Tag: applications of matrices and determinants

Questions Related to applications of matrices and determinants

Use matrix to solve the following system of equations
$x+ y +z = 3$

$x +2y+ 3z= 4$
$2x+3y +4z= 7$

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $x = 2 + k, :y = -1 - 2k, :z = -k $ where $k \in R$

  2. $x = 2 + k, :y = 1 - 2k, :z = k $ where $k \in R$

  3. $x = -2 - k, :y = 1 - 2k, :z = -k $ where $k \in R$

  4. $x = -2 + k, :y = -1 + 2k, :z = -k $ where $k \in R$

Show answer
Correct Option: B
Explanation:

Given system of equations can be written as
$AX=B$
where $A=\begin{bmatrix} { 1 } & { 1 } & { 1 } \ { 1 } & { 2 } & 3\ { 2 } & 3 & 4 \end{bmatrix}$ 
$X=\begin{bmatrix} x \ y \ z \end{bmatrix}$ ;$B=\begin{bmatrix} 3 \ 4 \ 7 \end{bmatrix}$

Here, $|A|=0$
Now, we will find $(adj A)B$

$adj A=C^{T}={\begin{bmatrix} { -1 } & { 2 } & { -1 } \ { -1 } & { 2 } & -1 \ { 1 } & -2 & 1 \end{bmatrix}}^T$

$\Rightarrow adj A=\begin{bmatrix} { -1 } & { -1 } & { 1 } \ { 2 } & { 2 } & -2 \ { -1 } & -1 & 1 \end{bmatrix}$

Now, $(adj A)B=\begin{bmatrix} { -1 } & { -1 } & { 1 } \ { 2 } & { 2 } & -2 \ { -1 } & -1 & 1 \end{bmatrix}\begin{bmatrix} 3 \ 4 \ 7 \end{bmatrix}$

$\Rightarrow (adj A)B=\begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}$
$\Rightarrow (adj A)B=O$

Hence,the system of equations has infinitely many solutions.
Let $z=k$ where $k\in R$
Then 
$x+y=3-k$
$x+2y=4-3k$
Solving these eqns, we get 
$y=1-2k ; x=2+k$

Investigate for what values of $\lambda, \mu$ the simultaneous equation $x+y+z=6; x+2y+3z=10$ & $x+2y+\lambda z=\mu$ have an infinite number of solutions

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $\lambda=4, \mu=11$

  2. $\lambda=3, \mu=10$

  3. $\lambda=2, \mu=8$

  4. $\lambda=1, \mu=11$

Show answer
Correct Option: B
Explanation:

$\Delta =\begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 2 & \lambda  \end{vmatrix}=1\left( 2\lambda -6 \right) -1\left( \lambda -3 \right) +1\left( 2-2 \right) =2\lambda -6-\lambda +3+0=\lambda -3\ \Delta _{ 1 }=\begin{vmatrix} 6 & 1 & 1 \ 10 & 2 & 3 \ \mu  & 2 & \lambda  \end{vmatrix}=6\left( 2\lambda -6 \right) -1\left( 10\lambda -3\mu  \right) +1\left( 20-2\mu  \right) \ =12\lambda -36-10\lambda +3\mu +20-2\mu =2\lambda +\mu -16$

For infinite solution $\Delta =0,{ \Delta  } _{ 1 }\Rightarrow \lambda =3\Rightarrow \mu =10$

The equations $x+4y-2z=3$, $3x+y+5z=7$ and $2x+3y+z=5$ have

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. a unique solution

  2. no solution

  3. two solutions

  4. infinite solutions

Show answer
Correct Option: B
Explanation:

$\Delta=\begin{vmatrix} 1 &4  &-2  \3  &1  &5  \  2&3  &1  \end{vmatrix}=1(1-15)-4(3-10)-2(9-2)=0$
and $\Delta _1$ is not zero
Therefore, it has no solution

Let $a,\ b,\ c$ be any real numbers. Suppose that there are real numbers $x, y, z$ not all zero such that $x=cy+bz,\ y=az+cx$ and $z=bx+ay$. Then $a^{2}+b^{2}+c^{2}+2abc$ is equal to 

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. 0

  2. 1

  3. 2

  4. $-1$

Show answer
Correct Option: B
Explanation:

For $x,y,z$ not to be simultaneously zero

determinant of the coefficients should be zero.
$\left| \begin{matrix}-1 & c & b\ c& -1 & a \ b & a & -1 \end{matrix}\right|=0$
$\Rightarrow a^2+b^2+c^2+2abc = 1$

One of the roots of $\begin{vmatrix} x+a & b & c\  a & x+b & c\  a & b & x+c \end{vmatrix}=0$ is :

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $abc$

  2. $a+b+c$

  3. $-(a+b+c)$

  4. $-abc$

Show answer
Correct Option: C
Explanation:
For the determinant
$ column\ 1=column\ 1+column\ 2+column \ 3\\ \left| \begin{matrix} x+a+b+c & b & c \\ x+a+b+c & \quad x+b & c \\ x+a+b+c & b & x+c \end{matrix} \right| =0\\ \\ row2=row2-row1\\ row3=row3-row1\\ \begin{vmatrix} x+a+b+c & \quad b & \quad c \\ 0 & \quad x & \quad 0 \\ 0 & \quad 0 & \quad x \end{vmatrix}=0$
Expanding the determinant 
$\Rightarrow  (x+a+b+c)(x.x)-b.0+c.0=0$ 
$\Rightarrow { x }^{ 2 }(x+a+b+c)=0$ 
Then, the roots are $x=0$ or $x=-(a+b+c)$

For the system of linear equations 2x + 3y + 5z = 9, 7x + 3y - 2z = 8 and 2x + 3y +$\lambda$z $=\mu$.Under what condition does the above system of equations have infinitely many solutions.

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $\lambda = 5$ and $\mu \neq 9$

  2. $\lambda = 5$ and $\mu = 9$

  3. $\lambda = 9$ and $\mu \neq 5$

  4. $\lambda = 9$ and $ \mu = 5$

Show answer
Correct Option: B
Explanation:

$2x+3y+5z=9$,

$ 7x+3y-2z=8$,
$ 2x+3y+λz=μ$ 
Now, for infinitely many solutions. If $2$ equations out of $3$ are same then we are left with only $2$ equations with $3$ variables, this will give infinite solutions. Now if $\lambda =5$, $\mu =9$, then equation (i) and (iii) will be same and the system will have infinite solutions. 
Hence, B is correct.

The system $2x+3y+z=5, 3x+y+5z=7, x+4y-2z=3$ has:

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. Unique Solution

  2. Finite number of solutions

  3. Infinite Solutions

  4. No solution

Show answer
Correct Option: D
Explanation:

${2}x+{3}y+z={5}$

${3}x+y+{5}z={7}$
$x+{4}y-{2}z={3}$
$A=\left[ \begin{matrix} 2 & 3 & 1 \ 3 & 1 & 5 \ 1 & 4 & -2 \end{matrix} \right] \quad \quad D=\left[ \begin{matrix} 5 \ 7 \ 3 \end{matrix} \right] \quad \quad AD=\left[ \begin{matrix}2&3&1&5\3&1&5&7\1&4&-2&3\end{matrix} \right] $

Rank of  $AD=\left[ \begin{matrix}2&3&1&5\3&1&5&7\1&4&-2&3\end{matrix} \right] $

$R _{2}\rightarrow{3}R _{1}-{2}R _{2}$
$R _{3}\rightarrow{2}R _{3}-R _{1}$

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

$R _{3}\rightarrow{7}R _{3}-{5}R _{2}$

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

Rank of $A=\left[ \begin{matrix} 2 & 3 & 1 \ 3 & 1 & 5 \ 1 & 4 & -2 \end{matrix} \right] $

$R _{2}\rightarrow{3}R _{1}-{2}R _{2}$
$R _{3}\rightarrow{2}R _{3}-R _{1}$


$A=\left[ \begin{matrix} 2 & 3 & 1 \ 0 & 7 & -7 \ 0 & 5 & -5 \end{matrix} \right] $

Clearly Rank of A $\neq$ Rank of AD.
Hence, no solution.


If AX = B where A is $3 \times 3$ and X and B are $3\times 1$ matrices then which of the following is correct?

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. If | A | = 0 then AX = B has infinite solutions

  2. If AX = B has infinite solutions then | A | = 0

  3. If (adj (A)) B = 0 and | A | $\neq$ 0 then AX = B has unique solution

  4. If (adj (A)) B $\neq$ 0 & |A| = 0 then AX = B has no solution

Show answer
Correct Option: A

The system of equations , $ ax+y+z = a-1 $ , $x+ay+z = a-1 $, $x+y+az = a-1 $has no solution, if a is 

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. either $-2\ or\ 1$

  2. $-2$

  3. $1$

  4. $not\ -2$

Show answer
Correct Option: A
Explanation:
Determinant of Coefficient Matrix , $\begin{bmatrix} a & 1 & 1 \\ 1 & a & 1 \\ 1 & 1 & a \end{bmatrix}=0$
$a(a^2-1)-1(a-1)+1(1-a)=0$
$a^3-a-a+1+1-a=0$
$a^3-3a+2=0\Rightarrow $ If $a=1$
$a=1|\begin{matrix} 1 & 0 & -3 & 2 \\ 0 & 1 & 1 & -2 \end{matrix}$
           |$\ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ \ _ $
            $\begin{matrix} 1 & 1 & -2 & 0 \end{matrix}$
$a^2+a-2=0\Rightarrow a^2+2a-a-2=0$
$a(a+2)-1(a+2)=0$
$a=-2,1$
$a$ is either $-2$ (or) $1$ 

The three distinct straight lines $ax+by+c=0$;$bx+cy+a=0$ and $cx+ay+b=0$ are concurrent then

business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

  1. $a+b+c=0$

  2. $a^{3}+b^{3}+c^{3}=3 abc$

  3. $a=b=c$

  4. $a^{2}+b^{2}+c^{2}=ab+bc+ca$

Show answer
Correct Option: A,B
Explanation:

The lines are concurrent,
$=>\begin{bmatrix}a&b& c\ b &c &a\ c & a& b\end{bmatrix}=0$
Applying,$ C _{1}=C _{1}+C _{2}+C _{3}$,
$=\begin{bmatrix}a+b+c&b& c\ a+b+c &c &a\ a+b+c & a& b\end{bmatrix}$
Applying, $R _{1}=R _{1}-R _{2}$ and $R _{2}=R _{2}-R _{3}$,
$=(a+b+c)\begin{bmatrix}0&b-c& c-a\ 0 &c-a &a-b\ 1 & a& b\end{bmatrix}$
Expanding by $C _{1}$,
$=(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$=> (a+b+c)=0$
or,
$(a+b+c)(-a^{2}-b^{2}-c^{2}+ab+bc+ac)=0$
$abc-a^{3}-b^{3}+abc+abc-c^{3}=0$
$a^{3}+b^{3}+c^{3}=3abc$
So options are A and B.