Question 1: Evaluate

\displaystyle \text{i) } \begin{bmatrix} 3 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix}       \displaystyle \text{ii) } \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} -2 & 3 \\ -1 & 4 \end{bmatrix}       \displaystyle \text{iii) } \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} -1 \\ 3 \end{bmatrix}       \displaystyle \text{iv) } \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} -1 & 3 \end{bmatrix}  

Answer:

\displaystyle \text{i) } \begin{bmatrix} 3 & 2 \end{bmatrix} \begin{bmatrix} 2 \\ 0 \end{bmatrix} = \begin{bmatrix} 6 \end{bmatrix}  

\displaystyle \text{ii) } \begin{bmatrix} 1 & -2 \end{bmatrix} \begin{bmatrix} -2 & 3 \\ -1 & 4 \end{bmatrix} = \begin{bmatrix} 0 & -5 \end{bmatrix}  

\displaystyle \text{iii) } \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 6 \\ -6 \end{bmatrix}  

\displaystyle \text{iv) } \begin{bmatrix} 6 & 4 \\ 3 & -1 \end{bmatrix} \begin{bmatrix} -1 & 3 \end{bmatrix}

This multiplication is not possible as the number of columns in the first matrix is not equal to the number of rows in the second matrix.

\displaystyle \\

\displaystyle \text{Question 2: If } A = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix}, B = \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}  \text{ and }  I \text{ is a unit matrix of the order } 2 \times 2 \text{, find: }

\displaystyle \text{i) } AB   \hspace{1.0cm}  \text{ii) } BA   \hspace{1.0cm}  \text{iii) } AI   \hspace{1.0cm}  \text{iv) } IB   \hspace{1.0cm}  \text{v) } A^2   \hspace{1.0cm}  \text{vi) } B^2A  

Answer:

\displaystyle I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}  

\displaystyle \text{i) } AB = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 4 \\ -1 & -9 \end{bmatrix}  

\displaystyle \text{ii) } BA = \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} = \begin{bmatrix} -5 & 4 \\ 10 & 2 \end{bmatrix}  

\displaystyle \text{iii) } AI = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix}  

\displaystyle \text{iv) } IB = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix}  

\displaystyle \text{v) } A^2 = \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} = \begin{bmatrix} 10 & -4 \\ -10 & 14 \end{bmatrix}  

\displaystyle \text{vi) } B^2A = \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} = \begin{bmatrix} -2 & -3 \\ 9 & 1 \end{bmatrix} \begin{bmatrix} 0 & 2 \\ 5 & -2 \end{bmatrix} = \begin{bmatrix} -15 & 2 \\ 5 & 16 \end{bmatrix}  

\displaystyle \\

\displaystyle \text{Question 3: If } M = \begin{bmatrix} 2 & 1 \\ 1 & -2 \end{bmatrix} \text{, find } M^2, \ M^3 \ and \ M^5

Answer:

\displaystyle M^2 = \begin{bmatrix} 2 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix}  

\displaystyle M^3 = \begin{bmatrix} 2 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ 1 & -2 \end{bmatrix} = \begin{bmatrix} 10 & 5 \\ 5 & -10 \end{bmatrix}  

\displaystyle M^5 = \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} \begin{bmatrix} 10 & 5 \\ 5 & -10 \end{bmatrix} = \begin{bmatrix} 50 & 25 \\ 25 & -50 \end{bmatrix}  

\displaystyle \\

Question 4: Find \displaystyle x \ and \ y if

\displaystyle \text{i) } \begin{bmatrix} 4 & 3x \\ x & -2 \end{bmatrix} \begin{bmatrix} 5 \\ 1 \end{bmatrix} = \begin{bmatrix} y \\ 8 \end{bmatrix}             \displaystyle \text{ii) } \begin{bmatrix} x & 0 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}  

Answer:

\displaystyle \text{i) } \begin{bmatrix} 4 & 3x \\ x & -2 \end{bmatrix} \begin{bmatrix} 5 \\ 1 \end{bmatrix} = \begin{bmatrix} y \\ 8 \end{bmatrix} \Rightarrow \begin{bmatrix} 20+3x \\ 5x-2 \end{bmatrix} = \begin{bmatrix} y \\ 8 \end{bmatrix}  

Therefore

 \displaystyle 20+3x = y  

 \displaystyle 5x-2=8 \Rightarrow x = 2  

Hence \displaystyle y = 26  

\displaystyle \text{ii) } \begin{bmatrix} x & 0 \\ -3 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 0 & y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix} \Rightarrow \begin{bmatrix} x & x \\ -3 & -3+y \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ -3 & -2 \end{bmatrix}  

Therefore

 \displaystyle x = 2  

 \displaystyle -3+y = - 2 \Rightarrow y = 1  

\displaystyle \\

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

\displaystyle \text{i) } (AB) C   \hspace{1.0cm}  \text{ii) } A (BC) \displaystyle \text{Is }  A(BC) = (AB)C?  

Answer:

\displaystyle \text{i) } (AB) C = \Bigg( \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} \Bigg)  \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 13 & 11 \\ 18 & 16 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 63 & 61 \\ 88 & 86 \end{bmatrix}

\displaystyle \text{ii) } A (BC) = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \Bigg( \begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 1 & 2 \end{bmatrix} \Bigg)  = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} 6 & 7 \\ 19 & 18 \end{bmatrix} = \begin{bmatrix} 63 & 61 \\ 88 & 86 \end{bmatrix}

\displaystyle \text{Is }  A(BC) = (AB)C? Yes.

\displaystyle \\

\displaystyle \text{Question 6: If } A = \begin{bmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{bmatrix}, B = \begin{bmatrix} 0 & -1 \\ -1 & 2 \\ -5 & -6 \end{bmatrix} \text{, calculate:}

\displaystyle \text{i) } AB   \hspace{1.0cm}  \text{ii) } BA   \hspace{1.0cm}  \text{iii) } A^2  

Answer:

\displaystyle \text{i) } AB = \begin{bmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ -1 & 2 \\ -5 & -6 \end{bmatrix} = \begin{bmatrix} -34 & -28 \\ 5 & 9 \end{bmatrix}  

\displaystyle \text{ii) } BA = \begin{bmatrix} 0 & -1 \\ -1 & 2 \\ -5 & -6 \end{bmatrix} \begin{bmatrix} 0 & 4 & 6 \\ 3 & 0 & -1 \end{bmatrix} = \begin{bmatrix} 3 & 0 & -1 \\ 6 & -4 & -8 \\ -18 & -20 & -24 \end{bmatrix}  

\displaystyle \text{iii) } A^2  

This multiplication is not possible as the number of columns in the first matrix is not equal to the number of rows in the second matrix.

\displaystyle \\

\displaystyle \text{Question 7: If } A = \begin{bmatrix} 2 & 1 \\ 0 & -2 \end{bmatrix}, B = \begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix} \text{ and } C = \begin{bmatrix} -3 & 2 \\ -1 & 4 \end{bmatrix} \\ \\ \text{Find } A^2 + AC-5B \hspace{1.0cm}  [2014]

Answer:

\displaystyle A^2 + AC-5B  

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

\displaystyle = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} + \begin{bmatrix} -7 & 8 \\ 2 & -8 \end{bmatrix} - \begin{bmatrix} 20 & 5 \\ -15 & -10 \end{bmatrix} = \begin{bmatrix} -23 & 3 \\ 17 & 14 \end{bmatrix}  

\displaystyle \\

\displaystyle \text{Question 8: If } M = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} \text{ and } I \text{ is the unit matrix of the same order as that of }  M \\ \\ \text{; show that: } M^2= 2M+3I  

Answer:

\displaystyle I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}  

\displaystyle M^2= 2M+3I  

\displaystyle LHS = M^2 = \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}  

\displaystyle RHS = 2M+3I = 2 \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} + 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 5 & 4 \\ 4 & 5 \end{bmatrix}  

Hence proved \displaystyle LHS = RHS  

\displaystyle \\

\displaystyle \text{Question 9: If } A = \begin{bmatrix} a & 0 \\ 0 & 2 \end{bmatrix}, B = \begin{bmatrix} 0 & -b \\ 1 & 0 \end{bmatrix} \ and \ M = \begin{bmatrix} 1 & -1 \\ \\  1 & 1 \end{bmatrix}  \text{ and }  BA=M^2 \text{, find the values of } a \ and \ b .

Answer:

\displaystyle BA=M^2  

\displaystyle \begin{bmatrix} 0 & -b \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a & 0 \\ 0 & 2 \end{bmatrix} =\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix}  

\displaystyle \begin{bmatrix} 0 & -2b \\ a & 0 \end{bmatrix} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix}  

Therefore \displaystyle b = 1 \ and \ a = 2  

\displaystyle \\

\displaystyle \text{Question 10: If } A = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} \text{, find}

\displaystyle \text{i) } A-B   \hspace{1.0cm}  \text{ii) } A^2   \hspace{1.0cm}  \text{iii) } AB   \hspace{1.0cm}  \text{iv) } A^2-Ab+2B  

Answer:

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

\displaystyle \text{ii) } A^2 = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 18 & 7 \\ 14 & 11 \end{bmatrix}  

\displaystyle \text{iii) } AB = \begin{bmatrix} 4 & 3 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ -4 & 3 \end{bmatrix}  

\displaystyle \text{iv) } A^2-AB+2B = \begin{bmatrix} 18 & 7 \\ 14 & 11 \end{bmatrix} - \begin{bmatrix} 1 & 1 \\ 4 & 3 \end{bmatrix} +2 \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}= \begin{bmatrix} 18 & 6 \\ 14 & 10 \end{bmatrix}

\displaystyle \\

\displaystyle \text{Question 11: If } A = \begin{bmatrix} 1 & 4 \\ 1 & -1 \end{bmatrix}  \text{ and }  B = \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \text{ find: }

\displaystyle \text{i) } (A+B)^2   \hspace{1.0cm}  \text{ii) } A^2+B^2  \hspace{1.0cm}  \text{iii) Is }  (A+B)^2 = A^2+B^2  

Answer:

\displaystyle \text{i) } (A+B)^2  

 \displaystyle = (\begin{bmatrix} 1 & 4 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} )^2  

\displaystyle = (\begin{bmatrix} 2 & 6 \\ 0 & -4 \end{bmatrix})^2 = \begin{bmatrix} 2 & 6 \\ 0 & -4 \end{bmatrix} \times \begin{bmatrix} 2 & 6 \\ 0 & -4 \end{bmatrix} = \begin{bmatrix} 4 & -12 \\ 0 & 16 \end{bmatrix}

\displaystyle \text{ii) } A^2+B^2  

\displaystyle = \begin{bmatrix} 1 & 4 \\ 1 & -1 \end{bmatrix} \times \begin{bmatrix} 1 & 4 \\ 1 & -1 \end{bmatrix} + \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix} \times \begin{bmatrix} 1 & 2 \\ -1 & -1 \end{bmatrix}

\displaystyle = \begin{bmatrix} 5 & -8 \\ -2 & 13 \end{bmatrix} + \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 4 & -8 \\ -2 & 12 \end{bmatrix}

iii) Is \displaystyle (A+B)^2 = A^2+B^2 : No

\displaystyle \\

\displaystyle \text{Question 13: If } A = \begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix}  \text{ and }  A^2=I \text{, find } a \ and \ b .

Answer:

\displaystyle \text{Given } I = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}  

Therefore

\displaystyle \begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix} \times \begin{bmatrix} -1 & 1 \\ a & b \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}  

 \displaystyle \begin{bmatrix} 1+a & -1+b \\ -a+ab & a+b^2 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}  

 \displaystyle \Rightarrow 1+a = a \Rightarrow a = 0  

Similarly \displaystyle -1+b=0 \Rightarrow b =1  

\displaystyle \\

\displaystyle \text{Question 14: If } A = \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix}, B = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}, \ and \ C = \begin{bmatrix} -1 & 4 \\ 0 & 2 \end{bmatrix} \text{, then show that }

\displaystyle \text{i) } A(B+C) = AB + AC   \hspace{1.0cm}  \text{ii) } (B-A)C=BC-AC  

Answer:

\displaystyle \text{i) } A(B+C) = AB + AC  

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

\displaystyle = \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} (\begin{bmatrix} 3 & 7 \\ 4 & 3 \end{bmatrix})  

\displaystyle = \begin{bmatrix} 10 & 17 \\ 0 & 0 \end{bmatrix}  

\displaystyle \text{RHS } = AB + AC  

\displaystyle = \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} . \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix}. \begin{bmatrix} -1 & 4 \\ 0 & 2 \end{bmatrix}  

\displaystyle = \begin{bmatrix} 8 & 7 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 2 & 10 \\ 0 & 0 \end{bmatrix}  

\displaystyle = \begin{bmatrix} 10 & 17 \\ 0 & 0 \end{bmatrix}  

Hence LHS = RHS

\displaystyle \text{ii) } (B-A)C=BC-AC  

\displaystyle \text{LHS } = (\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}- \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix}). \begin{bmatrix} -1 & 4 \\ 0 & 2 \end{bmatrix}  

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

\displaystyle =\begin{bmatrix} 0 & 4 \\ 4 & 18 \end{bmatrix}  

RHS \displaystyle = BC-AC  

\displaystyle = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix} . \begin{bmatrix} -1 & 4 \\ 0 & 2 \end{bmatrix} - \begin{bmatrix} 2 & 1 \\ 0 & 0 \end{bmatrix} . \begin{bmatrix} -1 & 4 \\ 0 & 2 \end{bmatrix}  

\displaystyle = \begin{bmatrix} 2 & 14 \\ 4 & 18 \end{bmatrix} - \begin{bmatrix} 2 & 10 \\ 0 & 0 \end{bmatrix}  

\displaystyle = \begin{bmatrix} 0 & 4 \\ 4 & 18 \end{bmatrix}  

\displaystyle \\

\displaystyle \text{Question 15: If } A = \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} , B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} \text{, and }  C = \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix} \text{, find the value of } \\ A^2+BC  

Answer:

\displaystyle A^2+BC  

\displaystyle = \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} . \begin{bmatrix} 1 & 4 \\ 2 & 1 \end{bmatrix} + \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} . \begin{bmatrix} 1 & 0 \\ 0 & 2 \end{bmatrix}  

\displaystyle = \begin{bmatrix} 9 & 8 \\ 4 & 9 \end{bmatrix} + \begin{bmatrix} -3 & 4 \\ 4 & 0 \end{bmatrix}  

\displaystyle = \begin{bmatrix} 6 & 12 \\ 8 & 9 \end{bmatrix}  

\displaystyle \\

Question 16: Solve for \displaystyle x \ and \ y  

\displaystyle \text{i) } \begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix} \hspace{1.0cm} \text{ii) } \begin{bmatrix} x+y & x-4 \end{bmatrix} . \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}  

\displaystyle \text{iii) } \begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} . \begin{bmatrix} -1 \\ 2x \end{bmatrix} +3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix} \hspace{1.0cm} [2014]  

Answer:

\displaystyle \text{i) } \begin{bmatrix} 2 & 5 \\ 5 & 2 \end{bmatrix} . \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}  

\displaystyle \Rightarrow \begin{bmatrix} 2x+5y & 5x+2y \end{bmatrix} = \begin{bmatrix} -7 \\ 14 \end{bmatrix}  

Therefore

\displaystyle 2x+5y = -7  

\displaystyle 5x+2y=14  

Solving the above two equations we get

\displaystyle x = 4  \text{ and }  y = -3  

\displaystyle \text{ii) } \begin{bmatrix} x+y & x-4 \end{bmatrix} . \begin{bmatrix} -1 & -2 \\ 2 & 2 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}  

\displaystyle \Rightarrow \begin{bmatrix} x-y-8 & -2y-8 \end{bmatrix} = \begin{bmatrix} -7 & 14 \end{bmatrix}  

Therefore

\displaystyle x-y-8=-7 \Rightarrow x-y=1  

Also \displaystyle -2y-8=-11 \Rightarrow y = \frac{3}{2}  

\displaystyle \text{Substituting we get } x = \frac{3}{2}+1 = \frac{5}{2}  

\displaystyle \text{iii) } \begin{bmatrix} -2 & 0 \\ 3 & 1 \end{bmatrix} . \begin{bmatrix} -1 \\ 2x \end{bmatrix} +3 \begin{bmatrix} -2 \\ 1 \end{bmatrix} = 2 \begin{bmatrix} y \\ 3 \end{bmatrix}  

\displaystyle \Rightarrow \begin{bmatrix} 2 \\ -3+2x \end{bmatrix} + \begin{bmatrix} -6 \\ 3 \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}  

\displaystyle \begin{bmatrix} -4 \\ 2x \end{bmatrix} = \begin{bmatrix} 2y \\ 6 \end{bmatrix}  

\displaystyle \text{Therefore } y = -2 \text{ and } x = \frac{6}{2}=3  

\displaystyle \\

Question 17: Find i) the order of matrix M ii) and find M

\displaystyle \text{i) } M \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix} \hspace{1.0cm} \text{ii) } \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \times M = \begin{bmatrix} 13 \\ 5 \end{bmatrix}  

Answer:

\displaystyle \text{i) } M \times \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}  

We know

\displaystyle M_{m \times n} \times B_{m \times n} = C_{m \times n}  

or \displaystyle M_{m \times n} \times B_{2 \times 2} = C_{1 \times 2}  

Hence \displaystyle m = 1 \text{ and } n = 2  

Therefore the order of \displaystyle M \ is \ 1 \times 2  

Let \displaystyle \begin{bmatrix} a & b \end{bmatrix}  

Therefore

\displaystyle \begin{bmatrix} a & b \end{bmatrix} . \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}  

\displaystyle \text{Or } \begin{bmatrix} a & a+2b \end{bmatrix} = \begin{bmatrix} 1 & 2 \end{bmatrix}  

\displaystyle \text{Hence } a = 1 \ and \ b =\frac{1}{2}  

\displaystyle \text{Therefore } M = \begin{bmatrix} 1 & \frac{1}{2} \end{bmatrix}  

\displaystyle \text{ii) } \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \times M = \begin{bmatrix} 13 \\ 5 \end{bmatrix}  

We know

\displaystyle A_{m \times n} \times M_{m \times n} = C_{m \times n}  

or \displaystyle A_{2 \times 2} \times M_{m \times n} = C_{2 \times 1}  

Hence \displaystyle n = 2 and p = 1  

Therefore the order of \displaystyle M \ is \ 2 \times 1  

\displaystyle \text{Let } \begin{bmatrix} a \\ b \end{bmatrix}  

\displaystyle \begin{bmatrix} 1 & 1 \\ 0 & 2 \end{bmatrix} \times \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 13 \\ 5 \end{bmatrix}  

\displaystyle \begin{bmatrix} a+4b \\ 2a+b \end{bmatrix} = \begin{bmatrix} 13 \\ 5 \end{bmatrix}  

Therefore

\displaystyle a+4b = 13  

\displaystyle 2a+b = 5  

Solving the above two equations

\displaystyle a =1 \text{ and } b = 3  

\displaystyle \\

\displaystyle text{Question 18: If } A = \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix}  \text{ and }  B = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix} \text{; find } x \text{ given } A^2=B  

Answer:

Given \displaystyle A^2=B  

Therefore

\displaystyle \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} . \begin{bmatrix} 2 & x \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}  

\displaystyle \begin{bmatrix} 4 & 3x \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 4 & 36 \\ 0 & 1 \end{bmatrix}  

\displaystyle \Rightarrow 3x = 36 \ or \ x = 12  

\displaystyle \\

\displaystyle \text{Question 19: Find positive integers } p \ and \ q \text{ such that } \begin{bmatrix} 2 & x \end{bmatrix} . \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} 25 \end{bmatrix}  

Answer:

\displaystyle \begin{bmatrix} 2 & x \end{bmatrix} . \begin{bmatrix} p \\ q \end{bmatrix} = \begin{bmatrix} 25 \end{bmatrix}  

\displaystyle \begin{bmatrix} p^2+q^2 \end{bmatrix}= \begin{bmatrix} 25 \end{bmatrix}  

\displaystyle p^2+q^2 = 25  

Hence the possible integer combinations are

\displaystyle (p = 3 \ and \ q = 4) or (p=4 \ and \ q=3)  

\displaystyle \\

Question 20: If \displaystyle A \ and \ B are any two \displaystyle 2 \times 2 matrices such that \displaystyle AB = BA = B  \text{ and }  B is not a zero matrix, what can you say about \displaystyle A matrix?

Answer:

Let \displaystyle A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}  \text{ and }  B = \begin{bmatrix} p & q \\ r & s \end{bmatrix}  

Given

\displaystyle \begin{bmatrix} a & b \\ c & d \end{bmatrix} . \begin{bmatrix} p & q \\ r & s \end{bmatrix} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}  

\displaystyle \begin{bmatrix} ap+br & aq+bs \\ cp+dr & cq+ds \end{bmatrix} = \begin{bmatrix} p & q \\ r & s \end{bmatrix}  

\displaystyle \Rightarrow  

\displaystyle ap+br=p  

\displaystyle aq+bs=q  

\displaystyle cp+dr=r  

\displaystyle cq+ds=s  

Solving we get \displaystyle a = 1, b = o, c= 0, \ and \ d=1  

\displaystyle \text{Hence } A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}