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If $A=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right], B=\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]_{\text {and }} C=\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]$ , verify:
(i) $(AB) C = A (BC)$
(ii) $A(B + C) = AB + AC$

Answers (1)

We have the given matrices A, B and C, such that

$A=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right], B=\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]_{\text {and }} C=\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]$

To multiply two given matrices A and B, the number of columns in A must equal the number of rows in B. Thus, if A is an m x n matrix and B is an r x s matrix, n = r.

(i). We need to verify: (AB)C = A(BC)

Take L.H.S = (AB)C

First, compute AB.

$AB=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right] \left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]$

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

$\\(1, 2)(2, 3) = (1 \times 2) + (2 \times 3) \\ \Rightarrow (1, 2)(2, 3) = 2 + 6 \\ \Rightarrow (1, 2)(2, 3) = 8$

$\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right] \left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\begin{bmatrix} 8 & \\ & \end{bmatrix}$

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

$\\(1, 2)(3, -4) = (1 \times 3) + (2 \times -4) \\ \Rightarrow (1, 2)(3, -4) = 3 - 8 \\ \Rightarrow (1, 2)(3, -4) = -5$

$\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right] \left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\begin{bmatrix} 8 &-5 \\ & \end{bmatrix}$

Similarly, let us repeat for the rest of the elements.

$\begin{aligned} &\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\left[\begin{array}{cc} 8 & -5 \\ (-2 \times 2)+(1 \times 3) & (-2 \times 3)+(1 \times-4) \end{array}\right]\\ &\Rightarrow\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\left[\begin{array}{cc} 8 & -5 \\ -4+3 & -6-4 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\\ &\text { Let } D=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\\ &\text { Now, compute for DC. }[\because(A B) C=D C]\\ &\mathrm{DC}=\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right] \end{aligned}$

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

$\\(8, -5)(1, -1) = (8 \times 1) + (-5 \times -1) \\ \Rightarrow (8, -5)(1, -1) = 8 + 5 \\ \Rightarrow (8, -5)(1, -1) = 13$

$\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right] = \begin{bmatrix} 13 & \\ & \end{bmatrix}$

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

$\\(8, -5)(0, 0) = (8 \times 0) + (-5 \times 0) \\ \Rightarrow (8, -5)(0, 0) = 0 + 0 \\ \Rightarrow (8, -5)(0, 0) = 0$

$\left[\begin{array}{cc} 8 & -5 \\ -1 & -10 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right] = \begin{bmatrix} 13 & 0\\ & \end{bmatrix}$

Similarly, let us repeat for the rest of the elements.

$\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ (-1 \times 1)+(-10 \times-1) & (-1 \times 0)+(-10 \times 0)\end{array}\right]\\ \\\Rightarrow\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ -1+10 & 0\end{array}\right]\\ \\\Rightarrow\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right] \\So, (\mathrm{AB}) \mathrm{C}=\left[\begin{array}{cc}13 & 0 \\ 9 & 0\end{array}\right] \\Take R.H.S: \mathrm{A}(\mathrm{BC}) \\First, compute BC. \\\mathrm{BC}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

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

$\\(2, 3)(1, -1) = (2 \times 1) + (3 \times -1) \\ \Rightarrow (2, 3)(1, -1) = 2 - 3 \\ \Rightarrow (2, 3)(1, -1) = -1$

$\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]= \begin{bmatrix} -1 & \\ & \end{bmatrix}$

Multiply the 1st row of matrix B by matching members of the 2nd column of matrix C, then finally sum them up.

$\\(2, 3)(0, 0) = (2 \times 0) + (3 \times 0) \\ \Rightarrow (2, 3)(0, 0) = 0 + 0 \\ \Rightarrow (2, 3)(0, 0) = 0$

$\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]= \begin{bmatrix} -1 & 0\\ & \end{bmatrix}$

Similarly, let us repeat for the rest of the elements.

$\begin{aligned} &\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ (3 \times 1)+(-4 \times-1) & (3 \times 0)+(-4 \times 0) \end{array}\right]\\ &\Rightarrow\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 3+4 & 0 \end{array}\right]\\ &\Rightarrow\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]=\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right]\\ &\text { Let } E=\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] .\\ &\text { Now, compute for AE. }\\ &A E=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] \end{aligned}$

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

$\\(1, 2)(-1, 7) = (1 \times -1) + (2 \times 7) \\ \Rightarrow (1, 2)(-1, 7) = -1 + 14 \\ \Rightarrow (1, 2)(-1, 7) = 13$

$\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] =\begin{bmatrix} 13 & \\ & \end{bmatrix}$

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

$\\(1, 2)(0, 0) = (1 \times 0) + (2 \times 0) \\ \Rightarrow (1, 2)(0, 0) = 0 + 0 \\ \Rightarrow (1, 2)(0, 0) = 0$

$\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right] =\begin{bmatrix} 13 &0 \\ & \end{bmatrix}$

Similarly, let us fill in other elements.

${\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right]=\left[\begin{array}{cc} 13 & 0 \\ (-2 \times-1)+(1 \times 7) & (-2 \times 0)+(1 \times 0) \end{array}\right]} \\ \Rightarrow\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right]=\left[\begin{array}{cc} 13 & 0 \\ 2+7 & 0 \end{array}\right] \\ \Rightarrow\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right]\left[\begin{array}{cc} -1 & 0 \\ 7 & 0 \end{array}\right]=\left[\begin{array}{cc} 13 & 0 \\ 9 & 0 \end{array}\right] \\ \text { So, } \\ A(B C)=\left[\begin{array}{cc} 13 & 0 \\ 9 & 0 \end{array}\right] \\ \text { Thus, }(A B) C=A(B C)$

We need to verify: A(B + C) = AB + AC

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

Now, by Adding B + C, we get:

$\\ \mathrm{B}+\mathrm{C}=\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]+\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right] \\\Rightarrow B+C=\left[\begin{array}{cc}2+1 & 3+0 \\ 3-1 & -4+0\end{array}\right] \\\Rightarrow \mathrm{B}+\mathrm{C}=\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right] \\Let \ B+C=F, \ such \ that \ F=\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right] \\Now, \text{by multiplying A and F}, we \ get,\\ A F=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]$

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

$\\(1, 2)(3, 2) = (1 \times 3) + (2 \times 2) \\ \Rightarrow (1, 2)(3, 2) = 3 + 4 \\ \Rightarrow (1, 2)(3, 2) = 7$

$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right] = \begin{bmatrix} 7 & \\ & \end{bmatrix}$

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

$\\(1, 2)(3, -4) = (1 \times 3) + (2 \times -4) \\ \Rightarrow (1, 2)(3, -4) = 3 - 8 \\ \Rightarrow (1, 2)(3, -4) = -5$

$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right] = \begin{bmatrix} 7 &-5 \\ & \end{bmatrix}$

Similarly, let us fill for other elements.

$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ (-2 \times 3)+(1 \times 2) & (-2 \times 3)+(1 \times-4)\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ -6+2 & -6-4\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}3 & 3 \\ 2 & -4\end{array}\right]=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right] \\So, A(B+C)=\left[\begin{array}{cc}7 & -5 \\ -4 & -10\end{array}\right] \\Now, \ take \ R.H.S: \mathrm{AB}+\mathrm{AC} \\Compute \ AB. \\A B=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]$

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

$\\(1, 2)(2, 3) = (1 \times 2) + (2 \times 3) \\ \Rightarrow (1, 2)(2, 3) = 2 + 6 \\ \Rightarrow (1, 2)(2, 3) = 8$

$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right] = \begin{bmatrix} 8 & \\ & \end{bmatrix}$

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

$\\(1, 2)(3, -4) = (1 \times 3) + (2 \times -4) \\ \Rightarrow (1, 2)(3, -4) = 3 - 8 \\ \Rightarrow (1, 2)(3, -4) = -5$

$\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right] = \begin{bmatrix} 8 &-5 \\ & \end{bmatrix}$

Similarly, let us fill for other elements.

$\\\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ (-2 \times 2)+(1 \times 3) & (-2 \times 3)+(1 \times-4)\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -4+3 & -6-4\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}2 & 3 \\ 3 & -4\end{array}\right]=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right] \\So, A B=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right] \\Now, \ compute \ AC. \\A C=\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]$

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

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

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

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

$\\(1, 2)(0, 0) = (1 \times 0) + (2 \times 0) \\ \Rightarrow (1, 2)(0, 0) = 0 + 0 \\ \Rightarrow (1, 2)(0, 0) = 0$

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

Similarly, let us fill in the other elements.

$\\\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ (-2 \times 1)+(1 \times-1) & (-2 \times 0)+(1 \times 0)\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ -2-1 & 0\end{array}\right] \\\Rightarrow\left[\begin{array}{cc}1 & 2 \\ -2 & 1\end{array}\right]\left[\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right]=\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right] \\So, \\A C=\left[\begin{array}{ll}-1 & 0 \\ -3 & 0\end{array}\right] \\Now, \ by \ Adding \ \ A B+A C. \\A B+A C=\left[\begin{array}{cc}8 & -5 \\ -1 & -10\end{array}\right]+\left[\begin{array}{cc}-1 & 0 \\ -3 & 0\end{array}\right]$

If two matrices have the same order, they can be added or subtracted.

$\begin{aligned} &\Rightarrow A B+A C=\left[\begin{array}{cc} 8-1 & -5+0 \\ -1-3 & -10+0 \end{array}\right]\\ &\Rightarrow \mathrm{AB}+\mathrm{AC}=\left[\begin{array}{cc} 7 & -5 \\ -4 & -10 \end{array}\right]\\ &\text { Hence proved, L.H.S }=\mathrm{R.H.S.}\\ &\text { Thus, } A(B+C)=A B+A C . \end{aligned}$

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If $A=\left[\begin{array}{cc} 1 & 2 \\ -2 & 1 \end{array}\right], B=\left[\begin{array}{cc} 2 & 3 \\ 3 & -4 \end{array}\right]_{\text {and }} C=\left[\begin{array}{cc} 1 & 0 \\ -1 & 0 \end{array}\right]$ , verify:
(i) $(AB) C = A (BC)$
(ii) $A(B + C) = AB + AC$