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  • If A = [2 1] Verify that A(B + C) = (AB + AC).

If \mathrm{A}=\left[\begin{array}{ll} 2 & 1 \end{array}\right], \mathrm{B}=\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right]_{\text {and }} \mathrm{C}=\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right] verify that A(B+C)=(AB+AC)

Answers (1)

We are given the following matrices A, B and C, such that

\mathrm{A}=\left[\begin{array}{ll} 2 & 1 \end{array}\right], \mathrm{B}=\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right]_{\text {and }} \mathrm{C}=\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]

We need to verify that, A(B + C) = AB + AC.

Take L.H.S: A(B + C)

By Solving (B + C).

\begin{aligned} &B+C=\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right]+\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]\\ &\text { since, the above matrices have the same order, they can be added. }\\ &\Rightarrow B+C=\left[\begin{array}{lll} 5-1 & 3+2 & 4+1 \\ 8+1 & 7+0 & 6+2 \end{array}\right]\\ &\Rightarrow B+C=\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right] \end{aligned}

Now, multiply A by (B + C).

Let (B + C) = D.

We get,

AD = A(B + C)

\Rightarrow \mathrm{AD}=\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right]

Order of A = 1 × 2

Order of D = 2 × 3

Then, order of the matrix is = 1 × 3

Multiply 1st row of matrix A by matching members of 1st column of matrix D, then finally sum them up.

\\(2, 1)(4, 9) = (2 $ \times $ 4) + (1 $ \times $ 9) \\$ \Rightarrow $ (2, 1)(4, 9) = 8 + 9 \\$ \Rightarrow $ (2, 1)(4, 9) = 17

\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right] = \begin{bmatrix} 17 & \end{bmatrix}

Multiply 1st row of matrix A by matching members of 2nd column of matrix D, then finally sum them up.

\\(2, 1)(5, 7) = (2 $ \times $ 5) + (1 $ \times $ 7) \\$ \Rightarrow $ (2, 1)(5, 7) = 10 + 7 \\$ \Rightarrow $ (2, 1)(5, 7) = 17

\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 4 & 5 & 5 \\ 9 & 7 & 8 \end{array}\right] = \begin{bmatrix} 17 & 17\end{bmatrix}

Multiply 1st row of matrix A by matching members of 3rd column of matrix D, then finally sum them up.

\\ (2,1)(5,8)=(2 \times 5)+(1 \times 8)$ \\$\Rightarrow(2,1)(5,8)=10+8$ \\$\Rightarrow(2,1)(5,8)=18$ \\$\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}4 & 5 & 5 \\ 9 & 7 & 8\end{array}\right]=\left[\begin{array}{lll}17 & 17 & 18\end{array}\right]$
So,
A(B+C)=\left[\begin{array}{lll}17 & 17 & 18\end{array}\right]$
Now, take R.H.S: $A B+A C$
Let us compute A B.
A B=\left[\begin{array}{ll}2 & 1\end{array}\right]\left[\begin{array}{lll}5 & 3 & 4 \\ 8 & 7 & 6\end{array}\right]$

Order of A = 1 × 2

Order of B = 2 × 3

Then, order of AB = 1 × 3

Multiply 1st row of matrix A by matching members of 1st column of matrix B, then finally sum them up.

\begin{aligned} &(2,1)(5,8)=(2 \times 5)+(1 \times 8)\\ &\Rightarrow(2,1)(5,8)=10+8\\ &\Rightarrow(2,1)(5,8)=18\\ &\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{lll} 5 & 3 & 4 \\ 8 & 7 & 6 \end{array}\right]=[18\\ \end{aligned}

Similarly, repeat steps to fill for the rest of the elements.

 \begin{bmatrix} 2 &1 \end{bmatrix}\begin{bmatrix} 5 &3 &4 \\8 &7 & 6 \end{bmatrix} = \begin{bmatrix} 18 &2*3+1*7 &2*4+1*6 \end{bmatrix}

\begin{bmatrix} 2 &1 \end{bmatrix}\begin{bmatrix} 5 &3 &4 \\8 &7 & 6 \end{bmatrix} = \begin{bmatrix} 18 &13 &14 \end{bmatrix}

Now, let us compute AC.

A C=\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right]

Order of AC = 1 × 3

Multiply 1st row of matrix A by matching members of 1st column of matrix C, then finally sum them up.

\\\\(2, 1)(-1, 1) = (2 $ \times $ -1) + (1 $ \times $ 1) \\$ \Rightarrow $ (2, 1)(-1, 1) = -2 + 1 \\$ \Rightarrow $ (2, 1)(-1, 1) = -1

\left[\begin{array}{ll} 2 & 1 \end{array}\right]\left[\begin{array}{ccc} -1 & 2 & 1 \\ 1 & 0 & 2 \end{array}\right] = \begin{bmatrix} -1 & & \end{bmatrix}

Similarly, repeat steps to fill for other elements.

 \begin{bmatrix} 2 &1 \end{bmatrix}\begin{bmatrix} -1 &2 &1 \\1 &0 & 2 \end{bmatrix} = \begin{bmatrix} -1 &2*2+1*0 &2*1+1*2 \end{bmatrix}

\begin{bmatrix} 2 &1 \end{bmatrix}\begin{bmatrix} -1 &2 &1 \\1 &0 & 2 \end{bmatrix} = \begin{bmatrix} -1 &4 &4 \end{bmatrix}

\begin{aligned} &\text { Now, } \mathrm{Add} , \mathrm{AB}+\mathrm{AC} .\\ &\begin{array}{lll} \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{lll} 18 & 13 & 14 \end{array}\right]+\left[\begin{array}{lll} -1 & 4 & 4 \end{array}\right] \end{array}\\ &\Rightarrow \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{lll} 18-1 & 13+4 & 14+4 \end{array}\right]\\ &\Rightarrow \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{lll} 17 & 17 & 18 \end{array}\right]\\ &\text { Thus, }\\ &\mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC} . \end{aligned}

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