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Matrices - Properties of scalar multiplication of a matrix

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Scalar multiplication involves multiplying every element of a matrix AA by a constant number kk, known as a scalar.

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If A=[aij]m×nA = [a_{ij}]_{m \times n}, then kA=[k⋅aij]m×nkA = [k \cdot a_{ij}]_{m \times n}. The order of the matrix remains the same after scalar multiplication.

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The property k(A+B)=kA+kBk(A + B) = kA + kB shows that scalar multiplication is distributive over matrix addition.

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The property (k+l)A=kA+lA(k + l)A = kA + lA shows that scalar multiplication is distributive over the addition of scalars.

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Multiplying a matrix by the scalar −1-1 results in the negative of the matrix: (−1)A=−A(-1)A = -A.

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Multiplying a matrix by the scalar 00 results in a zero matrix: 0⋅A=O0 \cdot A = O.

📐Formulae

kA=k[aij]m×n=[kaij]m×nk A = k [a_{ij}]_{m \times n} = [k a_{ij}]_{m \times n}

k(A+B)=kA+kBk(A + B) = kA + kB

(k+l)A=kA+lA(k + l)A = kA + lA

k(lA)=(kl)Ak(lA) = (kl)A

(−k)A=−(kA)=k(−A)(-k)A = -(kA) = k(-A)

💡Examples

Problem 1:

If A=[231−5]A = \begin{bmatrix} 2 & 3 \\ 1 & -5 \end{bmatrix} and B=[1−204]B = \begin{bmatrix} 1 & -2 \\ 0 & 4 \end{bmatrix}, find 3A−2B3A - 2B.

Solution:

First, find 3A3A and 2B2B separately: 3A=33A = 3 \begin{bmatrix} 2 & 3 \ 1 & -5 \end{bmatrix}==\begin{bmatrix} 3(2) & 3(3) \ 3(1) & 3(-5) \end{bmatrix}==\begin{bmatrix} 6 & 9 \ 3 & -15 \end{bmatrix} 2B = 2 \begin{bmatrix} 1 & -2 \ 0 & 4 \end{bmatrix}==\begin{bmatrix} 2(1) & 2(-2) \ 2(0) & 2(4) \end{bmatrix}==\begin{bmatrix} 2 & -4 \ 0 & 8 \end{bmatrix}Now, subtract $2B$ from $3A$:3A - 2B = \begin{bmatrix} 6 - 2 & 9 - (-4) \ 3 - 0 & -15 - 8 \end{bmatrix}=[4133−23] = \begin{bmatrix} 4 & 13 \\ 3 & -23 \end{bmatrix}

Explanation:

We use the definition of scalar multiplication to multiply each element of AA by 33 and each element of BB by 22. Finally, we perform matrix subtraction by subtracting corresponding elements.

Problem 2:

Let k=2k = 2 and l=3l = 3. Verify the property (k+l)A=kA+lA(k + l)A = kA + lA for A=[1234]A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}.

Solution:

Left Hand Side (LHS): (k+l)A=(2+3)(k + l)A = (2 + 3) \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}=5= 5\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}==\begin{bmatrix} 5 & 10 \ 15 & 20 \end{bmatrix}RightHandSide(RHS):Right Hand Side (RHS):kA + lA = 2 \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}+3+ 3\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \ 6 & 8 \end{bmatrix}++\begin{bmatrix} 3 & 6 \ 9 & 12 \end{bmatrix}==\begin{bmatrix} 2+3 & 4+6 \ 6+9 & 8+12 \end{bmatrix}=[5101520] = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} LHS = RHS.

Explanation:

This demonstrates the distributive property of scalar multiplication over the addition of scalars. We first add the scalars and then multiply, which gives the same result as multiplying the matrix by each scalar separately and then adding the results.