\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}   

Answer:

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.

Answer:

\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}

Answer:

\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

Answer:

\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

Answer:

\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   

Answer:

\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     

Answer:

\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). }

Answer:

\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.

Answer:

\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.

Answer:

\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. }

Answer:

\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]

Answer:

\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?

Answer:

\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.

Answer:

\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.

Answer:

\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}

Answer:

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 .

Answer:

(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

Answer:

(i) FALSE, as matrix multiplication is not commutative

(ii) TRUE, as matrix multiplication is always associative.

(iii) FALSE, as laws of algebra are not applicable to matrices

(iv) TRUE, as in the case of matrices the multiplication is always distributive over addition