Multiplying a Matrix by a Scalar Lesson

Over the course of the last two lessons, we have learned the basic principles involved with matrix algebra, along with how to perform addition and subtraction with matrices, and how to solve very basic matrix equations. Our next step is to learn about scalar multiplication, which is the product of a real number and a matrix.

Scalar Multiplication

When we work with matrices a real number is known as a scalar. This is done to distinguish our real number from a matrix.
For example, a 1 × 1 matrix with a single element of '3' is different from the real number '3' itself: $$[3] ≠ 3$$ The number 3 placed in front of a matrix implies we are multiplying the matrix by the scalar '3'. $$3\left[ \begin{array}{ccc}1&2 & 3\\ 4&5 & 6\end{array}\right]$$ To multiply a matrix by a scalar, we just multiply every element by the scalar. $$3\left[ \begin{array}{ccc}1&2 & 3\\ 4&5 & 6\end{array}\right] = \left[ \begin{array}{ccc}3 \cdot 1 & 3 \cdot 2 & 3 \cdot 3\\ 3 \cdot 4 & 3 \cdot 5 & 3 \cdot 6\end{array}\right] = \left[ \begin{array}{ccc}3 & 6 & 9\\ 12 & 15 & 18\end{array}\right]$$ In general, we can say that the product of some scalar k, which is a real number, and a matrix A is kA. Our matrix kA contains all the elements of A multiplied by the scalar k. Notice that we are using a lowercase letter k for our scalar and an uppercase letter A for the matrix name. Let's look at some examples.
Example #1: Find 2A. $$A=\left[\begin{array}{cc}9 & 4\\ 6 & 2\end{array}\right]$$ Multiply each element of matrix A by 2: $$2A = \left[\begin{array}{cc}2 \cdot 9 & 2 \cdot 4 \\ 2 \cdot 6 & 2 \cdot 2\end{array}\right] = \left[ \begin{array}{cc}18 & 8\\ 12 &4\end{array}\right]$$ Example #2: Find (1/2)A. $$A=\left[\begin{array}{cc}-8 & 2\\ 5 & -4 \\\frac{1}{2} & 16\end{array}\right]$$ Multiply each element of matrix A by (1/2): $$\frac{1}{2}A = \left[\begin{array}{cc}\frac{1}{2} \cdot -8 & \frac{1}{2} \cdot 2\\ \frac{1}{2} \cdot 5 & \frac{1}{2} \cdot -4 \\ \frac{1}{2} \cdot \frac{1}{2} & \frac{1}{2} \cdot 16\end{array}\right] = \left[\begin{array}{cc}-4 & 1 \\ \frac{5}{2} & -2 \\ \frac{1}{4} & 8 \end{array}\right]$$

Properties of Scalar Multiplication

A and B are m × n matrices, and c and d are scalars:

• 1A = A
  • Scalar identity property
• c(A + B) = cA + cB
  • Distributive property
• cA + dA = (c + d)A
  • Distributive property
• (cd)A = c(dA)
  • Associative property of scalar multiplication

Combining Matrix Operations

We previously learned how to perform addition and subtraction with matrices. Let's look at an example where we combine multiplication by a scalar with addition.
Example #3: Find 2A + 3B. $$A = \left[ \begin{array}{ccc}1 & 9 & 3 \\ -2 & -11 & -1\\ 9 & -4 & -7\end{array}\right]$$ $$B = \left[ \begin{array}{ccc}-8 & 7 & -2 \\ 8 & 7 & -8\\ 13 & 14 & 2\end{array}\right]$$ $$2A + 3B = 2\left[ \begin{array}{ccc}1 & 9 & 3 \\ -2 & -11 & -1\\ 9 & -4 & -7\end{array}\right] + 3\left[ \begin{array}{ccc}-8 & 7 & -2 \\ 8 & 7 & -8\\ 13 & 14 & 2\end{array}\right]$$ $$= \left[ \begin{array}{ccc}2 \cdot 1 & 2 \cdot 9 & 2 \cdot 3 \\ 2 \cdot -2 & 2 \cdot -11 & 2 \cdot -1\\ 2 \cdot 9 & 2 \cdot -4 & 2 \cdot -7\end{array}\right] + \left[ \begin{array}{ccc}3 \cdot -8 & 3 \cdot 7 & 3 \cdot -2 \\ 3 \cdot 8 & 3 \cdot 7 & 3 \cdot -8\\ 3 \cdot 13 & 3 \cdot 14 & 3 \cdot 2\end{array}\right]$$ $$= \left[ \begin{array}{ccc}2 & 18 & 6 \\ -4 & -22 & -2\\ 18 & -8 & -14\end{array}\right] + \left[ \begin{array}{ccc}-24 & 21 & -6 \\ 24 & 21 & -24\\ 39 & 42 & 6\end{array}\right]$$ $$= \left[ \begin{array}{ccc}2 + (-24) & 18 + 21 & 6 + (-6) \\ -4 + 24 & -22 + 21 & -2 + (-24)\\ 18 + 39 & -8 + 42 & -14 + 6\end{array}\right]$$ $$= \left[ \begin{array}{ccc}-22 & 39 & 0 \\ 20 & -1 & -26\\ 57 & 34 & -8\end{array}\right]$$

Solving Basic Matrix Equations with Scalars

In our previous lesson, we saw that we could solve basic matrix equations by treating matrices as variables and solving for X, the unknown matrix. Let's look at an example.
Example #4: Find matrix X. $$A = B - 2X$$ $$A = \left[ \begin{array}{cc}2 & 7 \\ -22 & -5 \\ 8 & 4\end{array}\right]$$ $$B = \left[ \begin{array}{cc}2 & 7 \\ -6 & -7 \\ 0 & 8\end{array}\right]$$ First, let's solve the matrix equation for X, our unknown matrix. $$A = B - 2X$$ Add 2X to both sides: $$2X + A = B$$ Subtract A from both sides: $$2X = B - A$$ Mutliply both sides by (1/2): $$X = \frac{1}{2}(B - A)$$ Now, we can perform our indicated operations to find X. $$X = \frac{1}{2}\left(\left[ \begin{array}{cc}2 & 7 \\ -6 & -7 \\ 0 & 8\end{array}\right] - \left[ \begin{array}{cc}2 & 7 \\ -22 & -5 \\ 8 & 4\end{array}\right]\right)$$ $$= \frac{1}{2}\left [ \begin{array}{cc}2 - 2 & 7 - 7 \\ -6 - (-22) & -7 - (-5) \\ 0 - 8 & 8 - 4\end{array}\right] $$ $$= \frac{1}{2}\left[ \begin{array}{cc}0 & 0 \\ 16 & -2 \\ -8 & 4\end{array}\right]$$ $$= \left[ \begin{array}{cc}\frac{1}{2} \cdot 0 & \frac{1}{2} \cdot 0 \\ \frac{1}{2} \cdot 16 & \frac{1}{2} \cdot -2 \\ \frac{1}{2} \cdot -8 & \frac{1}{2} \cdot 4\end{array}\right]$$ $$= \left[ \begin{array}{cc}0 & 0 \\ 8 & -1 \\ -4 & 2\end{array}\right]$$

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