KnowledgeBoat Logo
|

Mathematics

If A = [1324],B=[1243] and C=[4312]\begin{bmatrix}[r] 1 & 3 \ 2 & 4 \end{bmatrix}, B = \begin{bmatrix}[r] 1 & 2 \ 4 & 3 \end{bmatrix} \text{ and } C = \begin{bmatrix}[r] 4 & 3 \ 1 & 2 \end{bmatrix}, find :

(i) (AB)C

(ii) A(BC)

Is A(BC) = (AB)C ?

Matrices

18 Likes

Answer

(i) Substituting value of A, B and C in (AB)C we get,

⇒([1324][1243])[4312]=([1×1+3×41×2+3×32×1+4×42×2+4×3])[4312]=([1+122+92+164+12])[4312]=[13111816][4312]=[13×4+11×113×3+11×218×4+16×118×3+16×2]=[52+1139+2272+1654+32]=[63618886].\Rightarrow \Big(\begin{bmatrix}[r] 1 & 3 \ 2 & 4 \end{bmatrix}\begin{bmatrix}[r] 1 & 2 \ 4 & 3 \end{bmatrix}\Big)\begin{bmatrix}[r] 4 & 3 \ 1 & 2 \end{bmatrix} \\[1em] = \Big(\begin{bmatrix}[r] 1 \times 1 + 3 \times 4 & 1 \times 2 + 3 \times 3 \ 2 \times 1 + 4 \times 4 & 2 \times 2 + 4 \times 3 \end{bmatrix}\Big)\begin{bmatrix}[r] 4 & 3 \ 1 & 2 \end{bmatrix} \\[1em] = \Big(\begin{bmatrix}[r] 1 + 12 & 2 + 9 \ 2 + 16 & 4 + 12 \end{bmatrix}\Big)\begin{bmatrix}[r] 4 & 3 \ 1 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 13 & 11 \ 18 & 16 \end{bmatrix}\begin{bmatrix}[r] 4 & 3 \ 1 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 13 \times 4 + 11 \times 1 & 13 \times 3 + 11 \times 2 \ 18 \times 4 + 16 \times 1 & 18 \times 3 + 16 \times 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 52 + 11 & 39 + 22 \ 72 + 16 & 54 + 32 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 63 & 61 \ 88 & 86 \end{bmatrix}.

Hence, (AB)C = [63618886].\begin{bmatrix}[r] 63 & 61 \ 88 & 86 \end{bmatrix}.

(ii) Substituting value of A, B and C in A(BC) we get,

⇒[1324]([1243][4312])=[1324]([1×4+2×11×3+2×24×4+3×14×3+3×2])=[1324][4+23+416+312+6]=[1324][671918]=[1×6+3×191×7+3×182×6+4×192×7+4×18]=[6+577+5412+7614+72]=[63618886]\Rightarrow \begin{bmatrix}[r] 1 & 3 \ 2 & 4 \end{bmatrix}\Big(\begin{bmatrix}[r] 1 & 2 \ 4 & 3 \end{bmatrix}\begin{bmatrix}[r] 4 & 3 \ 1 & 2 \end{bmatrix}\Big) \\[1em] = \begin{bmatrix}[r] 1 & 3 \ 2 & 4 \end{bmatrix}\Big(\begin{bmatrix}[r] 1 \times 4 + 2\times 1 & 1 \times 3 + 2 \times 2 \ 4 \times 4 + 3 \times 1 & 4 \times 3 + 3 \times 2 \end{bmatrix}\Big) \\[1em] = \begin{bmatrix}[r] 1 & 3 \ 2 & 4 \end{bmatrix}\begin{bmatrix}[r] 4 + 2 & 3 + 4 \ 16 + 3 & 12 + 6 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 & 3 \ 2 & 4 \end{bmatrix}\begin{bmatrix}[r] 6 & 7 \ 19 & 18 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 6 + 3 \times 19 & 1 \times 7 + 3 \times 18 \ 2 \times 6 + 4 \times 19 & 2 \times 7 + 4 \times 18 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 6 + 57 & 7 + 54 \ 12 + 76 & 14 + 72 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 63 & 61 \ 88 & 86 \end{bmatrix}

Hence, A(BC) = [63618886]\begin{bmatrix}[r] 63 & 61 \ 88 & 86 \end{bmatrix}.

Hence, A(BC) = (AB)C.

Answered By

10 Likes

Related Questions