$\displaystyle \text{Question 1: Find the values of } x, y, a \text{ and, } b, \text{ if : } \begin{bmatrix} x-2 & y \\ \frac{a}{2} & b+1 \end{bmatrix} = \begin{bmatrix} 0 & 3 \\ 1 & 5 \end{bmatrix}$

If two matrices are equal, their corresponding elements are also equal.

$\displaystyle x-2= 0 \Rightarrow x = 2$

$\displaystyle y=3$

$\displaystyle \frac{a}{2} = 1 \Rightarrow a=2$

$\displaystyle b+1 = 5 \Rightarrow b = 4$

$\displaystyle \therefore x=2, y=3, a=2 \text{ and } b=4$

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$\displaystyle \text{Question 2: Let } A = \begin{bmatrix} 5 & 4 \\ 3 & -2 \end{bmatrix} , B= \begin{bmatrix} -3 & 0 \\ 1 & 4 \end{bmatrix} \text{ and } C= \begin{bmatrix} 1 & -3 \\ 0 & 2 \end{bmatrix}, \text{ find: }$

$\displaystyle \text{(i) A+B and B+A \ \ \ \ \ \ (ii) (A+B)+C and A+(B+C) } \\ \\ \text{ (iii) Is A+B=B+A? \ \ \ \ \ \ \ (iv) Is (A+B)+C=A+(B+C)? }$

In each case, write the conclusion (if any) that you can draw.

$\displaystyle \text{(i) } A + B = \begin{bmatrix} 5 & 4 \\ 3 & -2 \end{bmatrix} + \begin{bmatrix} -3 & 0 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 5-3 & 4+0 \\ 3+1 & -2+4 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}$

$\displaystyle B + A = \begin{bmatrix} -3 & 0 \\ 1 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 4 \\ 3 & -2 \end{bmatrix} = \begin{bmatrix} -3+5 & 0+4 \\ 1+3 & 4-2 \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}$

$\displaystyle \text{(ii) } \because A + B =\begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix}$

$\displaystyle \Rightarrow (A + B) + C =\begin{bmatrix} 2 & 4 \\ 4 & 2 \end{bmatrix} + \begin{bmatrix} 1 & -3 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 2+1 & 4-3 \\ 4+0 & 2+2 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 4 & 4 \end{bmatrix}$

$\displaystyle B + C = \begin{bmatrix} -3 & 0 \\ 1 & 4 \end{bmatrix} + \begin{bmatrix} 1 & -3 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} -3+1 & 0-3 \\ 1+0 & 4+2 \end{bmatrix} = \begin{bmatrix} -2 & -3 \\ 1 & 6 \end{bmatrix}$

$\displaystyle \Rightarrow A + (B + C) = \begin{bmatrix} 5 & 4 \\ 3 & -2 \end{bmatrix} + \begin{bmatrix} -2 & -3 \\ 1 & 6 \end{bmatrix} = \begin{bmatrix} 5-2 & 4-3 \\ 3+1 & -2+ 6 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 4 & 4 \end{bmatrix}$

(iii) Yes; $A+B=B+A$

Conclusion: Addition of matrices is commutative

(iv) Yes;  $(A+B) + C = A + ( B + C)$

Conclusion : Addition of matrices is associative.

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$\displaystyle \text{Question 3: Let } A = \begin{bmatrix} 5 & 4 \\ 3 & -1 \end{bmatrix} , B= \begin{bmatrix} 2 & 1 \\ 0 & 4 \end{bmatrix} \text{ and } C= \begin{bmatrix} -3 & 2 \\ 1 & 0 \end{bmatrix}, \text{ find: }$

$\displaystyle \text{(i) A + C \ \ \ \ \ \ (ii) B - A \ \ \ \ \ \ (iii) A+B - C}$

$\displaystyle \text{(i) } A + C = \begin{bmatrix} 5 & 4 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} -3 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 5-3 & 4+2 \\ 3+1 & -1+0 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ 4 & -1 \end{bmatrix}$

$\displaystyle \text{(ii) } B - A = \begin{bmatrix} 2 & 1 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} 5 & 4 \\ 3 & -1 \end{bmatrix} = \begin{bmatrix} 2-5 & 1-4 \\ 0-3 & 0+4 \end{bmatrix} = \begin{bmatrix} -3 & -3 \\ -3 & 5 \end{bmatrix}$

$\displaystyle \text{(iii) } A + B - C = \begin{bmatrix} 5 & 4 \\ 3 & -1 \end{bmatrix} +\begin{bmatrix} 2 & 1 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} -3 & 2 \\ 1 & 0 \end{bmatrix} \\ \\ \\ = \begin{bmatrix} 7 & 5 \\ 3 & 3 \end{bmatrix} - \begin{bmatrix} -3 & 2 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 7+3 & 5-2 \\ 3-1 & 3-0 \end{bmatrix} =\begin{bmatrix} 10 & 3 \\ 2 & 3 \end{bmatrix}$

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$\displaystyle \text{Question 4: If a Matrix } A = \begin{bmatrix} 2 & 1 & 3\\ 4 & -3 & 2 \end{bmatrix} \text{ and } B = \begin{bmatrix} 3 & -2 \\ 7 & 4 \end{bmatrix}; \\ \\ \text{ find transpose matrices } A^t \text{ and } B^t. \text{ If possible, find: (i) } A + A^t \text{ (ii) } B + B^t$

$\displaystyle A^t = \begin{bmatrix} 2 & 4 \\ 1 & -3 \\ 3 & 2 \end{bmatrix} \text{ and } B^t = \begin{bmatrix} 3 & 7 \\ -2 & 4 \end{bmatrix}$

(i) Since, the order of matrix $A$ is $2 \times 3$ and that of $A^t$ is $3 \times 2, A + A^t$  is not possible.

(ii) Since, the order of matrix $B$ is $2 \times 2$ and that of $B^t,$ is also $2 \times 2, B + B^t$  is possible.

$\displaystyle B + B^t = \begin{bmatrix} 3 & -2 \\ 7 & 4 \end{bmatrix} + \begin{bmatrix} 3 & 7 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 3+3 & -2+7 \\ 7-2 & 4+4 \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 5 & 8 \end{bmatrix}$

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$\displaystyle \text{Question 5: If } A = \begin{bmatrix} 8 & 6 \\ -2 & 4 \end{bmatrix} \text{ and } B = \begin{bmatrix} -3 & 5 \\ 1 & 0 \end{bmatrix}; \text{ then solve for } 2 \times 2 \text{ matrix } X \\ \\ \text{ such that: } \text{(i) } A + X = B \text{ (ii) } X - B = A$

$\displaystyle \text{(i) } A + X = B \Rightarrow X = B - A \\ \\ \\ { \hspace{3.0cm} = \begin{bmatrix} -3 & 5 \\ 1 & 0 \end{bmatrix} - \begin{bmatrix} 8 & 6 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} -3-8 & 5-6 \\ 1+2 & 0- 4 \end{bmatrix} = \begin{bmatrix} -11 & -1 \\ 3 & -4 \end{bmatrix}}$

$\displaystyle \text{(ii) } X - B = A \Rightarrow X = A+B = \begin{bmatrix} 8 & 6 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} -3 & 5 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 5 & 11 \\ -1 & 4 \end{bmatrix}$

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$\displaystyle \text{Question 6: Let } A = \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix} , B= \begin{bmatrix} -2 & -1 \\ 1 & 2 \end{bmatrix} \text{ and } C= \begin{bmatrix} 0 & 3 \\ 2 & -1 \end{bmatrix}, \\ \\ \text{ find: } A + 2B - 3C$

$\displaystyle A + 2B - 3C = \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix} + 2 \begin{bmatrix} -2 & -1 \\ 1 & 2 \end{bmatrix} - 3\begin{bmatrix} 0 & 3 \\ 2 & -1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & 2 \\ -2 & 3 \end{bmatrix} + \begin{bmatrix} -4 & -2 \\ 2 & 4 \end{bmatrix} - \begin{bmatrix} 0 & 9 \\ 6 & -3 \end{bmatrix} = \begin{bmatrix} -3 & 0 \\ 0 & 7 \end{bmatrix} - \begin{bmatrix} 0 & 9 \\ 6 & -3 \end{bmatrix} = \begin{bmatrix} -3 & -9 \\ -6 & 10 \end{bmatrix}$

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$\displaystyle \text{Question 7: Let } A = \begin{bmatrix} 5 \\ -3 \end{bmatrix} , \text{ and } B= \begin{bmatrix} -1 \\ 7 \end{bmatrix} \text{ find matrix } X \text{ such that:} \\ \\ A + 2X = B$

$\displaystyle A + 2X = B \Rightarrow 2X = B - A = \begin{bmatrix} -1 \\ 7 \end{bmatrix} - \begin{bmatrix} 5 \\ -3 \end{bmatrix} = \begin{bmatrix} -6 \\ 10 \end{bmatrix}$

$\displaystyle \Rightarrow X = \frac{1}{2} \begin{bmatrix} -6 \\ 10 \end{bmatrix} =\begin{bmatrix} \frac{1}{2} \times (-6) \\ \frac{1}{2} \times 10 \end{bmatrix} = \begin{bmatrix} -3 \\ 5 \end{bmatrix}$

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$\displaystyle \text{Question 8: Let } A = \begin{bmatrix} -2 & 3 \\ 4 & 1 \end{bmatrix} , \text{ and } B= \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} \\ \\ \text{find: (i) AB \ \ \ (ii) BA \ \ \ (iii) Is AB = BA } \\ \\ \text{ (iv) Write the conclusion that you draw from the result obtained above in (iii). }$

$\displaystyle \text{(i) } AB = \begin{bmatrix} -2 & 3 \\ 4 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} = \begin{bmatrix} -2 \times 1 + 3 \times 3 & -2 \times 2 + 3 \times 5 \\ 4 \times 1 + 1 \times 3 & 4 \times 2 + 1 \times 5 \end{bmatrix} = \begin{bmatrix} 7 & 11 \\ 7 & 13 \end{bmatrix}$

$\displaystyle \text{(ii) } BA = \begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix} \begin{bmatrix} -2 & 3 \\ 4 & 1 \end{bmatrix} = \begin{bmatrix} 1 \times (-2) + 2 \times 4 & 1 \times 3 + 2 \times 1 \\ 3 \times (-2) + 5 \times 4 & 3 \times 3 + 5 \times 1 \end{bmatrix} = \begin{bmatrix} 6 & 5 \\ 14 & 14 \end{bmatrix}$

(ii) No, $AB \neq BA$

(iv) Conclusion: Matrix multiplication is not commutative.

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$\displaystyle \text{Question 9: Let } A = \begin{bmatrix} -3 & 3 \\ 2 & -2 \end{bmatrix} , \text{ and } B= \begin{bmatrix} 4 & 6 \\ 4 & 6 \end{bmatrix}$

Find the matrix $AB.$ write the conclusion, if any, that you can draw from the result obtained.

$\displaystyle \text{(i) } AB = \begin{bmatrix} -3 & 3 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} 4 & 6 \\ 4 & 6 \end{bmatrix} = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

The result obtained is $AB =0,$ zero matrix.

Conclusion:  The product of two non-zero matrices can be a zero matrix.

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$\displaystyle \text{Question 10: Let } A = \begin{bmatrix} 4 & -4 \\ -3 & 3 \end{bmatrix} , B= \begin{bmatrix} 6 & 5 \\ 3 & 0 \end{bmatrix} \text{ and } C= \begin{bmatrix} -2 & 3 \\ -1 & -2 \end{bmatrix}$

Show that $AB = AC.$ Write the conclusion, if any, that you can draw from the result obtained above.

$\displaystyle AB = \begin{bmatrix} 4 & -4 \\ -3 & 3 \end{bmatrix} \begin{bmatrix} 6 & 5 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 24-12 & 20+0 \\ -18+9 & -15+0 \end{bmatrix} = \begin{bmatrix} 12 & 20 \\ -9 & -15 \end{bmatrix}$

$\displaystyle AC = \begin{bmatrix} 4 & -4 \\ -3 & 3 \end{bmatrix} \begin{bmatrix} -2 & 3 \\ -1 & -2 \end{bmatrix} = \begin{bmatrix} 8+4 & 12+8 \\ -6-3 & -9-6 \end{bmatrix} = \begin{bmatrix} 12 & 20 \\ -9 & -15 \end{bmatrix}$

From above, we get:  $AB = AC$

Conclusion :

$AB = AC \Rightarrow$  Matrices $B$ and $C$ are not equal and matrix $A$ is not a zero matrix, even then $AB=AC.$

Conversely, if $AB = AC,$ it does not imply that $B = C.$ That is in $AB = AC,$ we can not cancel matrix $A$ from both the sides.

In other words, cancellation law is not applicable in matrix multiplication.

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$\displaystyle \text{Question 11: Let } A = \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix}, \text{ evaluate } A^2 - 3A + 2I, \\ \\ \text{ where I is a unit matrix of order 2. }$

$\displaystyle A^2 - 3A + 2I= \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} - 3 \begin{bmatrix} 2 & -1 \\ -1 & 3 \end{bmatrix} + 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 5 & -5 \\ -5 & 10 \end{bmatrix} - \begin{bmatrix} 6 & -3 \\ -3 & 9 \end{bmatrix} + \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 5-6+2 & -5+3+0 \\ -5+3+0 & 10-9+2 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 1 & -2 \\ -2 & 3 \end{bmatrix}$

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$\displaystyle \text{Question 12: Let } A = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} , \text{ and } B= \begin{bmatrix} 2 \\ 4 \end{bmatrix}, \text{ is the product of } AB \text{ possible. }$

Give a reason. If yes, find $AB.$ [ICSE Board 2011]

$\displaystyle \text{Since order of matrix } A = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} \text{ is } 2 \times 2 \text{ i.e. has two rows and two columns. }$

$\displaystyle \text{And order of matrix } B= \begin{bmatrix} 2 \\ 4 \end{bmatrix} \text{ is } 2 \times 1 \text{ i.e. has two rows and two columns. }$

Hence, the number of columns in matrix $A$ is same as the number of rows in matrix $B$; therefore the product $AB$ is possible.

$\displaystyle AB = \begin{bmatrix} 3 & 5 \\ 4 & -2 \end{bmatrix} \begin{bmatrix} 2 \\ 4 \end{bmatrix} = \begin{bmatrix} 3 \times 2 + 5 \times 4 \\ 4 \times 2 - 2 \times 4 \end{bmatrix}= \begin{bmatrix} 26 \\ 0 \end{bmatrix}$

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$\displaystyle \text{Question 13: Let } A = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} , \text{ and } B= \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} ; \text{ find }:$

$\displaystyle \text{(i) } ( A + B) (A - B) \ \ \ \text{(ii) } A^2 - B^2 \text{ is } ( A + B)(A - B) = A^2 - B^2?$

$\displaystyle \text{(i) }A + B = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 3+1 & 2+0 \\ 0+1 & 5+2 \end{bmatrix} = \begin{bmatrix} 4 & 2 \\ 1 & 7 \end{bmatrix}$

$\displaystyle A - B = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 3-1 & 2-0 \\ 0-1 & 5-2 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -1 & 3 \end{bmatrix}$

$\displaystyle \therefore ( A + B) (A - B) = \begin{bmatrix} 4 & 2 \\ 1 & 7 \end{bmatrix} \begin{bmatrix} 2 & 2 \\ -1 & 3 \end{bmatrix}$

$\displaystyle = \begin{bmatrix} 4 \times 2 + 2 \times (-1) & 4 \times 2 + 2 \times 3 \\ 1 \times 2 + 7 \times ( -1) & 1 \times 2 + 7 \times 3 \end{bmatrix} = \begin{bmatrix} 6 & 14 \\ -5 & 23 \end{bmatrix}$

$\displaystyle \therefore A^2 = \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 3 & 2 \\ 0 & 5 \end{bmatrix} = \begin{bmatrix} 3 \times 3 + 2 \times 0 & 3 \times 2 + 2 \times 5 \\ 0 \times 3 + 5 \times 0 & 0 \times 2 + 5 \times 5 \end{bmatrix} = \begin{bmatrix} 9 & 16 \\ 0 & 25 \end{bmatrix}$

$\displaystyle \therefore B^2 = \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 \times 1 + 0 \times 1 & 1 \times 0 + 0 \times 2 \\ 1 \times 1 + 2 \times 1 & 1 \times 0 + 2 \times 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix}$

$\displaystyle A^2 - B^2 = \begin{bmatrix} 9 & 16 \\ 0 & 25 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 8 & 16 \\ -3 & 21 \end{bmatrix}$

From the results of parts (i) and (ii) it is clear that : $( A + B)(A - B) \neq A^2 - B^2$

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$\displaystyle \text{Question 14: Given } \begin{bmatrix} 3 & -8 \\ 9 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -2 \\ 8 \end{bmatrix} , \text{ find } x \text{ and } y.$

$\displaystyle \begin{bmatrix} 3 & -8 \\ 9 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -2 \\ 8 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} 3x-8y \\ 9x+4y \end{bmatrix} = \begin{bmatrix} -2 \\ 8 \end{bmatrix}$

$\displaystyle \Rightarrow 3x - 8y = -2 \text{ and } 9x +4y = 8$

$\displaystyle \text{Solving the two equations above we get } x = \frac{2}{3} \text{ and } y = \frac{1}{2}$

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$\displaystyle \text{Question 15: If } B \text{ and } C \text{ are two matrices such that, } \\ \\ B = \begin{bmatrix} 1 & 3 \\ -2 & 0 \end{bmatrix} \text{ and } C = \begin{bmatrix} 17 & 7 \\ -4 & -8 \end{bmatrix}, \text{ find the matrix } A \text{ so that } BA = C.$

$\displaystyle \text{Let order of matrix } A = m \times n$

$\displaystyle \therefore BA = C \Rightarrow B_{2 \times 2} \cdot A_{m \times n} = C_{2 \times 2}$

$\displaystyle \Rightarrow \text{ order of matrix } A = m \times n = 2 \times 2$

$\displaystyle \text{Now take matrix } A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

$\displaystyle \text{Given } BA = C \Rightarrow \begin{bmatrix} 1 & 3 \\ -2 & 0 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 17 & 7 \\ -4 & -8 \end{bmatrix}$

$\displaystyle \Rightarrow \begin{bmatrix} a+3c & b+3d \\ -2a & -2b \end{bmatrix} = \begin{bmatrix} 17 & 7 \\ -4 & -8 \end{bmatrix}$

$\displaystyle \Rightarrow a + 3c = 17 \text{ ... ... ... ... ... i)} \hspace{1.0cm} -2 a = - 4 \text{ ... ... ... ... ... ii)} \\ \\ \Rightarrow b+3d = 7 \text{ ... ... ... ... ... iii)} \hspace{1.0cm} -2b = -8 \text{ ... ... ... ... ... iv)}$

On solving equations i) and ii), we get  a = 2 and c = 5

On solving equations iii)  and iv) we get  b = 4 and d = 1

$\displaystyle \therefore A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} 2 & 4 \\ 5 & 1 \end{bmatrix}$

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$\displaystyle \text{Question 16: Find the matrix } M, \text{ such that } M \times \begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix} = \begin{bmatrix} -2 & 16 \end{bmatrix}$

First of all, we must find the order of matrix $M$.

Let the order of matrix $M$ be $a \times b$

$\displaystyle \Rightarrow M_{a \times b} \times \begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix} = \begin{bmatrix} -2 & 16 \end{bmatrix}$

Since the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

$\therefore b = 2$

Also, the no. of rows of product (resulting) matrix is equal to the no. of rows of first matrix.

$\therefore a = 1 \Rightarrow \text{ Order of matrix } M = a \times b = 1 \times 2$

$\text{Let } M = \begin{bmatrix} x & y \end{bmatrix}$

$\therefore \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 3 & 6 \\ -2 & -8 \end{bmatrix} = \begin{bmatrix} -2 & 16 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 3x-2y & 6x-8y \end{bmatrix} = \begin{bmatrix} -2 & 16 \end{bmatrix}$

$\Rightarrow 3x-2y = - 2 \text{ and } 6x - 8y = 16$

On solving the above equations we get $x = - 4$ and $y = - 5$

$\therefore M =\begin{bmatrix} -4 & -5 \end{bmatrix}$

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$\displaystyle \text{Question 17: } \begin{bmatrix} 8 & 4 \\ 1 & -8 \end{bmatrix} \cdot X= \begin{bmatrix} 12 \\ 10 \end{bmatrix}$

Write down :  (i) the order of the matrix $X$      (ii) the matrix $X$.

(i) Let the order of matrix $X \ be \ a \times b$

$\displaystyle \therefore {\begin{bmatrix} 8 & 4 \\ 1 & -8 \end{bmatrix}}_{2 \times 2} \cdot X_{a \times b}= {\begin{bmatrix} 12 \\ 10 \end{bmatrix}}_{2 \times 1}$

Therefore the order of the matrix $X = a \times b = 2 \times 1$

$\text{(ii) Let } X = \begin{bmatrix} x \\ y \end{bmatrix}$

$\therefore \begin{bmatrix} 8 & -2 \\ 1 & 4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 12 \\ 10 \end{bmatrix}$

$\Rightarrow \begin{bmatrix} 8x-2y \\ x+4y \end{bmatrix} = \begin{bmatrix} 12 \\ 10 \end{bmatrix}$

$\Rightarrow 8x-2y = 12 \text{ and } x+4y = 10$

$\text{On solving, we get : } x = 2 \text{ and } y = 2$

$\text{Therefore the matrix } X = \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}$

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Question 18:  State with reason, whether the following are TRUE or FALSE. A, B, C are matrices of order 2 \times 2.

$\text{(i) } A \cdot B = B \cdot A \hspace{1.0cm} \text{(ii) } A \cdot (B \cdot C) = ( A \cdot B) \cdot C \hspace{1.0cm} \\ \\ \text{(iii) } (A+B)^2 = A^2 + 2 AB + B^2 \hspace{1.0cm} \text{(iv) } A \cdot ( B + C) = A \cdot B + A \cdot C$