Solution: The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element $a_{j i}$ shifted to new position $a_{j i}$. The symmetric matrix is defined...
Solution: We have $A=\left(\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right)$. Now addition/subtraction of two matrices is possible if order of both the matrices are same and multiplication...
Solution: We have $A$ and $B$ are symmetric matrices. Therefore $A^{T}=A$ and $B^{T}=B$ The transpose of the matrix is an operation of making interchange of elements by the rule on positioned...
Solution: We have $A=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right)$. The transpose of the matrix is an operation of making interchange of...
Solution: We have $A=\left(\begin{array}{ll}4 & 2 \\ 1 & 3\end{array}\right), B=\left(\begin{array}{cc}-2 & 1 \\ 3 & 2\end{array}\right)$ and $3 A-2 B+X=0$ We can have the addition...
Solution: We have $A=\left(\begin{array}{cc}2 & -3 \\ 4 & 5\end{array}\right), B=\left(\begin{array}{cc}-1 & 2 \\ 0 & 3\end{array}\right)$ and $A+2 B+X=0$. We can have the addition...
Solution: We have $\left(\begin{array}{ll}2 & 3 \\ 4 & 5\end{array}\right)$. The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element...
Solution: We have $\left(\begin{array}{ll}4 & 5 \\ 1 & 8\end{array}\right)$. The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element...
Solution: We have $\left(\begin{array}{cc}x & y \\ 3 y & x\end{array}\right)\left(\begin{array}{l}1 \\ 2\end{array}\right)=\left(\begin{array}{l}3 \\ 5\end{array}\right)$. Use the...
Solution: We have $\left(\begin{array}{cc}2 & -3 \\ 1 & 1\end{array}\right)\left(\begin{array}{l}x \\ y\end{array}\right)=\left(\begin{array}{l}1 \\ 3\end{array}\right)$. Use the...
Solution: We have $A=\left(\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right)$ Use the addition rule and get $A+A^{T}=I_{2}$ as follow:...
Solution: We have $A=\left(\begin{array}{cc}1 & -5 \\ -3 & 2 \\ 4 & -2\end{array}\right) ; B=\left(\begin{array}{cc}3 & 1 \\ 2 & -1 \\ -2 & 3\end{array}\right) .$ and...
Solution: We have $\cos \theta\left(\begin{array}{cc}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta\end{array}\right)+\sin \theta\left(\begin{array}{cc}\sin \theta & -\cos...
Solution: We have $A=\operatorname{diag}(3-25)=\left(\begin{array}{ccc}3 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 5\end{array}\right) ; B=$...
Solution: We have $\left(\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right)+\left(\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right)=3\left(\begin{array}{cc}x & y \\ z &...
Solution: We have $\left(\begin{array}{cc}x & 3 x-y \\ 2 x+z & 3 y-w\end{array}\right)=\left(\begin{array}{ll}3 & 2 \\ 4 & 7\end{array}\right)$. Now from the equality of matrices we...
Solution: We have $x\left(\begin{array}{l}2 \\ 3\end{array}\right)+y\left(\begin{array}{c}-1 \\ 1\end{array}\right)=\left(\begin{array}{c}10 \\ 5\end{array}\right)$. Use the addition rule and get...
Solution: We have $2\left(\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right)+\left(\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right)=\left(\begin{array}{ll}5 & 6 \\ 1 &...
Solution: We have $\left(\begin{array}{cc}x+2 y & -y \\ 3 x & 4\end{array}\right)=\left(\begin{array}{cc}-4 & 3 \\ 6 & 4\end{array}\right)$ Now from the equality of matrices we can...
Solution: We have $a_{i j}=\frac{1}{2}|-3 i+j|^{2}$ Now $\begin{array}{l} a_{11}=\frac{|-3(1)+1|}{2}=1, a_{12}=\frac{|-3(1)+2|}{2}=\frac{9}{2}, a_{13}=\frac{|-3(1)+3|}{2}=\frac{9}{2} \\...
Solution: We have $a_{i j}=\frac{1}{2}(i-2 j)^{2}$ Now $\begin{array}{l} a_{11}=\frac{(1-2(1))^{2}}{2}=\frac{1}{2}, a_{12}=\frac{(1-2(2))^{2}}{2}=\frac{9}{2} \\ a_{21}=\frac{(2-2(1))^{2}}{2}=0...
Solution: We have $A=\left(\begin{array}{ccc}-1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
Solution: We have $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
Solution: We have $A=\left(\begin{array}{ccc}1 & 2 & 3 \\ 2 & 5 & 7 \\ -2 & -4 & -5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary...
Solution: We have $A=\left(\begin{array}{ccc}1 & 3 & -2 \\ -3 & 0 & -1 \\ 2 & 1 & 0\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
Solution: We have $A=\left(\begin{array}{ccc}3 & -1 & -2 \\ 2 & 0 & -1 \\ 3 & -5 & 0\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
Solution: We have $A=\left(\begin{array}{ccc}1 & 2 & -3 \\ 2 & 3 & 2 \\ 3 & -3 & -4\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
Solution: We have $A=\left(\begin{array}{lll}3 & 0 & 2 \\ 1 & 5 & 9 \\ 6 & 4 & 7\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row...
Solution: We have $A=\left(\begin{array}{ccc}2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary...
Solution: We have $A=\left(\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row...
Solution: We have $A=\left(\begin{array}{ll}6 & 7 \\ 8 & 9\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
Solution: We have $A=\left(\begin{array}{ll}4 & 0 \\ 2 & 5\end{array}\right)$. To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
Solution: We have $A=\left(\begin{array}{cc}2 & -3 \\ 4 & 5\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
Solution: We have $A=\left(\begin{array}{cc}2 & 5 \\ -3 & 1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
Solution: We have $A=\left(\begin{array}{cc}1 & 2 \\ 2 & -1\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
Solution: We have $A=\left(\begin{array}{ll}1 & 2 \\ 3 & 7\end{array}\right)$ To get the inverse we will proceed by augmented matrix with elementary row transformation process is as follow:...
Solution: Given that $A=\left[\begin{array}{ll}1 & 23\end{array}\right]$ We will find $A$ ' to calculate AA': $\mathrm{A}^{\prime}=\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$ Now...
Solution: Given that $A=\left[\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \theta & \cos \alpha\end{array}\right]$. We wil find $A$ $A^{\prime}=\left[\begin{array}{cc}\cos \alpha...
Solution: Take $\mathrm{C}=\mathrm{AB}$ $\begin{array}{l} C=\left[\begin{array}{ccc} -1 & 2 & -3 \\ 4 & -5 & 6 \end{array}\right]\left[\begin{array}{cc} 3 & -4 \\ 2 & 1 \\ -1...
Solution: Take $C=A B$ $\begin{array}{l} C=\left[\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right]\left[\begin{array}{lll} -2 & -1 & -4 \end{array}\right] \\...
Solution: Take $C=A B$ $\begin{array}{l} C=\left[\begin{array}{rr} 3 & -1 \\ 2 & -2 \end{array}\right]\left[\begin{array}{ll} 1 & -3 \\ 2 & -1 \end{array}\right] \\...
Solution: Take $\mathrm{C}=\mathrm{A} 8$ $\begin{array}{l} C=\left[\begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right]\left[\begin{array}{ll} 1 & 4 \\ 2 & 5 \end{array}\right] \\...
Solution: Given that $A=\left[\begin{array}{lll}3 & 2 & 5 \\ 4 & 1 & 3 \\ 0 & 6 & 7\end{array}\right]$, to express as sum of symmetric matrix $P$ and skew symmetric matrix...
Solution: Given that $A=\left[\begin{array}{ccc}3 & -1 & 0 \\ 2 & 0 & 3 \\ 1 & -1 & 2\end{array}\right]$, to express as sum of symmetric matrix $P$ and skew symmetric matrix...
Solution: Given that $\mathrm{A}=\left[\begin{array}{ccc}-1 & 5 & 1 \\ 2 & 3 & 4 \\ 7 & 0 & 9\end{array}\right]$, to express as sum of symmetric matrix $\mathrm{P}$ and skew...
Solution: Given that $\mathrm{A}=\left[\begin{array}{rr}3 & -4 \\ 1 & -1\end{array}\right]$,to express as the sum of symmetric matrix $\mathrm{P}$ and skew symmetric matrix Q. $A=P+Q$ Where...
Solution: Given that $A=\left[\begin{array}{cc}2 & 3 \\ -1 & 4\end{array}\right]$, As for a symmetric matrix $A^{\prime}=A$ hence $\begin{array}{l} A+A^{\prime}=2 A \\...
Solution: We have $A=\left(\begin{array}{ccc}0 & a & b \\ -a & 0 & c \\ -b & -c & 0\end{array}\right)$. The transpose of the matrix is an operation of making interchange of...
Solution: We have $A=\left(\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right)$ The transpose of the matrix is an operation of making interchange of elements by the rule on positioned...
Solution: We have $A=\left(\begin{array}{ll}4 & 1 \\ 5 & 8\end{array}\right)$. The transpose of the matrix is an operation of making interchange of elements by the rule on positioned element...
Solution: We have $P=\left(\begin{array}{cc}3 & 4 \\ 2 & -1 \\ 0 & 5\end{array}\right)$ and $Q=\left(\begin{array}{cc}7 & -5 \\ -4 & 0 \\ 2 & 6\end{array}\right)$. The...
Solution: We have $A=\left(\begin{array}{ccc}3 & 2 & -1 \\ -5 & 0 & -6\end{array}\right)$ and $B=\left(\begin{array}{ccc}-4 & -5 & -2 \\ 3 & 1 & 8\end{array}\right)$....
Solution: We have $A=\left(\begin{array}{cc}3 & 5 \\ -2 & 0 \\ 4 & -6\end{array}\right)$. Thus $2 A=\left(\begin{array}{cc}6 & 10 \\ -4 & 0 \\ 8 & -12\end{array}\right)$ The...
Solution: We have $A=\left(\begin{array}{ccc}2 & -3 & 5 \\ 0 & 7 & -4\end{array}\right)$. The transpose of the matrix is an operation of making interchange of elements by the rule on...
Solution: We have $A=\left(\begin{array}{ll}2 & 3 \\ 5 & 7\end{array}\right), B=\left(\begin{array}{cc}1 & -3 \\ -2 & 4\end{array}\right)$ and $C=\left(\begin{array}{cc}-4 & 6 \\...
Solution: We have $A=\left(\begin{array}{cc}1 & 0 \\ -1 & 7\end{array}\right), B=\left(\begin{array}{cc}0 & 4 \\ -1 & 7\end{array}\right)$. (i) Let's compute first $A^{2}$ $A^{2}=A...
Solution: We have $\boldsymbol{A B}=\boldsymbol{A} \boldsymbol{C}$ but $\boldsymbol{B} \neq \boldsymbol{C}$. WE need to find: $\boldsymbol{A}, \boldsymbol{B}$. Let's take...
Solution: We have $\boldsymbol{A} \neq \boldsymbol{O}, \boldsymbol{B} \neq \boldsymbol{O}, \boldsymbol{A B}=\boldsymbol{O}$ and $\boldsymbol{B A} \neq \boldsymbol{O}$. We need to find:...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{1} & \mathbf{1} \\ \mathbf{0} & \mathbf{1}\end{array}\right)$. We need to show: $A^{n}=\left(\begin{array}{ll}1 & n \\ 0...
Solution: We have $A=\left(\begin{array}{cc}3 & 4 \\ -4 & -3\end{array}\right)$ and equation $f(x)=x^{2}-5 x+7$. (i) Let us compute first $A^{2}$ $A^{2}=A A=\left(\begin{array}{cc} 3 & 4...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\boldsymbol{a} & \boldsymbol{b} \\ -\boldsymbol{a} & \boldsymbol{2} \boldsymbol{b}\end{array}\right),...
Solution: We have $A=\left(\begin{array}{ccc}2 & 1 & 2 \\ 1 & 0 & 2 \\ 0 & 2 & -4\end{array}\right), B=\left(\begin{array}{lll}x & 4 & 1\end{array}\right)$ and...
Solution: We have $A=\left(\begin{array}{lll}1 & 2 & 3 \\ 4 & 5 & 6 \\ 3 & 2 & 5\end{array}\right), B=\left(\begin{array}{lll}1 & x & 1\end{array}\right)$ and...
Solution: We have $A=\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha\end{array}\right)$ and to show $A^{2}=$ $\left(\begin{array}{cc}\cos ^2 \alpha & \sin...
Solution: We have $\boldsymbol{F}(\boldsymbol{X})=\left(\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right)$ and to show...
Solution: We have $A=\left(\begin{array}{ll}1 & -1 \\ 2 & -1\end{array}\right), B=\left(\begin{array}{ll}a & -1 \\ b & -1\end{array}\right)$ and $(A+B)^{2}=\left(A^{2}+\right.$...
Solution: We have $\boldsymbol{B}=\left(\begin{array}{ll}\mathbf{2} & \mathbf{3} \\ \mathbf{4} & \mathbf{5}\end{array}\right)$ and $\boldsymbol{C}=\left(\begin{array}{cc}\mathbf{0} &...
Solution: We have $\boldsymbol{B}=\left(\begin{array}{cc}\mathbf{5} & -\mathbf{7} \\ -\mathbf{2} & \mathbf{3}\end{array}\right)$ and $\boldsymbol{C}=\left(\begin{array}{cc}-\mathbf{1 6}...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{3} & \mathbf{2} \\ \mathbf{1} & \mathbf{1}\end{array}\right)$. To find $\boldsymbol{a}, \boldsymbol{b}$ such that...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{3} & \mathbf{1} \\ \mathbf{7} & \mathbf{5}\end{array}\right)$. To find $\boldsymbol{x}, \boldsymbol{y}$ such that...
Solution: We have $A=\left(\begin{array}{cc}3 & -4 \\ 1 & 2\end{array}\right), B=\left(\begin{array}{c}3 \\ 11\end{array}\right)$ and $X=\left(\begin{array}{l}x \\ y\end{array}\right)$. We...
Solution: We have $A=\left(\begin{array}{cc}2 & -3 \\ 1 & 1\end{array}\right), B=\left(\begin{array}{l}1 \\ 3\end{array}\right)$ and $X=\left(\begin{array}{l}x \\ y\end{array}\right)$. To...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{1} & \mathbf{2} \\ \mathbf{4} & \mathbf{- 3}\end{array}\right)$. Now addition/subtraction of two matrices is possible if...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}-1 & 2 \\ 3 & 1\end{array}\right)$. Now addition/subtraction of two matrices is possible if order of both the matrices are same and...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{3} & -\mathbf{2} \\ \mathbf{4} & -\mathbf{2}\end{array}\right)$. Now addition/subtraction of two matrices is possible if...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{2} & \mathbf{3} \\ \mathbf{1} & \mathbf{2}\end{array}\right)$. Now addition of two matrices is possible if order of both the...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{3} & \mathbf{1} \\ -\mathbf{1} & \mathbf{2}\end{array}\right)$. Now addition of two matrices is possible if order of both the...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{2} & \mathbf{- 2} \\ -\mathbf{3} & \mathbf{4}\end{array}\right)$. Now addition of two matrices is possible if order of both...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{2} & \mathbf{- 1} \\ \mathbf{3} & \mathbf{2}\end{array}\right)$ and $\boldsymbol{B}=\left(\begin{array}{cc}\mathbf{0} &...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{4} & -\mathbf{1} & -\mathbf{4} \\ \mathbf{3} & \mathbf{0} & -\mathbf{4} \\ \mathbf{3} & -\mathbf{1} &...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{2} & \mathbf{- 2} & -\mathbf{4} \\ \mathbf{- 1} & \mathbf{3} & \mathbf{4} \\ \mathbf{1} & -\mathbf{2} &...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\boldsymbol{a} \boldsymbol{b} & \boldsymbol{b}^{2} \\ -\boldsymbol{a}^{2} & -\boldsymbol{a} \boldsymbol{b}\end{array}\right) .$ To...
Solution: We have $A=\left(\begin{array}{ccc}1 & 0 & 2 \\ 3 & -1 & 0 \\ -2 & 1 & 1\end{array}\right), B=\left(\begin{array}{ccc}0 & 5 & -4 \\ -2 & 1 & 3 \\ 1...
Solution: We have $A=\left(\begin{array}{cc}2 & 3 \\ -1 & 4 \\ 0 & 1\end{array}\right), B=\left(\begin{array}{cc}5 & -3 \\ 2 & 1\end{array}\right)$. and...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ll}\mathbf{1} & \mathbf{2} \\ \mathbf{3} & \mathbf{4}\end{array}\right), \boldsymbol{B}=\left(\begin{array}{cc}\mathbf{2} &...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{2} & \mathbf{3} & \mathbf{- 1} \\ \mathbf{3} & \mathbf{0} & \mathbf{2}\end{array}\right),...
Solution: We have $A=\left(\begin{array}{lll}1 & 2 & 5 \\ 0 & 1 & 3\end{array}\right), B=\left(\begin{array}{ccc}2 & 3 & 0 \\ 1 & 0 & 4 \\ 1 & -1 &...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{0} & \boldsymbol{c} & -\boldsymbol{b} \\ -\boldsymbol{c} & \mathbf{0} & \boldsymbol{a} \\ \boldsymbol{b} &...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{2} & -\mathbf{3} & -\mathbf{5} \\ -\mathbf{1} & \mathbf{4} & \mathbf{5} \\ \mathbf{1} & -\mathbf{3} &...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{1} & \mathbf{3} & -\mathbf{1} \\ \mathbf{2} & \mathbf{2} & -\mathbf{1} \\ \mathbf{3} & \mathbf{0} &...
Solution: We have $A=\left(\begin{array}{lll}1 & 2 & 1 \\ 3 & 4 & 2 \\ 1 & 3 & 2\end{array}\right)$ and $B=\left(\begin{array}{ccc}10 & -4 & -1 \\ -11 & 5 & 0...
Solution: We have $A=\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta\end{array}\right)$ and $B=\left(\begin{array}{cc}\cos \phi & -\sin \phi \\ \sin \phi...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{lll}\mathbf{1} & \mathbf{2} & \mathbf{3} \\ \mathbf{0} & \mathbf{1} & \mathbf{0} \\ \mathbf{1} & \mathbf{1} &...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{5} & -\mathbf{1} \\ \mathbf{6} & \mathbf{7}\end{array}\right)$ and $\boldsymbol{B}=\left(\begin{array}{ll}\mathbf{2} &...
Solution: We have $\boldsymbol{B}=\left(\begin{array}{ccc}\mathbf{1} & \mathbf{0} & \mathbf{1} \\ -\mathbf{1} & \mathbf{2} & \mathbf{1}\end{array}\right)$ and...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{llll}\mathbf{1} & 2 & 3 & 4\end{array}\right)$ and $\boldsymbol{B}=\left(\begin{array}{l}\mathbf{1} \\ \mathbf{2} \\ \mathbf{3} \\...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{ccc}\mathbf{0} & \mathbf{1} & -\mathbf{5} \\ \mathbf{2} & \mathbf{4} & \mathbf{0}\end{array}\right)$ and...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}-\mathbf{1} & \mathbf{1} \\ -\mathbf{2} & \mathbf{2} \\ -\mathbf{3} & \mathbf{3}\end{array}\right)$ and...
Solution: We have $\boldsymbol{A}=\left(\begin{array}{cc}\mathbf{2} & \mathbf{- 1} \\ \mathbf{3} & \mathbf{0} \\ \mathbf{- 1} & \mathbf{4}\end{array}\right)$ and...
Solution: $\text { If }\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\left[\begin{array}{ll} e & f \\ g & h \end{array}\right]$ Therefore $a=e, b=f, c=g, d=h$ It is given...
Solution: It is given that $\begin{array}{l} 2\left[\begin{array}{ll} 1 & 3 \\ 0 & x \end{array}\right]+\left[\begin{array}{ll} y & 0 \\ 1 & 2...
Solution: (i) Given $2\left(\begin{array}{lc} x & 5 \\ 7 y-3 \end{array}\right)+\left(\begin{array}{c} 3-4 \\ 12 \end{array}\right)=\left(\begin{array}{cc} 7 & 6 \\ 1514 \end{array}\right)$...
Solution: (i) Given $\left(\begin{array}{l}\boldsymbol{x}+\boldsymbol{y} \\ \boldsymbol{x}-\boldsymbol{y}\end{array}\right)=\left(\begin{array}{l}8 \\ \mathbf{4}\end{array}\right)$. By equality of...
Solution: If $Z=\operatorname{diag}[a, b, c]$, then it can be written as $Z=\left[\begin{array}{lll} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array}\right]$ Therefore, $A+2...
Solution: It is given that $A+B+C$ is zero matrix i.e $A+B+C=0$ $\begin{array}{l} {\left[\begin{array}{ccc} 1 & -3 & 2 \\ 2 & 0 & 2 \end{array}\right]+\left[\begin{array}{ccc} 2...
Solution: It is given that $2 A-B+X=0$ $\begin{array}{l} 2\left(\left[\begin{array}{ll} 3 & 1 \\ 0 & 2 \end{array}\right]\right)-\left[\begin{array}{cc} -2 & 1 \\ 0 & 3...
Solution: It is given that $A+B-C=0$ $\begin{array}{c} {\left[\begin{array}{cc} -2 & 3 \\ 4 & 5 \\ 1 & -6 \end{array}\right]+\left[\begin{array}{cc} 5 & 2 \\ -7 & 3 \\ 6 & 4...
Solution: It is given that $\left[\begin{array}{ccc}3 & 5 & -9 \\ -1 & 4 & -7\end{array}\right]+x=\left[\begin{array}{lll}6 & 2 & 3 \\ 4 & 8 & 6\end{array}\right]$...
Solution: $\operatorname{Add} 2(2 A-B)$ and $(2 B+A)$ $\begin{array}{l} 2(2 A-B)+(2 B+A)=2\left(\left[\begin{array}{ccc} 6 & -6 & 0 \\ -4 & 2 & 1...
Solution: Add $(A+B)$ and $(A-B)$ We obtain $(A+B)+(A-B)=\left[\begin{array}{ccc}1 & 0 & 2 \\ 5 & 4 & -6 \\ 7 & 3 & 8\end{array}\right]+\left[\begin{array}{ccc}-5 & -4...
Solution: $5 A=\left[\begin{array}{ccc}5 & 10 & -15 \\ 2 & 3 & 4 \\ 1 & 0 & -5\end{array}\right]$ $A=\left[\begin{array}{ccc}\frac{5}{5} & \frac{10}{5} &...
Solution: $\begin{array}{l} \left.5 A-3 B+4 C=5\left(\left[\begin{array}{ccc} 0 & 1 & -2 \\ 5 & -1 & -4 \end{array}\right]\right)-3\left(\begin{array}{ccc} 1 & -3 & -1 \\ 0...
Solution: $\begin{array}{l} \text { i. } A-2 B+3 C \\ \text { A- } 2 B+3 C=\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]-2\left(\left[\begin{array}{cc} 1 & 3 \\ -2 & 5...
Solution: i. $\begin{array}{l} A+2 B=\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]+2\left(\left[\begin{array}{cc} 1 & 3 \\ -2 & 5 \end{array}\right]\right) \\...
Solution: $\begin{array}{l} 2 A=2\left(\left[\begin{array}{ccc} 3 & 1 & 2 \\ 1 & 2 & -3 \end{array}\right]\right) \\ =\left[\begin{array}{ccc} 6 & 2 & 4 \\ 2 & 4 & -6...
Solution: $\begin{array}{l} (A+B)+C=\left(\left[\begin{array}{cc} 3 & 5 \\ -2 & 0 \\ 6 & -1 \end{array}\right]+\left[\begin{array}{cc} -1 & -3 \\ 4 & 2 \\ -2 & 3...
Solution: $\begin{array}{l} A+B=\left[\begin{array}{ccc} 2 & -3 & 5 \\ -1 & 0 & 3 \end{array}\right]+\left[\begin{array}{ccc} 3 & 2 & -2 \\ 4 & -3 & 1...
Solution: It is a (3 $x 4)$ matrix. Therefore, it has 3 rows and 4 columns. Given that $a_{i j}=\frac{|-a \|+| l}{2}$ Therefore, $a_{11}=1 . a_{12}=\frac{1}{2}, a_{13}=0, a_{13}=\frac{1}{2}$...
Solution: It is a $(2 \times 3)$ matrix. Therefore, it has 2 rows and 3 columns. Given that $a_{i j}=\frac{\left(t-2 \rho^{2}\right.}{2}$ Therefore, $a_{11}=\frac{1}{2}, a_{12}=\frac{9}{2},...
Solution: It is a $(2 \times 2)$ matrix. So, it has 2 rows and 2 columns. Given that $a_{i j}=\frac{(i+2 j)^{2}}{2}$ Therefore, $a_{11}=\frac{9}{2}, a_{12}=\frac{25}{2}$. $a_{21}=8 . a_{22}=18$...
Solution: It is $(4 \times 3)$ matrix. Therefore it has 4 rows and 3 columns Given that$a_{i j}=\frac{i}{j}$ Therefore, $a_{11}=1 . a_{12}=\frac{1}{2}, a_{13}=\frac{1}{3}$ $\begin{array}{l}...
Solution: It is given that: $a_{i j}=(2 \mid-j)$ Now, $a_{11}=(2 \times 1-1)=2-1=1$ $\begin{array}{l} a_{12}=2 \times 1-2=2-2=0 \\ a_{21}=2 \times 2-1=4-1=3 \\ a_{22}=2 \times 2-2=4-2=2 \\ a_{31}=2...
Solution: No. of entries $=($ No. of rows) $x$ (No. of columns) $=7$ If order is $(\mathrm{a} \times \mathrm{b})$ then, No. of entries = $\mathrm{a} \times \mathrm{b}$ Therefore now a $x b=7$ (in...
Solution: No. of entries $=$ (No. of rows) $x$ (No. of columns) $=18$ If order is $(a \times b)$ then, No. of entries = $a \times b$ Therefore now $a \times b=18$ (in this case) Possible cases are...
Solution: i. $E=\left[\begin{array}{c}-2 \\ 3 \\ 0\end{array}\right]$ Order of matrix $=$ Number of rows $x$ Number of columns $\begin{array}{l} =(3 \times 1) \\ \text { ii, } F=[6] \end{array}$...
Solution: i. $C=[7-\sqrt{2} \quad 5 \quad 0]$ Order of matrix $=$ Number of rows $x$ Number of columns $=(1 \times 4)$ ii. $D=[8-3]$ Order of matrix $=$ Number of rows $x$ Number of columns $=(1...
Solution: i. $A=\left[\begin{array}{cccc}3 & 5 & 4 & -2 \\ 0 & \sqrt{3} & -1 & \frac{4}{9}\end{array}\right]$ Order of matrix $=$ Number of rows $x$ Number of columns $=(2...
Solution: (i) $a_{i j}=$ element of $t^{\text {th }}$ row and $j^{\text {th }}$ column $\begin{array}{l} a_{23}=8 \\ a_{31}=\sqrt{2} \\ a_{14}=1 \\ a_{33}=4 \\ a_{22}=0 \end{array}$
Solution: (i) Order of matrix $=$ Number of rows $x$ Number of columns $=(3 \times 4)$ (ii) Number of entries = (Number of rows) $x$ (Number of columns) $=3 \times 4$ $=12$
Solution: (i) Number of rows $=3$ (ii) Number of columns = 4