\displaystyle \textbf{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}.
\displaystyle \text{Answer:}
\displaystyle \text{If two matrices are equal, then their corresponding elements are 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 \textbf{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\text{ and }B+A\qquad\text{(ii) }(A+B)+C\text{ and }A+(B+C)
\displaystyle \text{(iii) Is }A+B=B+A?\qquad\text{(iv) Is }(A+B)+C=A+(B+C)?\text{ In each case, write the conclusion.}
\displaystyle \text{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 \therefore (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 \therefore 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}
\displaystyle \text{(iii) Yes, }A+B=B+A.
\displaystyle \therefore \text{Addition of matrices is commutative.}
\displaystyle \text{(iv) Yes, }(A+B)+C=A+(B+C).
\displaystyle \therefore \text{Addition of matrices is associative.}
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\displaystyle \textbf{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\qquad\text{(ii) }B-A\qquad\text{(iii) }A+B-C
\displaystyle \text{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&4-(-1)\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}
\displaystyle =\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 \textbf{Question 4: If }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 \text{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}
\displaystyle \text{(i) Since the order of }A\text{ is }2\times3\text{ and that of }A^t\text{ is }3\times2,\ A+A^t\text{ is not possible.}
\displaystyle \text{(ii) Since the order of }B\text{ and }B^t\text{ is }2\times2,\ B+B^t\text{ 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 \textbf{Question 5: If }A=\begin{bmatrix}8&6\\-2&4\end{bmatrix}\text{ and }B=\begin{bmatrix}-3&5\\1&0\end{bmatrix},\text{ solve for the }2\times2\text{ matrix }X\text{ such that:}
\displaystyle \text{(i) }A+X=B\qquad\text{(ii) }X-B=A
\displaystyle \text{Answer:}
\displaystyle \text{(i) }A+X=B\Rightarrow X=B-A
\displaystyle X=\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
\displaystyle X=\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 \textbf{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 \text{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}
\displaystyle =\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 \textbf{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 \text{Answer:}
\displaystyle A+2X=B\Rightarrow 2X=B-A
\displaystyle 2X=\begin{bmatrix}-1\\7\end{bmatrix}-\begin{bmatrix}5\\-3\end{bmatrix}=\begin{bmatrix}-6\\10\end{bmatrix}
\displaystyle X=\frac{1}{2}\begin{bmatrix}-6\\10\end{bmatrix}=\begin{bmatrix}\frac{1}{2}\times(-6)\\\frac{1}{2}\times10\end{bmatrix}=\begin{bmatrix}-3\\5\end{bmatrix}
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\displaystyle \textbf{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:}
\displaystyle \text{(i) }AB\qquad\text{(ii) }BA\qquad\text{(iii) Is }AB=BA?\qquad\text{(iv) Write the conclusion.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }AB=\begin{bmatrix}-2&3\\4&1\end{bmatrix}\begin{bmatrix}1&2\\3&5\end{bmatrix}
\displaystyle =\begin{bmatrix}-2\times1+3\times3&-2\times2+3\times5\\4\times1+1\times3&4\times2+1\times5\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}
\displaystyle =\begin{bmatrix}1\times(-2)+2\times4&1\times3+2\times1\\3\times(-2)+5\times4&3\times3+5\times1\end{bmatrix}=\begin{bmatrix}6&5\\14&14\end{bmatrix}
\displaystyle \text{(iii) No, }AB\neq BA.
\displaystyle \text{(iv) Therefore, matrix multiplication is not commutative.}
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\displaystyle \textbf{Question 9: Let }A=\begin{bmatrix}-3&3\\2&-2\end{bmatrix}\text{ and }B=\begin{bmatrix}4&6\\4&6\end{bmatrix}.\text{ Find the matrix }AB.\text{ Write the conclusion, if any.}
\displaystyle \text{Answer:}
\displaystyle 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}
\displaystyle \therefore AB=\begin{bmatrix}0&0\\0&0\end{bmatrix},\text{ the zero matrix.}
\displaystyle \therefore \text{The product of two non-zero matrices can be a zero matrix.}
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\displaystyle \textbf{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}.\text{ Show that }AB=AC.\text{ Write the conclusion, if any.}
\displaystyle \text{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}
\displaystyle \therefore AB=AC.
\displaystyle \text{Here }B\neq C\text{ and }A\neq O,\text{ yet }AB=AC.
\displaystyle \therefore AB=AC\text{ does not imply }B=C.
\displaystyle \text{Hence, the cancellation law is not applicable in matrix multiplication.}
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\displaystyle \textbf{Question 11: Let }A=\begin{bmatrix}2&-1\\-1&3\end{bmatrix},\text{ evaluate }A^2-3A+2I,\text{ where }I\text{ is the unit matrix of order }2.
\displaystyle \text{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}=\begin{bmatrix}1&-2\\-2&3\end{bmatrix}
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\displaystyle \textbf{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 }AB\text{ possible? Give a reason. If yes, find }AB.\hfill\text{[ICSE Board 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{The order of matrix }A\text{ is }2\times2,\text{ i.e., it has two rows and two columns.}
\displaystyle \text{The order of matrix }B\text{ is }2\times1,\text{ i.e., it has two rows and one column.}
\displaystyle \text{Since the number of columns of }A\text{ equals the number of rows of }B,\text{ the product }AB\text{ is possible.}
\displaystyle AB=\begin{bmatrix}3&5\\4&-2\end{bmatrix}\begin{bmatrix}2\\4\end{bmatrix}=\begin{bmatrix}3\times2+5\times4\\4\times2-2\times4\end{bmatrix}=\begin{bmatrix}26\\0\end{bmatrix}
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\displaystyle \textbf{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)\qquad\text{(ii) }A^2-B^2\text{ Is }(A+B)(A-B)=A^2-B^2?
\displaystyle \text{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\times2+2\times(-1)&4\times2+2\times3\\1\times2+7\times(-1)&1\times2+7\times3\end{bmatrix}=\begin{bmatrix}6&14\\-5&23\end{bmatrix}
\displaystyle \text{(ii) }A^2=\begin{bmatrix}3&2\\0&5\end{bmatrix}\begin{bmatrix}3&2\\0&5\end{bmatrix}=\begin{bmatrix}3\times3+2\times0&3\times2+2\times5\\0\times3+5\times0&0\times2+5\times5\end{bmatrix}=\begin{bmatrix}9&16\\0&25\end{bmatrix}
\displaystyle B^2=\begin{bmatrix}1&0\\1&2\end{bmatrix}\begin{bmatrix}1&0\\1&2\end{bmatrix}=\begin{bmatrix}1\times1+0\times1&1\times0+0\times2\\1\times1+2\times1&1\times0+2\times2\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}
\displaystyle \therefore (A+B)(A-B)\neq A^2-B^2.
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\displaystyle \textbf{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 \text{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 these equations, }x=\frac{2}{3}\text{ and }y=\frac{1}{2}.
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\displaystyle \textbf{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{Answer:}
\displaystyle \text{Let the order of matrix }A\text{ be }m\times n.
\displaystyle \therefore BA=C\Rightarrow B_{2\times2}\cdot A_{m\times n}=C_{2\times2}
\displaystyle \Rightarrow \text{The order of matrix }A=2\times2.
\displaystyle \text{Let }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\qquad\ldots\text{(i)}\qquad-2a=-4\qquad\ldots\text{(ii)}
\displaystyle \Rightarrow b+3d=7\qquad\ldots\text{(iii)}\qquad-2b=-8\qquad\ldots\text{(iv)}
\displaystyle \text{From (i) and (ii), }a=2\text{ and }c=5.
\displaystyle \text{From (iii) and (iv), }b=4\text{ 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 \textbf{Question 16: Find the matrix }M\text{ such that }M\begin{bmatrix}3&6\\-2&-8\end{bmatrix}=\begin{bmatrix}-2&16\end{bmatrix}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the order of matrix }M\text{ be }a\times b.
\displaystyle \Rightarrow M_{a\times b}\begin{bmatrix}3&6\\-2&-8\end{bmatrix}=\begin{bmatrix}-2&16\end{bmatrix}
\displaystyle \text{Since matrix multiplication is possible, the number of columns of }M\text{ equals the number of rows of the second matrix.}
\displaystyle \therefore b=2.
\displaystyle \text{Also, the number of rows of the product equals the number of rows of }M.
\displaystyle \therefore a=1\Rightarrow \text{The order of matrix }M=1\times2.
\displaystyle \text{Let }M=\begin{bmatrix}x&y\end{bmatrix}.
\displaystyle \therefore \begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}3&6\\-2&-8\end{bmatrix}=\begin{bmatrix}-2&16\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}3x-2y&6x-8y\end{bmatrix}=\begin{bmatrix}-2&16\end{bmatrix}
\displaystyle \Rightarrow 3x-2y=-2\text{ and }6x-8y=16
\displaystyle \text{Solving these equations, }x=-4\text{ and }y=-5.
\displaystyle \therefore M=\begin{bmatrix}-4&-5\end{bmatrix}.
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\displaystyle \textbf{Question 17: }\begin{bmatrix}8&4\\1&-8\end{bmatrix}\cdot X=\begin{bmatrix}12\\10\end{bmatrix}.
\displaystyle \text{Write down: (i) the order of the matrix }X\qquad\text{(ii) the matrix }X.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let the order of matrix }X\text{ be }a\times b.
\displaystyle \therefore \begin{bmatrix}8&4\\1&-8\end{bmatrix}_{2\times2}\cdot X_{a\times b}=\begin{bmatrix}12\\10\end{bmatrix}_{2\times1}
\displaystyle \therefore \text{The order of matrix }X=a\times b=2\times1.
\displaystyle \text{(ii) Let }X=\begin{bmatrix}x\\y\end{bmatrix}.
\displaystyle \therefore \begin{bmatrix}8&4\\1&-8\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}12\\10\end{bmatrix}
\displaystyle \Rightarrow \begin{bmatrix}8x+4y\\x-8y\end{bmatrix}=\begin{bmatrix}12\\10\end{bmatrix}
\displaystyle \Rightarrow 8x+4y=12\text{ and }x-8y=10
\displaystyle \text{On solving, we get }x=2\text{ and }y=-1.
\displaystyle \therefore X=\begin{bmatrix}2\\-1\end{bmatrix}.
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\displaystyle \textbf{Question 18: State with reason whether the following are TRUE or FALSE. }A,\ B,\ C\text{ are matrices of order }2\times2.
\displaystyle \text{(i) }AB=BA\qquad\text{(ii) }A(BC)=(AB)C\qquad\text{(iii) }(A+B)^2=A^2+2AB+B^2\qquad\text{(iv) }A(B+C)=AB+AC
\displaystyle \text{Answer:}
\displaystyle \text{(i) FALSE, since matrix multiplication is not commutative.}
\displaystyle \text{(ii) TRUE, since matrix multiplication is associative.}
\displaystyle \text{(iii) FALSE, since matrix multiplication is not commutative. In general, }(A+B)^2=A^2+AB+BA+B^2\neq A^2+2AB+B^2.
\displaystyle \text{(iv) TRUE, since matrix multiplication is distributive over matrix addition.}
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