Try the fastest way to create flashcards
Question

Verify the distributive \mid law AA (BB+CC)= ABAB+ACAC when

A=(1234),B=(21103121),C=(11122201)\mathbf{A}=\left(\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right), \quad \mathbf{B}=\left(\begin{array}{rrrr} 2 & -1 & 1 & 0 \\ 3 & -1 & 2 & 1 \end{array}\right), \quad \mathbf{C}=\left(\begin{array}{rrrr} -1 & 1 & 1 & 2 \\ -2 & 2 & 0 & -1 \end{array}\right)

Solution

Verified
Answered 2 years ago
Answered 2 years ago
Step 1
1 of 3

In order to verify the distributive law, we will compute A(B+C)\boldsymbol{A}(\boldsymbol{B}+\boldsymbol{C}) and AB+AC\boldsymbol{A}\boldsymbol{B}+\boldsymbol{A}\boldsymbol{C} separately, then we will compare them. Let's begin by A(B+C)\boldsymbol{A}(\boldsymbol{B}+\boldsymbol{C}) :

B+C=(21103121)+(11122201)=(211+11+10+2321+22+011)=(10221120)\begin{aligned} \boldsymbol{B}+\boldsymbol{C}&= \begin{pmatrix} 2&-1&1&0\\3&-1&2&1 \end{pmatrix}+ \begin{pmatrix} -1&1&1&2\\-2&2&0&-1 \end{pmatrix}\\ &=\begin{pmatrix} 2-1&-1+1&1+1&0+2\\3-2&-1+2&2+0&1-1 \end{pmatrix}\\ &=\begin{pmatrix} 1&0&2&2\\1&1&2&0 \end{pmatrix} \end{aligned}

then,

A(B+C)=(1234)(10221120)=(11+2110+2112+2212+2031+4130+4132+4232+40)=(326274146)\begin{aligned} \boldsymbol{A}(\boldsymbol{B}+\boldsymbol{C})&= \begin{pmatrix} 1&2\\3&4 \end{pmatrix} \begin{pmatrix} 1&0&2&2\\1&1&2&0 \end{pmatrix}\\ &= \begin{pmatrix} 1\cdot 1+2\cdot 1&1\cdot 0+2\cdot 1&1\cdot 2+2\cdot 2&1\cdot 2+2\cdot 0\\ 3\cdot 1+4\cdot 1&3\cdot 0+4\cdot 1&3\cdot 2+4\cdot 2&3\cdot 2+4\cdot 0\\ \end{pmatrix}\\ &= \begin{pmatrix} 3&2&6&2\\7&4&14&6 \end{pmatrix} \end{aligned}

Create a free account to view solutions

Create a free account to view solutions

Recommended textbook solutions

Mathematics for Business and Personal Finance 1st Edition by Lange, Rousos

Mathematics for Business and Personal Finance

1st EditionISBN: 9780078805059 (1 more)Lange, Rousos
4,857 solutions
Essential Mathematics for Economic Analysis 4th Edition by Arne Strom, Knut Sydsaeter, Peter Hammond

Essential Mathematics for Economic Analysis

4th EditionISBN: 9780273760689Arne Strom, Knut Sydsaeter, Peter Hammond
1,164 solutions
Financial Algebra 1st Edition by Richard Sgroi, Robert Gerver

Financial Algebra

1st EditionISBN: 9780538449670 (1 more)Richard Sgroi, Robert Gerver
2,606 solutions
Financial Algebra: Advanced Algebra with Financial Applications 2nd Edition by Richard Sgroi, Robert Gerver

Financial Algebra: Advanced Algebra with Financial Applications

2nd EditionISBN: 9781337271790Richard Sgroi, Robert Gerver
3,016 solutions

More related questions

1/4

1/7