Chapter 3: Matrices

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. Therefore, here the order is \[3\times 4\].

Ans: Since the order of the given matrix is \[3\times 4\] therefore, the number of elements in it is \[3\times 4=12\].

Ans: The elements are given as \[{{a}_{mn}}\] . Therefore, here \[{{a}_{13}}=19\] , \[{{a}_{21}}=35\] , \[{{a}_{33}}=-5\] , \[{{a}_{24}}=12\] , \[{{a}_{23}}=\dfrac{5}{2}\].

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. To find the possible orders of a matrix, we have to find all the ordered pairs of natural numbers whose product is \[24\] .

\[\therefore \left( 1\times 24 \right),\left( 24\times 1 \right),\left( 2\times 12 \right),\left( 12\times 2 \right),\left( 3\times 8 \right),\left( 8\times 3 \right),\left( 4\times 6 \right),\left( 6\times 4 \right)\] are all the possible ordered pairs here.

If the matrix had \[13\] elements, then the ordered pairs would be \[\left( 1\times 13 \right)\] and \[\left( 13\times 1 \right)\].

Ans: The order of a matrix is \[m\times n\] where \[m\] is the number of rows and \[n\] is the number of columns. To find the possible orders of a matrix, we have to find all the ordered pairs of natural numbers whose product is \[18\] .

\[\therefore \left( 1\times 18 \right),\left( 18\times 1 \right),\left( 2\times 9 \right),\left( 9\times 2 \right),\left( 3\times 6 \right),\left( 6\times 3 \right)\] are all the possible ordered pairs here.

If the matrix had \[5\] elements, then the ordered pairs would be \[\left( 1\times 5 \right)\] and \[\left( 5\times 1 \right)\].

Ans:

(i) ${a_{ij}}\, = \,\frac{{{{\left( {i + j} \right)}^2}}}{2}$

Elements for $2 \times 2$ matrix are: ${a_{11}},{a_{12}},{a_{21}},{a_{22}}$

${a_{11}} = \frac{{{{\left( {1 + 1} \right)}^2}}}{2}\, = \,\frac{{{{\left( 2 \right)}^2}}}{2}\, = \,2$

${a_{12}}\, = \,\frac{{{{\left( {1 + 2} \right)}^2}}}{2}\, = \,\frac{{{{\left( 3 \right)}^2}}}{2}\, = \,\frac{9}{2}$

${a_{21}}\, = \,\frac{{{{\left( {2 + 1} \right)}^2}}}{2}\, = \,\frac{{{{\left( 3 \right)}^2}}}{2}\, = \,\frac{9}{2}$

${a_{22}}\, = \,\frac{{{{\left( {2 + 2} \right)}^2}}}{2}\, = \,\frac{{{{\left( 4 \right)}^2}}}{2}\, = \,8$

So, the required matrix is:  $\begin{pmatrix} 2& {\frac{9}{2}} \\ \frac{9}{2} & 8 \\ \end{pmatrix}$.

(ii) ${a_{ij}} = \frac{i}{j}$

Elements for $2 \times 2$ matrix are: ${a_{11}},{a_{12}},{a_{21}},{a_{22}}$

${a_{11}} = \frac{1}{1}\,\, = \,1$

${a_{12}}\, = \,\frac{1}{2}\,$

${a_{21}}\, = \,\frac{2}{1}\, = \,2$

${a_{22}}\, = \,\frac{2}{2}\, = \,1$

So, the required matrix is: $\begin{pmatrix} 1 & \frac{1}{2}\\ 2& 1\\ \end{pmatrix}$.

(iii) ${a_{ij}} = \frac{{{{\left( {i + 2j} \right)}^2}}}{2}$

Elements for $2 \times 2$ matrix are: ${a_{11}},{a_{12}},{a_{21}},{a_{22}}$

${a_{11}} = \frac{{{{\left( {1 + 2} \right)}^2}}}{2}\, = \,\frac{{{{\left( 3 \right)}^2}}}{2}\, = \,\frac{9}{2}$

${a_{12}}\, = \,\frac{{{{\left( {1 + 4} \right)}^2}}}{2}\, = \,\frac{{{{\left( 5 \right)}^2}}}{2}\, = \,\frac{{25}}{2}$

${a_{21}}\, = \,\frac{{{{\left( {2 + 2} \right)}^2}}}{2}\, = \,\frac{{{{\left( 4 \right)}^2}}}{2}\, = 8$

${a_{22}}\, = \,\frac{{{{\left( {2 + 4} \right)}^2}}}{2}\, = \,\frac{{{{\left( 6 \right)}^2}}}{2}\, = 1\,8$

So, the required matrix is:$\begin{pmatrix} \frac{9}{2}& \frac{25}{2}\\ 8& 18\\ \end{pmatrix}$.

Ans: Given that \[{{a}_{ij}}=\dfrac{1}{2}\left| -3i+j \right|\] ,

\[\therefore {{a}_{11}}=\dfrac{1}{2}\left| -3\times 1+1 \right|=1\]

\[{{a}_{21}}=\dfrac{1}{2}\left| -3\times 2+1 \right|=\dfrac{5}{2}\]

\[{{a}_{31}}=\dfrac{1}{2}\left| -3\times 3+1 \right|=4\]

\[{{a}_{12}}=\dfrac{1}{2}\left| -3\times 1+2 \right|=\dfrac{1}{2}\]

\[{{a}_{22}}=\dfrac{1}{2}\left| -3\times 2+2 \right|=2\]

\[{{a}_{32}}=\dfrac{1}{2}\left| -3\times 3+2 \right|=\dfrac{7}{2}\]

\[{{a}_{13}}=\dfrac{1}{2}\left| -3\times 1+3 \right|=0\]

\[{{a}_{23}}=\dfrac{1}{2}\left| -3\times 2+3 \right|=\dfrac{3}{2}\]

\[{{a}_{33}}=\dfrac{1}{2}\left| -3\times 3+3 \right|=3\]

\[{{a}_{14}}=\dfrac{1}{2}\left| -3\times 1+4 \right|=\dfrac{1}{2}\]

\[{{a}_{24}}=\dfrac{1}{2}\left| -3\times 2+4 \right|=1\]

\[{{a}_{34}}=\dfrac{1}{2}\left| -3\times 3+4 \right|=\dfrac{5}{2}\]

Thus, the required matrix is \[A=\left[ \begin{matrix} 1 & \dfrac{1}{2} & 0 & \dfrac{1}{2}  \\ \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1  \\ 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}  \\ \end{matrix} \right]\].

Ans: A \[3\times 4\] matrix is given by \[A=\left[ \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} & {{a}_{14}}  \\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} & {{a}_{24}}  \\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} & {{a}_{34}}  \\ \end{matrix} \right]\]

Given that \[{{a}_{ij}}=2i-j\] ,

\[\therefore {{a}_{11}}=2\times 1-1=1\]

\[{{a}_{21}}=2\times 2-1=3\]

\[{{a}_{31}}=2\times 3-1=5\]

\[{{a}_{12}}=2\times 1-2=0\]

\[{{a}_{22}}=2\times 2-2=4\]

\[{{a}_{32}}=2\times 3-2=4\]

\[{{a}_{13}}=2\times 1-3=-1\]

\[{{a}_{23}}=2\times 2-3=1\]

\[{{a}_{33}}=2\times 3-3=3\]

\[{{a}_{14}}=2\times 1-4=-2\]

\[{{a}_{24}}=2\times 2-4=0\]

\[{{a}_{34}}=2\times 3-4=2\]

Thus, the required matrix is \[A=\left[ \begin{matrix} 1 & 0 & -1 & -2  \\ 3 & 2 & 1 & 0  \\ 5 & 4 & 3 & 2  \\ \end{matrix} \right]\].

Ans: Given \[\left[ \begin{matrix} 4 & 3  \\  x & 5  \\ \end{matrix} \right]=\left[ \begin{matrix} y & z  \\ 1 & 5  \\ \end{matrix} \right]\] 

Comparing the corresponding elements we get,

\[x=1,y=4,z=3\]

Ans: Given \[\left[ \begin{matrix} x+y & 2  \\ 5+z & xy  \\ \end{matrix} \right]=\left[ \begin{matrix} 6 & 2  \\ 5 & 8  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[x+y=6,xy=8,5+z=5\]

Now, \[\because 5+z=5\]

\[\Rightarrow z=0\]

We know that, \[{{\left( x-y \right)}^{2}}={{\left( x+y \right)}^{2}}-4xy\]

\[\Rightarrow {{\left( x-y \right)}^{2}}=36-32\]

\[\Rightarrow \left( x-y \right)=\pm 2\]

When \[\left( x-y \right)=2\] and \[\left( x+y \right)=6\],

We get \[x=4,y=2\]

When \[\left( x-y \right)=-2\] and \[\left( x+y \right)=6\],

We get \[x=2,y=4\]

\[\therefore x=4,y=2,z=0\] or \[\therefore x=2,y=4,z=0\]

Ans: Given \[\left[ \begin{matrix} x+y+z  \\ x+z  \\ y+z  \\ \end{matrix} \right]=\left[ \begin{matrix} 9  \\ 5  \\ 7  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[x+y+z=9\]                      …(1)

\[x+z=5\]                              …(2)

\[y+z=7\]                              …(3)

From equation (1) and (2),

\[y+5=9\]

\[\Rightarrow y=4\]

From equation (3) we have,

\[4+z=7\]

\[\Rightarrow z=3\]

\[x+z=5\]

\[\Rightarrow x=2\]

\[\therefore x=2,y=4,z=3\]

Ans: Given \[\left[ \begin{matrix} a-b & 2a+c  \\ 2a-b & 3c+d  \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 5  \\ 0 & 13  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[a-b=-1\]                              …(1)

\[2a-b=0\]                              …(2)

\[2a+c=5\]                              …(3)

\[3c+d=13\]                                        …(4)

From equation (2),

\[b=2a\]

From equation (1),

\[a-2a=-1\]

\[\Rightarrow a=1\]

\[\Rightarrow b=2\]

From equation (3),

\[2\times 1+c=5\]

\[\Rightarrow c=3\]

From equation (4),

\[3\times 3+d=13\]

\[\Rightarrow d=4\]

\[\therefore a=1,b=2,c=3,d=4\]

Ans: A given matrix is said to be a square matrix if the number of rows is equal to the number of columns.

\[\therefore A={{\left[ {{a}_{y}} \right]}_{m\times n}}\] is a square matrix if, \[m=n\].

Thus, option (C) is correct.

Ans: Given \[\left[ \begin{matrix} 3x+7 & 5  \\ y+1 & 2-3x  \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & y-2  \\  8 & 4  \\ \end{matrix} \right]\]

Comparing the corresponding elements we get,

\[3x+7=0\]

\[\Rightarrow x=-\dfrac{7}{3}\]

\[y-2=5\]

\[\Rightarrow y=7\]

\[y+1=8\]

\[\Rightarrow y=7\]

\[2-3x=4\]

\[\Rightarrow x=-\dfrac{2}{3}\]

Since we get two different values of \[x\] ,which is not possible. It is not possible to find the values of \[x\] and \[y\] for which the given matrices are equal.

Thus, the correct option is (B).

Ans: Given a matrix of the order \[3\times 3\] has nine elements and each of these elements can be either \[0\] or \[1\] .

Now, each of the nine elements can be filled in two possible ways.

Therefore, the required number of possible matrices is \[{{2}^{9}}=512\].