix} 32&16 &8 \ 8&36 &24 \end{bmatrix}$

$\begin{bmatrix} 32&8 \ 8&36 \ 12&1 \end{bmatrix}$

$\begin{bmatrix}8&32 \ 36&8 \ 1&12 \end{bmatrix}$

$\begin{bmatrix} 16\-28 \ 11 \end{bmatrix}$

Explanation

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 $\tiny \tiny \small 2 eq3$ , we cannot multiply these two matrices together

is a $2 \times 2$ nonsingular matrix; is a $3 \times 2$ matrix; is a $2 \times 2$ singular matrix.

Which of the following is defined?

$BA^{-1}C$

$C^{-1}AB^{T}$

$CA^{-1}B^{-1}$

$A^{-1}B^{T}C$

Explanation

$CA^{-1}B^{-1}$ can be eliminated as a choice; is not a square matrix, so its inverse, $B^{-1}$ , is undefined.

$C^{-1}AB^{T}$ can be eliminated as a choice, since it is given that is singular - that is, $C^{-1}$ does not exist.

For the product of three matrices to be defined, they must be, in order, an $m \times n$ matrix, an $n \times p$ matrix, and a $p \times q$ matrix. We examine the three remaining choices.

$A^{-1}B^{T}C$ :

is a $2 \times 2$ matrix, so $A^{-1}$ is as well; is a $3 \times 2$ matrix, so its transpose $B^{T}$ is a $2 \times 3$ matrix; is a $2 \times 2$ matrix. These are incompatible, so $A^{-1}B^{T}C$ is undefined.

is a $2 \times 2$ matrix; is a $2 \times 2$ matrix; is a $3 \times 2$ matrix; These are also incompatible, so is undefined.

$BA^{-1}C$ :

is a $3 \times 2$ matrix; $A^{-1}$ is a $2 \times 2$ matrix, as stated before; is a $2 \times 2$ matrix. These are compatible, so $BA^{-1}C$ is defined. This is the correct choice.

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

True or false:

is an example of an idempotent matrix.

True

False

Explanation

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.

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 $A\cdot B$ ?

The number of columns in matrix must be equal to the number of rows in matrix .

The number of rows in matrix must be equal to the number of rows in matrix .

The number of rows in matrix must be equal to the number of columns in matrix .

The number of columns in matrix must be equal to the number of columns in matrix .

and must both be square matrices, otherwise you cannot multiply them.

Explanation

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.

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