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Linear Algebra : Matrices

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Example Question #31 : Matrix Matrix Product

Let $L = \begin{bmatrix} 1& 0\\ 4& 2 \end{bmatrix}$ , $U = \begin{bmatrix} 2&-5 \\ 0& 1 \end{bmatrix}$ , and $A = \begin{bmatrix} 2&-5\\ 8& -18 \end{bmatrix}$ .

True or false: is an example of a valid -factorization of .

Possible Answers:

False, because is not the right kind of matrix.

False, because is not a factorization of .

True

Correct answer:

True

Explanation:

An -factorization is a way of expressing a matrix as a product of two matrices and . For the factorization to be valid:

1) must be a Lower triangular matrix - all elements above its main diagonal (upper left corner to lower right corner) must be "0".

2) must be an Upper triangular matrix - all elements below its main diagonal must be "0".

3) LU = A

The factorization can be seen to satisfy the first two criteria - is lower triangular in that there are no nonzero elements above its main diagonal, and is, analogously, upper triangular. It remains to be shown that LU = A .

Multiply each row in by each column in - add the products of each element in the former by the corresponding element in the latter - as follows:

$LU = \begin{bmatrix} 1& 0\\ 4& 2 \end{bmatrix}\begin{bmatrix} 2&-5 \\ 0& 1 \end{bmatrix}$

$= \begin{bmatrix} 1(2)+ 0(0) & 1(-5)+ 0(1)\\ 4(2)+ 2(0) & 4(-5)+ 2(1) \end{bmatrix}$

$= \begin{bmatrix} 2&-5\\ 8& -18 \end{bmatrix}$

All three criteria are met, and gives a valid -factorization of .

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Example Question #712 : Linear Algebra

Let $L = \begin{bmatrix} 1& 0\\ -3& 2 \end{bmatrix}$ , $U = \begin{bmatrix} 2&2 \\ 0& -3 \end{bmatrix}$ , and $A = \begin{bmatrix} 2& 0\\ 0& -6 \end{bmatrix}$

True or false: is an example of a valid -factorization of .

Possible Answers:

False, because is not a factorization of .

True

False, because is not the right kind of matrix.

Correct answer:

False, because is not a factorization of .

Explanation:

An -factorization is a way of expressing a matrix as a product of two matrices and . For the factorization to be valid:

1) must be a Lower triangular matrix - all elements above its main diagonal (upper left corner to lower right corner) must be "0".

2) must be an Upper triangular matrix - all elements below its main diagonal must be "0".

3) LU = A

Multiply each row in by each column in - add the products of each element in the former by the corresponding element in the latter - as follows:

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

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

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

$LU \ne A$ , so does not give a valid -factorization of .

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Example Question #713 : Linear Algebra

True or False: If is a square matrix, and A^2 = 0 , then is either or .

Possible Answers:

True

False

Correct answer:

False

Explanation:

For example, if $A = \begin{bmatrix} 0&1 \\ 0&0 \end{bmatrix}$ , then $A^2 = \begin{bmatrix} 0& 1\\ 0& 0 \end{bmatrix}\begin{bmatrix} 0& 1\\ 0&0 \end{bmatrix} = \begin{bmatrix} 0&0 \\ 0& 0 \end{bmatrix}$ , but itself is not or .

(If represented a single real number, then the question would be true, but since is a matrix, the question is not true anymore.)

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Example Question #714 : Linear Algebra

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

Possible Answers:

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

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

Correct answer:

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

Explanation:

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

$(A\times B)$ can only be multiplied if n=r , 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.

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Example Question #31 : Matrix Matrix Product

If is a $3 \times 2$ matrix and is a $2 \times 4$ matrix, can the product $A \times B$ be multiplied? What about $B \times A$ ?

Possible Answers:

$A \times B$ can be multiplied

$B \times A$ cannot be multiplied

$A \times B$ can be multiplied

$B \times A$ can be multiplied

$A \times B$ cannot be multiplied

$B \times A$ can be multiplied

$A \times B$ cannot be multiplied

$B \times A$ cannot be multiplied

Correct answer:

$A \times B$ can be multiplied

$B \times A$ cannot be multiplied

Explanation:

If is an $(m\times n)$ matrix and is an $(r\times s)$ matrix, $(A\times B)$ can only be multiplied if n=r .

Since is a $3 \times 2$ matrix and is a $2 \times 4$ matrix, n=r and $A \times B$ can be multiplied.

has rows and has columns, therefore $B \times A$ cannot be multiplied.

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Example Question #31 : Matrices

If is a $3 \times 4$ matrix and is a $4 \times 4$ matrix, what are the dimensions of the product $A \times B$ ?

Possible Answers:

$3 \times 3$

$A \times B$ cannot be multiplied

$4 \times 4$

$3 \times 4$

Correct answer:

$3 \times 4$

Explanation:

If is an $(m\times n)$ matrix and is an $(n\times s)$ matrix, the dimensions of $(A\times B)$ are $(m\times s)$ .

In this problem, If is a $3 \times 4$ matrix and is a $4 \times 4$ matrix, so

the dimensions of $(A\times B)$ are $(3\times 4)$ .

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Example Question #33 : Matrices

If is a $2 \times 5$ matrix and is a $5 \times 2$ matrix, what are the dimensions of the product $A \times B$ ?

Possible Answers:

$(2\times 5)$

$(5\times 2)$

$(2\times 2)$

$A \times B$ cannot be multiplied

Correct answer:

$(2\times 2)$

Explanation:

If is an $(m\times n)$ matrix and is an $(n\times s)$ matrix, the dimensions of $(A\times B)$ are $(m\times s)$ .

In this problem, If is a $2 \times 5$ matrix and is a $5 \times 2$ matrix, so

the dimensions of $(A\times B)$ are $(2\times 2)$ .

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Example Question #34 : Matrices

Find the product $A \times B$ .

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

Possible Answers:

$A \times B=\begin{bmatrix} -1 &5&5\\ 4&-4&4 \end{bmatrix}$

$A \times B=\begin{bmatrix} -2 &-5&5\\ 0 &28&4 \end{bmatrix}$

$A \times B=\begin{bmatrix} -2& -34& 15\\ 8& 24&4 \end{bmatrix}$

$A \times B$ cannot be multiplied

Correct answer:

$A \times B=\begin{bmatrix} -2& -34& 15\\ 8& 24&4 \end{bmatrix}$

Explanation:

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

$(A\times B)$ can only be multiplied if n=r , 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. $(A\times B)$ will be an $(m\times s)$ matrix

Since is a $2\times2$ matrix and is a $2\times3$ matrix, then $(A\times B)$ can be multiplied and will have the dimensions $2\times3$ .

$A \times B=\begin{bmatrix} ab_{11}& ab_{12}& ab_{13}\\ ab_{21}& ab_{22}& ab_{23} \end{bmatrix}$

To find the product $A \times B$ , you must find the dot product of the rows of and the columns of

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

$A \times B=\begin{bmatrix} ab_{11}& ab_{12}& ab_{13}\\ ab_{21}& ab_{22}& ab_{23} \end{bmatrix}$

We find $ab_{11}$ by finding the dot product of the row of and column of .

$ab_{11}=(-1)(2)+(5)(0)=-2+0=-2$

We find $ab_{12}$ by finding the dot product of the row of and column of .

$ab_{12}=(-1)(-1)+(5)(-7)=1-35=-34$

We use the same method to find the rest of the matrix values

$ab_{13}=(-1)(5)+(5)(4)=-5+20=15$

$ab_{21}=(4)(2)+(-4)(0)=8+0=8$

$ab_{22}=(4)(-1)+(-4)(-7)=-4+28=24$

$ab_{23}=(4)(5)+(-4)(4)=20-16=4$

$A \times B=\begin{bmatrix} -2& -34& 15\\ 8& 24&4 \end{bmatrix}$

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Example Question #35 : Matrices

Find the product $A \times B$ .

$A=\begin{bmatrix} -1& 2\\ 4& 3 \end{bmatrix}$ , $B=\begin{bmatrix} 7& -1\\ -2& 5 \end{bmatrix}$

Possible Answers:

$A \times B=\begin{bmatrix} 6& 1\\ 2& 8 \end{bmatrix}$

$A \times B=\begin{bmatrix} -11& 11\\ 22& 11 \end{bmatrix}$

$A \times B=\begin{bmatrix} -7& -2\\ -8& 15 \end{bmatrix}$

$A\times B$ cannot be multiplied

Correct answer:

$A \times B=\begin{bmatrix} -11& 11\\ 22& 11 \end{bmatrix}$

Explanation:

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

Since is a $2\times2$ matrix and is a $2\times2$ matrix, then $(A\times B)$ can be multiplied and will have the dimensions $2\times2$ .

$A \times B=\begin{bmatrix} ab_{11}& ab_{12}\\ ab_{21}& ab_{22} \end{bmatrix}$

To find the product $A \times B$ , you must find the dot product of the rows of and the columns of

$A=\begin{bmatrix} -1& 2\\ 4& 3 \end{bmatrix}$ , $B=\begin{bmatrix} 7& -1\\ -2& 5 \end{bmatrix}$

$A \times B=\begin{bmatrix} ab_{11}& ab_{12}\\ ab_{21}& ab_{22} \end{bmatrix}$

We find $ab_{11}$ by finding the dot product of the row of and column of .

$ab_{11}=(-1)(7)+(2)(-2)=-7-4=-11$

We find $ab_{12}$ by finding the dot product of the row of and column of .

$ab_{12}=(-1)(-1)+(2)(5)=1+10=11$

We use the same method to find the rest of the matrix values

$ab_{21}=(4)(7)+(3)(-2)=28-6=22$

$ab_{22}=(4)(-1)+(3)(5)=-4+15=11$

$A \times B=\begin{bmatrix} -11& 11\\ 22& 11 \end{bmatrix}$

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Example Question #36 : Matrices

Find the product $A \times B$ .

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

Possible Answers:

$A \times B=\begin{bmatrix} -1& 2\\ -2& 5 \end{bmatrix}$

$A\times B$ cannot be multiplied

$A \times B=\begin{bmatrix} 2& 10\\ 4& 9 \end{bmatrix}$

$A \times B=\begin{bmatrix} -11& 11\\ 28& 29 \end{bmatrix}$

Correct answer:

$A \times B=\begin{bmatrix} -11& 11\\ 28& 29 \end{bmatrix}$

Explanation:

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

Since is a $2\times3$ matrix and is a $3\times2$ matrix, then $(A\times B)$ can be multiplied and will have the dimensions $2\times2$ .

$A \times B=\begin{bmatrix} ab_{11}& ab_{12}\\ ab_{21}& ab_{22} \end{bmatrix}$

To find the product $A \times B$ , you must find the dot product of the rows of and the columns of

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

$A \times B=\begin{bmatrix} ab_{11}& ab_{12}\\ ab_{21}& ab_{22} \end{bmatrix}$

We find $ab_{11}$ by finding the dot product of the row of and column of .

$ab_{11}=(-1)(7)+(2)(-2)+(0)(1)=-7-4+0=-11$

We find $ab_{12}$ by finding the dot product of the row of and column of .

$ab_{12}=(-1)(-1)+(2)(5)+(0)(3)=1+10+0=11$

We use the same method to find the rest of the matrix values

$ab_{21}=(4)(7)+(3)(-2)+(6)(1)=28-6+6=28$

$ab_{22}=(4)(-1)+(3)(5)+(6)(3)=-4+15+18=29$

$A \times B=\begin{bmatrix} -11& 11\\ 28& 29 \end{bmatrix}$

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Linear Algebra : Matrices

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #31 : Matrix Matrix Product

Example Question #712 : Linear Algebra

Example Question #713 : Linear Algebra

Example Question #714 : Linear Algebra

Example Question #31 : Matrix Matrix Product

Example Question #31 : Matrices

Example Question #33 : Matrices

Example Question #34 : Matrices

Example Question #35 : Matrices

Example Question #36 : Matrices

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