Question

Q 7. The sum of three numbers is -1. If we multiply the second number by 2, third number by 3 and add them, we get 5. If we subtract the third number from the sum of first and second numbers, we get -1. Answer the following: Represent the above information algebraically. Convert the system of linear equations in matrix form. If the matrix formed by coefficients of variables is written by A, then find |A|. Find adj (A); where adjoint of A is the transpose of the matrix of the co factors. Find the numbers using matrix method.

Answer 1

Why is Hamlet seen as a procrastinator? And how do his delayed actions contribute to the way the story plays out? 4

Answer 2

$\large\underline{\sf{Solution-}}$

Let us assume that

➢ First number = x

➢ Second number = y

➢ Third number = z

According to first condition

➢ The sum of three numbers is -1.

$\rm :\longmapsto\:x + y + z = - 1$

According to second condition,

➢ If we multiply the second number by 2, third number by 3 and add them, we get 5.

$\rm :\longmapsto\:2y + 3z = 5$

According to third condition,

➢ If we subtract the third number from the sum of first and second numbers, we get -1.

$\rm :\longmapsto\:x + y - z = - 1$

So, we get the three linear equation as

$\bf :\longmapsto\:x + y + z = - 1$

$\bf :\longmapsto\:2y + 3z = 5$

$\bf :\longmapsto\:x + y - z = - 1$

Now, The equations can be represented in matrix form as

$\begin{gathered}\rm :\longmapsto\:\rm A=\left[\begin{array}{ccc}1&1&1\\0&2&3\\1&1& - 1\end{array}\right]\end{gathered}$

$\begin{gathered}\sf \: \rm :\longmapsto\: B=\left[\begin{array}{c} - 1\\5\\ - 1\end{array}\right]\end{gathered}$

$\begin{gathered}\sf \: \rm :\longmapsto\: X=\left[\begin{array}{c}x\\y\\z\end{array}\right]\end{gathered}$

So matrix form is

$\rm :\longmapsto\:AX = B$

Evaluation of determinant of A,

$\begin{gathered}\rm :\longmapsto\:\rm A=\left[\begin{array}{ccc}1&1&1\\0&2&3\\1&1& - 1\end{array}\right]\end{gathered}$

$\rm :\longmapsto\: |A| = 1( - 2 - 3) - 0 + 1(3 - 2)$

$\rm :\longmapsto\: |A| = - 5 + 1$

$\rm :\longmapsto\: |A| = - 4$

$\bf\implies \: |A| = 0$

So, system of equations is consistent having unique solution.

Calculation of adjA

Here we have to first find the co factor matrix of A and then the transpose of it.

$\begin{gathered}\sf \: \rm :\longmapsto\: adj\:A=\left[\begin{array}{ccc}A_{11}&A_{21}&A_{31}\\A_{12}&A_{22}&A_{32}\\A_{13}&A_{23}&A_{33}\end{array}\right]\end{gathered}$

$\begin{gathered}\sf \: \rm :\longmapsto\: adj\:A=\left[\begin{array}{ccc} - 2-3&-(-1 - 1)&3 - 2\\-(0 - 3)& - 1-1&-(3-0)\\0 - 2&-(1 - 1)&2 - 0\end{array}\right]\end{gathered}$

$\begin{gathered}\rm :\longmapsto\: adj\:A=\left[\begin{array}{ccc} - 5&2&1\\3&-2& - 3\\-2&0&2\end{array}\right]\end{gathered}$

Hence,

$\rm :\longmapsto\: {A}^{ - 1} = \dfrac{1}{ |A| } \: adjA$

$\rm :\longmapsto\: {A}^{ - 1} \: = \dfrac{1}{ - 4} \begin{gathered}\rm \: \left[\begin{array}{ccc} - 5&2&1\\3&-2& - 3\\-2&0&2\end{array}\right]\end{gathered}$

Now,

we know that ,

Solution is given by

$\bf :\longmapsto\:X = {A}^{ - 1} B$

Thus, on substituting the values, we get

$\rm :\longmapsto\:X \: = \dfrac{1}{ - 4} \begin{gathered}\rm \: \left[\begin{array}{ccc} - 5&2&1\\3&-2& - 3\\-2&0&2\end{array}\right]\end{gathered}\begin{gathered}\sf \: \left[\begin{array}{c} - 1\\5\\ - 1\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:X \: = \dfrac{1}{ - 4}\begin{gathered}\sf \: \rm \left[\begin{array}{c} 5 + 10 - 1\\ - 3 - 10 + 3\\2 + 0 - 2\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:X \: = \dfrac{1}{ - 4}\begin{gathered}\sf \: \rm \left[\begin{array}{c} 14\\ - 10\\ 0\end{array}\right]\end{gathered}$

$\rm :\longmapsto\:X \: = \begin{gathered}\sf \: \rm \left[\begin{array}{c} \dfrac{ - 7}{2} \\ \\ \dfrac{5}{2} \\ \\ 0\end{array}\right]\end{gathered}$

So,

$\rm :\longmapsto\:x = - \dfrac{7}{2}$

$\rm :\longmapsto\:y = \dfrac{5}{2}$

$\rm :\longmapsto\:z = 0$