{"id":1788,"date":"2015-11-12T18:30:45","date_gmt":"2015-11-12T18:30:45","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1788"},"modified":"2015-11-12T18:30:45","modified_gmt":"2015-11-12T18:30:45","slug":"finding-scalar-multiples-of-a-matrix","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/finding-scalar-multiples-of-a-matrix\/","title":{"raw":"Finding Scalar Multiples of a Matrix","rendered":"Finding Scalar Multiples of a Matrix"},"content":{"raw":"

Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar<\/strong> is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication<\/strong> involves multiplying each entry in a matrix by a scalar. A scalar multiple<\/strong> is any entry of a matrix that results from scalar multiplication.\n\nConsider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that 15% more equipment is needed in both labs. The school\u2019s current inventory is displayed in the table below.\n<\/p>\n
Lab A<\/th>\nLab B<\/th>\n<\/tr><\/thead>
Computers<\/strong><\/td>\n15<\/td>\n27<\/td>\n<\/tr>
Computer Tables<\/strong><\/td>\n16<\/td>\n34<\/td>\n<\/tr>
Chairs<\/strong><\/td>\n16<\/td>\n34<\/td>\n<\/tr><\/tbody><\/table>\nConverting the data to a matrix, we have\n
[latex]{C}_{2013}=\\left[\\begin{array}{c}15\\\\ 16\\\\ 16\\end{array}\\begin{array}{c}27\\\\ 34\\\\ 34\\end{array}\\right][\/latex]<\/div>\nTo calculate how much computer equipment will be needed, we multiply all entries in matrix [latex]C[\/latex] by 0.15.\n
[latex]\\left(0.15\\right){C}_{2013}=\\left[\\begin{array}{c}\\left(0.15\\right)15\\\\ \\left(0.15\\right)16\\\\ \\left(0.15\\right)16\\end{array}\\begin{array}{c}\\left(0.15\\right)27\\\\ \\left(0.15\\right)34\\\\ \\left(0.15\\right)34\\end{array}\\right]=\\left[\\begin{array}{c}2.25\\\\ 2.4\\\\ 2.4\\end{array}\\begin{array}{c}4.05\\\\ 5.1\\\\ 5.1\\end{array}\\right][\/latex]<\/div>\nWe must round up to the next integer, so the amount of new equipment needed is\n
[latex]\\left[\\begin{array}{c}3\\\\ 3\\\\ 3\\end{array}\\begin{array}{c}5\\\\ 6\\\\ 6\\end{array}\\right][\/latex]<\/div>\nAdding the two matrices as shown below, we see the new inventory amounts.\n
[latex]\\left[\\begin{array}{c}15\\\\ 16\\\\ 16\\end{array}\\begin{array}{c}27\\\\ 34\\\\ 34\\end{array}\\right]+\\left[\\begin{array}{c}3\\\\ 3\\\\ 3\\end{array}\\begin{array}{c}5\\\\ 6\\\\ 6\\end{array}\\right]=\\left[\\begin{array}{c}18\\\\ 19\\\\ 19\\end{array}\\begin{array}{c}32\\\\ 40\\\\ 40\\end{array}\\right][\/latex]<\/div>\nThis means\n
[latex]{C}_{2014}=\\left[\\begin{array}{c}18\\\\ 19\\\\ 19\\end{array}\\begin{array}{c}32\\\\ 40\\\\ 40\\end{array}\\right][\/latex]<\/div>\nThus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs.\n
\n

A General Note: Scalar Multiplication<\/h3>\nScalar multiplication involves finding the product of a constant by each entry in the matrix. Given\n
[latex]A=\\left[\\begin{array}{cccc}{a}_{11}& & & {a}_{12}\\\\ {a}_{21}& & & {a}_{22}\\end{array}\\right][\/latex]<\/div>\nthe scalar multiple [latex]cA[\/latex] is\n
[latex]\\begin{array}{l}cA=c\\left[\\begin{array}{ccc}{a}_{11}& & {a}_{12}\\\\ {a}_{21}& & {a}_{22}\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{ccc}c{a}_{11}& & c{a}_{12}\\\\ c{a}_{21}& & c{a}_{22}\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\nScalar multiplication is distributive. For the matrices [latex]A,B[\/latex], and [latex]C[\/latex] with scalars [latex]a[\/latex] and [latex]b[\/latex],\n
[latex]\\begin{array}{l}\\\\ \\begin{array}{c}a\\left(A+B\\right)=aA+aB\\\\ \\left(a+b\\right)A=aA+bA\\end{array}\\end{array}[\/latex]<\/div>\n<\/div>\n
\n

Example 6: Multiplying the Matrix by a Scalar<\/h3>\nMultiply matrix [latex]A[\/latex] by the scalar 3.\n
[latex]A=\\left[\\begin{array}{cc}8& 1\\\\ 5& 4\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
\n

Solution<\/h3>\nMultiply each entry in [latex]A[\/latex] by the scalar 3.\n
[latex]\\begin{array}{l}3A=3\\left[\\begin{array}{rr}\\hfill 8& \\hfill 1\\\\ \\hfill 5& \\hfill 4\\end{array}\\right]\\hfill \\\\ = \\left[\\begin{array}{rr}\\hfill 3\\cdot 8& \\hfill 3\\cdot 1\\\\ \\hfill 3\\cdot 5& \\hfill 3\\cdot 4\\end{array}\\right]\\hfill \\\\ = \\left[\\begin{array}{rr}\\hfill 24& \\hfill 3\\\\ \\hfill 15& \\hfill 12\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
\n

Try It 2<\/h3>\nGiven matrix [latex]B,\\text{}[\/latex] find [latex]-2B[\/latex] where\n
[latex]B=\\left[\\begin{array}{cc}4& 1\\\\ 3& 2\\end{array}\\right][\/latex]<\/div>\n
Solution<\/a><\/div>\n<\/div>\n
\n

Example 7: Finding the Sum of Scalar Multiples<\/h3>\nFind the sum [latex]3A+2B[\/latex].\n
[latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill -1& \\hfill 2\\\\ \\hfill 4& \\hfill 3& \\hfill -6\\end{array}\\right]\\text{ and }B=\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 2& \\hfill 1\\\\ \\hfill 0& \\hfill -3& \\hfill 2\\\\ \\hfill 0& \\hfill 1& \\hfill -4\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
\n

Solution<\/h3>\nFirst, find [latex]3A,\\text{}[\/latex] then [latex]2B[\/latex].\n
[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ 3A=\\left[\\begin{array}{lll}3\\cdot 1\\hfill & 3\\left(-2\\right)\\hfill & 3\\cdot 0\\hfill \\\\ 3\\cdot 0\\hfill & 3\\left(-1\\right)\\hfill & 3\\cdot 2\\hfill \\\\ 3\\cdot 4\\hfill & 3\\cdot 3\\hfill & 3\\left(-6\\right)\\hfill \\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill 3& \\hfill -6& \\hfill 0\\\\ \\hfill 0& \\hfill -3& \\hfill 6\\\\ \\hfill 12& \\hfill 9& \\hfill -18\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n
[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ 2B=\\left[\\begin{array}{lll}2\\left(-1\\right)\\hfill & 2\\cdot 2\\hfill & 2\\cdot 1\\hfill \\\\ 2\\cdot 0\\hfill & 2\\left(-3\\right)\\hfill & 2\\cdot 2\\hfill \\\\ 2\\cdot 0\\hfill & 2\\cdot 1\\hfill & 2\\left(-4\\right)\\hfill \\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -2& \\hfill 4& \\hfill 2\\\\ \\hfill 0& \\hfill -6& \\hfill 4\\\\ \\hfill 0& \\hfill 2& \\hfill -8\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\nNow, add [latex]3A+2B[\/latex].\n
[latex]\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ 3A+2B=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill -6& \\hfill 0\\\\ \\hfill 0& \\hfill -3& \\hfill 6\\\\ \\hfill 12& \\hfill 9& \\hfill -18\\end{array}\\right]+\\left[\\begin{array}{rrr}\\hfill -2& \\hfill 4& \\hfill 2\\\\ \\hfill 0& \\hfill -6& \\hfill 4\\\\ \\hfill 0& \\hfill 2& \\hfill -8\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rrr}\\hfill 3 - 2& \\hfill -6+4& \\hfill 0+2\\\\ \\hfill 0+0& \\hfill -3 - 6& \\hfill 6+4\\\\ \\hfill 12+0& \\hfill 9+2& \\hfill -18 - 8\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill -2& \\hfill 2\\\\ \\hfill 0& \\hfill -9& \\hfill 10\\\\ \\hfill 12& \\hfill 11& \\hfill -26\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<\/div>","rendered":"

Besides adding and subtracting whole matrices, there are many situations in which we need to multiply a matrix by a constant called a scalar. Recall that a scalar<\/strong> is a real number quantity that has magnitude, but not direction. For example, time, temperature, and distance are scalar quantities. The process of scalar multiplication<\/strong> involves multiplying each entry in a matrix by a scalar. A scalar multiple<\/strong> is any entry of a matrix that results from scalar multiplication.<\/p>\n

Consider a real-world scenario in which a university needs to add to its inventory of computers, computer tables, and chairs in two of the campus labs due to increased enrollment. They estimate that 15% more equipment is needed in both labs. The school\u2019s current inventory is displayed in the table below.\n<\/p>\n\n\n\n\n\n\n\n
\n<\/th>\nLab A<\/th>\nLab B<\/th>\n<\/tr>\n<\/thead>\n
Computers<\/strong><\/td>\n15<\/td>\n27<\/td>\n<\/tr>\n
Computer Tables<\/strong><\/td>\n16<\/td>\n34<\/td>\n<\/tr>\n
Chairs<\/strong><\/td>\n16<\/td>\n34<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Converting the data to a matrix, we have<\/p>\n

[latex]{C}_{2013}=\\left[\\begin{array}{c}15\\\\ 16\\\\ 16\\end{array}\\begin{array}{c}27\\\\ 34\\\\ 34\\end{array}\\right][\/latex]<\/div>\n

To calculate how much computer equipment will be needed, we multiply all entries in matrix [latex]C[\/latex] by 0.15.<\/p>\n

[latex]\\left(0.15\\right){C}_{2013}=\\left[\\begin{array}{c}\\left(0.15\\right)15\\\\ \\left(0.15\\right)16\\\\ \\left(0.15\\right)16\\end{array}\\begin{array}{c}\\left(0.15\\right)27\\\\ \\left(0.15\\right)34\\\\ \\left(0.15\\right)34\\end{array}\\right]=\\left[\\begin{array}{c}2.25\\\\ 2.4\\\\ 2.4\\end{array}\\begin{array}{c}4.05\\\\ 5.1\\\\ 5.1\\end{array}\\right][\/latex]<\/div>\n

We must round up to the next integer, so the amount of new equipment needed is<\/p>\n

[latex]\\left[\\begin{array}{c}3\\\\ 3\\\\ 3\\end{array}\\begin{array}{c}5\\\\ 6\\\\ 6\\end{array}\\right][\/latex]<\/div>\n

Adding the two matrices as shown below, we see the new inventory amounts.<\/p>\n

[latex]\\left[\\begin{array}{c}15\\\\ 16\\\\ 16\\end{array}\\begin{array}{c}27\\\\ 34\\\\ 34\\end{array}\\right]+\\left[\\begin{array}{c}3\\\\ 3\\\\ 3\\end{array}\\begin{array}{c}5\\\\ 6\\\\ 6\\end{array}\\right]=\\left[\\begin{array}{c}18\\\\ 19\\\\ 19\\end{array}\\begin{array}{c}32\\\\ 40\\\\ 40\\end{array}\\right][\/latex]<\/div>\n

This means<\/p>\n

[latex]{C}_{2014}=\\left[\\begin{array}{c}18\\\\ 19\\\\ 19\\end{array}\\begin{array}{c}32\\\\ 40\\\\ 40\\end{array}\\right][\/latex]<\/div>\n

Thus, Lab A will have 18 computers, 19 computer tables, and 19 chairs; Lab B will have 32 computers, 40 computer tables, and 40 chairs.<\/p>\n

\n

A General Note: Scalar Multiplication<\/h3>\n

Scalar multiplication involves finding the product of a constant by each entry in the matrix. Given<\/p>\n

[latex]A=\\left[\\begin{array}{cccc}{a}_{11}& & & {a}_{12}\\\\ {a}_{21}& & & {a}_{22}\\end{array}\\right][\/latex]<\/div>\n

the scalar multiple [latex]cA[\/latex] is<\/p>\n

[latex]\\begin{array}{l}cA=c\\left[\\begin{array}{ccc}{a}_{11}& & {a}_{12}\\\\ {a}_{21}& & {a}_{22}\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{ccc}c{a}_{11}& & c{a}_{12}\\\\ c{a}_{21}& & c{a}_{22}\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n

Scalar multiplication is distributive. For the matrices [latex]A,B[\/latex], and [latex]C[\/latex] with scalars [latex]a[\/latex] and [latex]b[\/latex],<\/p>\n

[latex]\\begin{array}{l}\\\\ \\begin{array}{c}a\\left(A+B\\right)=aA+aB\\\\ \\left(a+b\\right)A=aA+bA\\end{array}\\end{array}[\/latex]<\/div>\n<\/div>\n
\n

Example 6: Multiplying the Matrix by a Scalar<\/h3>\n

Multiply matrix [latex]A[\/latex] by the scalar 3.<\/p>\n

[latex]A=\\left[\\begin{array}{cc}8& 1\\\\ 5& 4\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
\n

Solution<\/h3>\n

Multiply each entry in [latex]A[\/latex] by the scalar 3.<\/p>\n

[latex]\\begin{array}{l}3A=3\\left[\\begin{array}{rr}\\hfill 8& \\hfill 1\\\\ \\hfill 5& \\hfill 4\\end{array}\\right]\\hfill \\\\ = \\left[\\begin{array}{rr}\\hfill 3\\cdot 8& \\hfill 3\\cdot 1\\\\ \\hfill 3\\cdot 5& \\hfill 3\\cdot 4\\end{array}\\right]\\hfill \\\\ = \\left[\\begin{array}{rr}\\hfill 24& \\hfill 3\\\\ \\hfill 15& \\hfill 12\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
\n

Try It 2<\/h3>\n

Given matrix [latex]B,\\text{}[\/latex] find [latex]-2B[\/latex] where<\/p>\n

[latex]B=\\left[\\begin{array}{cc}4& 1\\\\ 3& 2\\end{array}\\right][\/latex]<\/div>\n
Solution<\/a><\/div>\n<\/div>\n
\n

Example 7: Finding the Sum of Scalar Multiples<\/h3>\n

Find the sum [latex]3A+2B[\/latex].<\/p>\n

[latex]A=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill -2& \\hfill 0\\\\ \\hfill 0& \\hfill -1& \\hfill 2\\\\ \\hfill 4& \\hfill 3& \\hfill -6\\end{array}\\right]\\text{ and }B=\\left[\\begin{array}{rrr}\\hfill -1& \\hfill 2& \\hfill 1\\\\ \\hfill 0& \\hfill -3& \\hfill 2\\\\ \\hfill 0& \\hfill 1& \\hfill -4\\end{array}\\right][\/latex]<\/div>\n<\/div>\n
\n

Solution<\/h3>\n

First, find [latex]3A,\\text{}[\/latex] then [latex]2B[\/latex].<\/p>\n

[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ 3A=\\left[\\begin{array}{lll}3\\cdot 1\\hfill & 3\\left(-2\\right)\\hfill & 3\\cdot 0\\hfill \\\\ 3\\cdot 0\\hfill & 3\\left(-1\\right)\\hfill & 3\\cdot 2\\hfill \\\\ 3\\cdot 4\\hfill & 3\\cdot 3\\hfill & 3\\left(-6\\right)\\hfill \\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill 3& \\hfill -6& \\hfill 0\\\\ \\hfill 0& \\hfill -3& \\hfill 6\\\\ \\hfill 12& \\hfill 9& \\hfill -18\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n
[latex]\\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ 2B=\\left[\\begin{array}{lll}2\\left(-1\\right)\\hfill & 2\\cdot 2\\hfill & 2\\cdot 1\\hfill \\\\ 2\\cdot 0\\hfill & 2\\left(-3\\right)\\hfill & 2\\cdot 2\\hfill \\\\ 2\\cdot 0\\hfill & 2\\cdot 1\\hfill & 2\\left(-4\\right)\\hfill \\end{array}\\right]\\hfill \\end{array}\\hfill \\\\ =\\left[\\begin{array}{rrr}\\hfill -2& \\hfill 4& \\hfill 2\\\\ \\hfill 0& \\hfill -6& \\hfill 4\\\\ \\hfill 0& \\hfill 2& \\hfill -8\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n

Now, add [latex]3A+2B[\/latex].<\/p>\n

[latex]\\begin{array}{l}\\hfill \\\\ \\hfill \\\\ 3A+2B=\\left[\\begin{array}{rrr}\\hfill 3& \\hfill -6& \\hfill 0\\\\ \\hfill 0& \\hfill -3& \\hfill 6\\\\ \\hfill 12& \\hfill 9& \\hfill -18\\end{array}\\right]+\\left[\\begin{array}{rrr}\\hfill -2& \\hfill 4& \\hfill 2\\\\ \\hfill 0& \\hfill -6& \\hfill 4\\\\ \\hfill 0& \\hfill 2& \\hfill -8\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rrr}\\hfill 3 - 2& \\hfill -6+4& \\hfill 0+2\\\\ \\hfill 0+0& \\hfill -3 - 6& \\hfill 6+4\\\\ \\hfill 12+0& \\hfill 9+2& \\hfill -18 - 8\\end{array}\\right]\\hfill \\\\ \\text{ }=\\left[\\begin{array}{rrr}\\hfill 1& \\hfill -2& \\hfill 2\\\\ \\hfill 0& \\hfill -9& \\hfill 10\\\\ \\hfill 12& \\hfill 11& \\hfill -26\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n\n\t\t\t
\n\t\t\t

Candela Citations<\/h3>\n\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t\t
CC licensed content, Specific attribution<\/div>