Chapter 3: Matrices

Ans: \[A\] and \[B\] are symmetric matrix, therefore, we have:

\[A'=A\] and \[B'=B\]                                                    …(1)

Here, \[\left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)'\]

\[\Rightarrow \left( AB-BA \right)'=B'A'-A'B'\]

From (1),

\[\Rightarrow \left( AB-BA \right)'=BA-AB\]

\[\Rightarrow \left( AB-BA \right)'=-\left( AB-BA \right)\]                           …(2)

From equation (2),

\[AB-BA\] is a skew symmetric matrix.

Ans: Let \[A\] be a symmetric matrix, then \[A'=A\]               …(1)

\[\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'\]

\[\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'\]

\[\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)\]

From (1),

\[\Rightarrow \left( B'AB \right)'=B'\left( AB \right)\]

Thus, if \[A\] is a symmetric matrix, then \[B'AB\] is a symmetric matrix.

Let \[A\] be a skew symmetric matrix, then \[A'=-A\]               …(2)

\[\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'\]

\[\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'\]

\[\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)\]

From (2),

\[\Rightarrow \left( B'AB \right)'=B'\left( -AB \right)\]

\[\Rightarrow \left( B'AB \right)'=-B'AB\]

Thus, if \[A\] is a skew-symmetric matrix, then \[B'AB\] is a skew-symmetric matrix.

Therefore, the matrix \[B'AB\] is symmetric or skew symmetric accordingly as \[A\] is symmetric or skew symmetric.

Ans: Given matrix A=$\begin{bmatrix} 0 & 2y& z \\ x & y & -z \\ x& -y& z \\ \end{bmatrix}$

Transpose of A = ${A}'=\begin{bmatrix} 0 &x &x \\ 2y& y &-y \\ z& -z &z \\ \end{bmatrix}$

Given ${A}'A=I$

= $\begin{bmatrix} 0 & 2y &z \\ x& y&-z \\ x& -y &z \\ \end{bmatrix}\begin{bmatrix} 0 &x &x \\ 2y& y &-y \\ z &-z &z \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 &0 \\ 0& 1& 0 \\ 0& 0 &1 \\ \end{bmatrix}$

${A}'$=$\begin{bmatrix} 0+x^{2}+x^{2} &0+xy-xy &0-xz+xz \\ 0+xy-xy & 4y^{2}+y^{2}+y^{2} &2yz-yz-yz \\ 0-zx+zx& 2yz-yz-yz &z^{2}+z^{2}+z^{2} \\ \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0& 1& 0\\ 0& 0 &1 \\ \end{bmatrix}$

=$\begin{bmatrix} 2x^{2} &0 &0 \\ 0& 6y^{2}&0 \\ 0& 0 & 3z^{2} \\ \end{bmatrix}$ =$\begin{pmatrix} 1 &0 &0 \\ 0& 1& 0\\ 0& 0 &1 \\ \end{pmatrix}$

To find the values of x, y, z we have to equate the entries with corresponding matrix

$2x^{2}=1$

$\Rightarrow x^{2}$=$\dfrac{1}{2}$

$\Rightarrow$ x= $\pm \dfrac{1}{\sqrt{2}}$

$6y^{2}=1$

$\Rightarrow y^{2}$ = $\dfrac{1}{6}$

$\Rightarrow$ y = $\pm \dfrac{1}{\sqrt{6}}$ 

And

$3z^{2} = 1$

$\Rightarrow z^{2} = \dfrac{1}{3}$

$\Rightarrow$ z= $\pm \dfrac{1}{\sqrt{3}}$

Ans: Given \[\left[ \begin{matrix}   1 & 2 & 1  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 0  \\   2 & 0 & 1  \\   1 & 0 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ \begin{matrix}   1+4+1 & 2+0+0 & 0+2+2  \\ \end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ \begin{matrix}   6 & 2 & 4  \\\end{matrix} \right]\left[ \begin{matrix}   0  \\   2  \\   x  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ 6\left( 0 \right)+2\left( 2 \right)+4\left( x \right) \right]=O\]

\[\Rightarrow \left[ 4+4x \right]=\left[ 0 \right]\]

\[\Rightarrow 4+4x=0\]

\[\therefore x=-1\]

Thus, the required value of \[x\] is \[-1\] .

Ans: Given \[A=\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\]

\[\therefore {{A}^{2}}=A\cdot A\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\left[ \begin{matrix}   3 & 1  \\   -1 & 2  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   3\left( 3 \right)+1\left( -1 \right) & 3\left( 1 \right)+1\left( 2 \right)  \\   -1\left( 3 \right)+2\left( -1 \right) & -1\left( 1 \right)+2\left( 2 \right)  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   9-1 & 3+2  \\   -3-2 & -1+4  \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]\]

LHS: \[{{A}^{2}}-5A+7I\]

\[\therefore \left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]-5\left[ \begin{matrix}  3 & 1  \\   -1 & 2  \\ \end{matrix} \right]+7\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   8 & 5  \\   -5 & 3  \\ \end{matrix} \right]-\left[ \begin{matrix}   15 & 5  \\   -5 & 10  \\ \end{matrix} \right]+\left[ \begin{matrix}   7 & 0  \\   0 & 7  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   0 & 0  \\   0 & 0  \\ \end{matrix} \right]\] 

\[\Rightarrow O\]

RHS: \[\Rightarrow O\]

\[LHS=RHS\]

\[\therefore {{A}^{2}}-5A+7I=O\] hence proved.

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

\[\Rightarrow \left[ \begin{matrix}   x-2 & -10 & 2x-8  \\ \end{matrix} \right]\left[ \begin{matrix}   x  \\   4  \\   1  \\ \end{matrix} \right]=O\]

\[\Rightarrow \left[ x\left( x-2 \right)-40+2x-8 \right]=O\]

\[\Rightarrow \left[ {{x}^{2}}-48 \right]=\left[ 0 \right]\]

\[\Rightarrow {{x}^{2}}-48=0\]

\[\therefore x=\pm 4\sqrt{3}\]

Thus, the required value of \[x\] is \[\pm 4\sqrt{3}\] .

Ans: Here the total revenue in market \[I\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   10000 & 2000 & 18000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.50  \\   1.50  \\   1.00  \\ \end{matrix} \right]\]

\[\Rightarrow 1000\times 2.50+2000\times 1.50+18000\times 1.00\]

\[\Rightarrow 46000\]

And, the total revenue in market \[II\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   6000 & 20000 & 8000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.50  \\   1.50  \\   1.00  \\ \end{matrix} \right]\]

\[\Rightarrow 6000\times 2.50+20000\times 1.50+8000\times 1.00\]

\[\Rightarrow 53000\]

Thus, the total revenue in market \[I\] is Rs \[46000\] and the same in market \[II\] is Rs \[53000\].

Ans: Here, the total cost prices of all the products in the market \[I\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   10000 & 2000 & 18000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.00  \\   1.00  \\   0.50  \\ \end{matrix} \right]\]

\[\Rightarrow 1000\times 2.00+2000\times 1.00+18000\times 0.50\]

\[\Rightarrow 31000\]

As the total revenue in market \[I\] is Rs \[46000\] , the gross profit in this market is \[Rs46000-Rs31000=Rs15000\] .

And, the total revenue in market \[II\] can be represented in the form of a matrix as:

\[\left[ \begin{matrix}   6000 & 20000 & 8000  \\ \end{matrix} \right]\left[ \begin{matrix}   2.00  \\   1.00  \\   0.50  \\ \end{matrix} \right]\]

\[\Rightarrow 6000\times 2.00+20000\times 1.00+8000\times 0.50\]

\[\Rightarrow 36000\]

As the total revenue in market \[II\] is Rs \[53000\] , the gross profit in this market is \[Rs53000-Rs36000=Rs17000\] .

Ans: Given \[X\left[ \begin{matrix}   1 & 2 & 3  \\   4 & 5 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9  \\   2 & 4 & 6  \\ \end{matrix} \right]\]

Here, \[X\] has to be a \[2\times 2\] matrix.

Now, let \[X=\left[ \begin{matrix}   a & c  \\   b & d  \\ \end{matrix} \right]\]

Thus, we have:

\[\left[ \begin{matrix}   a & c  \\   b & d  \\ \end{matrix} \right]\left[ \begin{matrix}   1 & 2 & 3  \\   4 & 5 & 6  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9 \\   2 & 4 & 6  \\ \end{matrix} \right]\]

\[\Rightarrow \left[ \begin{matrix}   a+4c & 2a+5c & 3a+6c  \\   b+4d & 2b+5d & 3b+6d  \\ \end{matrix} \right]=\left[ \begin{matrix}   -7 & -8 & -9  \\   2 & 4 & 6  \\ \end{matrix} \right]\]

Comparing the corresponding elements of two matrices, we have:

\[a+4c=-7\]

\[2a+5c=-8\]

\[3a+6c=-9\]

\[b+4d=2\]

\[2b+5d=4\]

\[3b+6d=6\]

Now, solving the above equations we get,

\[\therefore a=1,b=2,c=-2,d=0\]

Thus, the required matrix \[X\] is \[\left[ \begin{matrix}   1 & -2  \\   2 & 0  \\ \end{matrix} \right]\].

Ans: Given \[A=\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha  \\ \end{matrix} \right]\] 

\[\therefore {{A}^{2}}=A\cdot A\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha   \\ \end{matrix} \right]\left[ \begin{matrix}   \alpha  & \beta   \\   \gamma  & -\alpha   \\ \end{matrix} \right]\]

\[\Rightarrow {{A}^{2}}=\left[ \begin{matrix}   {{\alpha }^{2}}+\beta \gamma  & \alpha \beta -\alpha \beta   \\   \alpha \gamma -\alpha \gamma  & \beta \gamma +{{\alpha }^{2}}  \\ \end{matrix} \right]\]

Now, \[{{A}^{2}}=I\]

\[\Rightarrow \left[ \begin{matrix}   {{\alpha }^{2}}+\gamma  & 0  \\   0 & \beta \gamma +{{\alpha }^{2}}  \\ \end{matrix} \right]=\left[ \begin{matrix}   1 & 0  \\   0 & 1  \\ \end{matrix} \right]\]

Equating the corresponding elements, we get:

\[\beta \gamma +{{\alpha }^{2}}=1\]

\[\Rightarrow {{\alpha }^{2}}+\beta \gamma -1=0\]

\[\Rightarrow 1-{{\alpha }^{2}}-\beta \gamma =0\]

Ans: If a matrix \[A\] is both symmetric and skew symmetric, then

\[A'=A\] and \[A'=-A\]

\[\Rightarrow A=-A\]

\[\Rightarrow A+A=O\]

\[\Rightarrow A=O\]

Thus, option (B) is correct.