{"id":1371,"date":"2025-07-24T19:13:55","date_gmt":"2025-07-24T19:13:55","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1371"},"modified":"2026-03-23T20:20:15","modified_gmt":"2026-03-23T20:20:15","slug":"matrices-and-matrix-operations-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/matrices-and-matrix-operations-learn-it-3\/","title":{"raw":"Matrices and Matrix Operations: Learn It 3","rendered":"Matrices and Matrix Operations: Learn It 3"},"content":{"raw":"

Scalar Multiples of a Matrix<\/h2>\r\nBesides 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.\r\n\r\nThe 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.\r\n\r\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 from the fall 2013 semester to the fall of 2014. They estimate that [latex]15\\%[\/latex] more equipment is needed in both labs. The school\u2019s current inventory is displayed in the table below.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
<\/th>\r\nLab A<\/th>\r\nLab B<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
Computers<\/strong><\/td>\r\n[latex]15[\/latex]<\/td>\r\n[latex]27[\/latex]<\/td>\r\n<\/tr>\r\n
Computer Tables<\/strong><\/td>\r\n[latex]16[\/latex]<\/td>\r\n[latex]34[\/latex]<\/td>\r\n<\/tr>\r\n
Chairs<\/strong><\/td>\r\n[latex]16[\/latex]<\/td>\r\n[latex]34[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nConverting the data to a matrix, we have the computer inventory in fall 2013 given by\r\n

[latex]{C}_{2013}=\\left[\\begin{array}{c}15\\\\ 16\\\\ 16\\end{array}\\begin{array}{c}27\\\\ 34\\\\ 34\\end{array}\\right][\/latex]<\/p>\r\nTo calculate how much computer equipment will be needed in 2014, we multiply all entries in matrix [latex]C[\/latex] by [latex]0.15[\/latex].\r\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]<\/p>\r\nWe must round up to the next integer, so the amount of new equipment needed is\r\n

[latex]\\left[\\begin{array}{c}3\\\\ 3\\\\ 3\\end{array}\\begin{array}{c}5\\\\ 6\\\\ 6\\end{array}\\right][\/latex]<\/p>\r\nAdding the two matrices as shown below, we see the new inventory amounts.\r\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]<\/p>\r\nThis means\r\n

[latex]{C}_{2014}=\\left[\\begin{array}{c}18\\\\ 19\\\\ 19\\end{array}\\begin{array}{c}32\\\\ 40\\\\ 40\\end{array}\\right][\/latex]<\/p>\r\nThus, Lab A will have [latex]18[\/latex] computers, [latex]19[\/latex] computer tables, and [latex]19[\/latex] chairs; Lab B will have [latex]32[\/latex] computers, [latex]40[\/latex] computer tables, and [latex]40[\/latex] chairs.\r\n\r\n<\/section>

\r\n

scalar multiplication<\/h3>\r\nScalar multiplication involves finding the product of a constant by each entry in the matrix. Given\r\n

[latex]A=\\left[\\begin{array}{cccc}{a}_{11}& & & {a}_{12}\\\\ {a}_{21}& & & {a}_{22}\\end{array}\\right][\/latex]<\/p>\r\nthe scalar multiple [latex]cA[\/latex] is\r\n

[latex]\\begin{array}{ll}cA & =c\\left[\\begin{array}{ccc}{a}_{11}& & {a}_{12}\\\\ {a}_{21}& & {a}_{22}\\end{array}\\right]\\hfill \\\\ & =\\left[\\begin{array}{ccc}c{a}_{11}& & c{a}_{12}\\\\ c{a}_{21}& & c{a}_{22}\\end{array}\\right]\\hfill \\end{array}[\/latex]<\/p>\r\nScalar multiplication is distributive. For the matrices [latex]A,B[\/latex], and [latex]C[\/latex] with scalars [latex]a[\/latex] and [latex]b[\/latex],\r\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]<\/p>\r\n\r\n<\/section>

Multiply matrix [latex]A[\/latex] by the scalar [latex]3[\/latex].\r\n

[latex]A=\\left[\\begin{array}{cc}8& 1\\\\ 5& 4\\end{array}\\right][\/latex]<\/p>\r\n[reveal-answer q=\"90744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"90744\"]\r\n\r\nMultiply each entry in [latex]A[\/latex] by the scalar [latex]3[\/latex].\r\n

[latex]\\begin{array}{ll}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]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

[ohm_question hide_question_numbers=1]321715[\/ohm_question]<\/section>
Given:\r\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]<\/p>\r\nFind the sum [latex]3A+2B[\/latex].\r\n\r\n[reveal-answer q=\"755068\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"755068\"]\r\n\r\nFirst, find [latex]3A,\\text{}[\/latex] then [latex]2B[\/latex].\r\n

[latex]\\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 \\\\ & =\\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]\r\n[latex]\\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 \\\\ & =\\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]<\/p>\r\nNow, add [latex]3A+2B[\/latex].\r\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 \\\\ & =\\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 \\\\ & =\\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]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

[ohm_question hide_question_numbers=1]321716[\/ohm_question]<\/section>","rendered":"

Scalar Multiples of a Matrix<\/h2>\n

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.<\/p>\n

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 from the fall 2013 semester to the fall of 2014. They estimate that [latex]15\\%[\/latex] more equipment is needed in both labs. The school\u2019s current inventory is displayed in the table below.<\/p>\n\n\n\n\n\n\n\n
<\/th>\nLab A<\/th>\nLab B<\/th>\n<\/tr>\n<\/thead>\n
Computers<\/strong><\/td>\n[latex]15[\/latex]<\/td>\n[latex]27[\/latex]<\/td>\n<\/tr>\n
Computer Tables<\/strong><\/td>\n[latex]16[\/latex]<\/td>\n[latex]34[\/latex]<\/td>\n<\/tr>\n
Chairs<\/strong><\/td>\n[latex]16[\/latex]<\/td>\n[latex]34[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Converting the data to a matrix, we have the computer inventory in fall 2013 given by<\/p>\n

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

To calculate how much computer equipment will be needed in 2014, we multiply all entries in matrix [latex]C[\/latex] by [latex]0.15[\/latex].<\/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]<\/p>\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]<\/p>\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]<\/p>\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]<\/p>\n

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

\n

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]<\/p>\n

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

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

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

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