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To prove these results, we will use the definitions of matrix addition and scalar multiplication. Let A and B be m x n matrices, and let g and k be scalars.

Mathematics

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2 parts
a

Proven\text{Proven}

b

Proven\text{Proven}

Step-by-step solution

6 steps

Here are the proofs for the given properties of scalar multiplication and matrix addition.

To prove these results, we will use the definitions of matrix addition and scalar multiplication. Let AA and BB be m×nm \times n matrices, and let gg and kk be scalars.
Let A=[aij]A = [a_{ij}] and B=[bij]B = [b_{ij}], where aija_{ij} and bijb_{ij} are the elements in the ii-th row and jj-th column of matrices AA and BB, respectively.

Part a· 3 steps

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

  1. Consider the left-hand side (LHS), k(A+B)k(A+B).

    First, calculate A+BA+B.

    A+B=[aij]+[bij]=[aij+bij]A+B = [a_{ij}] + [b_{ij}] = [a_{ij} + b_{ij}]

    Now, multiply by the scalar kk.

    k(A+B)=k[aij+bij]=[k(aij+bij)]k(A+B) = k[a_{ij} + b_{ij}] = [k(a_{ij} + b_{ij})]

    Using the distributive property of scalars over real numbers, k(aij+bij)=kaij+kbijk(a_{ij} + b_{ij}) = ka_{ij} + kb_{ij}.
    So,

    k(A+B)=[kaij+kbij]k(A+B) = [ka_{ij} + kb_{ij}]
  2. Consider the right-hand side (RHS), kA+kBkA + kB.

    First, calculate kAkA.

    kA=k[aij]=[kaij]kA = k[a_{ij}] = [ka_{ij}]

    Next, calculate kBkB.

    kB=k[bij]=[kbij]kB = k[b_{ij}] = [kb_{ij}]

    Now, add kAkA and kBkB.

    kA+kB=[kaij]+[kbij]=[kaij+kbij]kA + kB = [ka_{ij}] + [kb_{ij}] = [ka_{ij} + kb_{ij}]
  3. Compare LHS and RHS.

    Since [kaij+kbij]=[kaij+kbij][ka_{ij} + kb_{ij}] = [ka_{ij} + kb_{ij}], the LHS equals the RHS.
    Therefore, k(A+B)=kA+kBk(A+B) = kA + kB is proven.
    Proven\boxed{\text{Proven}}

Part b

Proven\text{Proven}

Prove (g+k)A=gA+kA(g+k)A = gA + kA

  1. Consider the left-hand side (LHS), (g+k)A(g+k)A.

    (g+k)A=(g+k)[aij]=[(g+k)aij](g+k)A = (g+k)[a_{ij}] = [(g+k)a_{ij}]

    Using the distributive property of scalars over real numbers, (g+k)aij=gaij+kaij(g+k)a_{ij} = ga_{ij} + ka_{ij}.
    So,

    (g+k)A=[gaij+kaij](g+k)A = [ga_{ij} + ka_{ij}]
  2. Consider the right-hand side (RHS), gA+kAgA + kA.

    First, calculate gAgA.

    gA=g[aij]=[gaij]gA = g[a_{ij}] = [ga_{ij}]

    Next, calculate kAkA.

    kA=k[aij]=[kaij]kA = k[a_{ij}] = [ka_{ij}]

    Now, add gAgA and kAkA.

    gA+kA=[gaij]+[kaij]=[gaij+kaij]gA + kA = [ga_{ij}] + [ka_{ij}] = [ga_{ij} + ka_{ij}]
  3. Compare LHS and RHS.

    Since [gaij+kaij]=[gaij+kaij][ga_{ij} + ka_{ij}] = [ga_{ij} + ka_{ij}], the LHS equals the RHS.
    Therefore, (g+k)A=gA+kA(g+k)A = gA + kA is proven.
    Proven\boxed{\text{Proven}}

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