Chapter 7.4, Problem 106E

a.

To determine

To find: The values of a andb for which the given augmented matrix has exactly one solution.

Answer to Problem 106E

The required values of a and b are $a\in R-\left\{0\right\}$ and $b\in R$ .

Explanation of Solution

Given: The given augmented matrix is $\left[\begin{array}{cc}1& 2\\ 0& a\end{array}|\begin{array}{c}-4\\ b\end{array}\right]$ .

Concept Used: An augmented matrix is a matrix representation of a standard system of simultaneous linear equations. It is a matrix which consists of two parts with a line separating both parts. On the left hand side of the line is the coefficient matrix (generally denoted by A ) of the system of equations and on the right hand side of the line are the constant terms (generally denoted by B ), written in order. Multiplying the coefficient matrix part to the variable column matrix (generally denoted by X) gives the algebraic expressions which are then equated to the constant terms to get the system of equations. Thus, the system of equations can be found using the matrix equation-

  $AX=B$

If a system consists of two linear equations, say-

  $a_{1}x+b_{1}y=c_1$

  $a_{2}x+b_{2}y=c_2$

The associated coefficient matrix and constant matrix are-

  $A=\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$ and $B=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$ respectively.

Then, the above system has-

1. Exactly one solution if $\det (A)\ne 0$ .
2. Infinitely many solutions if $\det (A)=0$ and $(adjA)B=O$ .
3. No solution if $\det (A)=0$ and $(adjA)B\ne O$ .

where $O$ denotes the null matrix.

Calculation: The given augmented matrix is $\left[\begin{array}{cc}1& 2\\ 0& a\end{array}|\begin{array}{c}-4\\ b\end{array}\right]$ .

The associated coefficient matrix and constant matrix are-

  $A=\left[\begin{array}{cc}1& 2\\ 0& a\end{array}\right]$ and $B=\left[\begin{array}{c}-4\\ b\end{array}\right]$ respectively.

The given system will have exactly one solution if-

  $\det (A)\ne 0$

  $\Rightarrow \left|\begin{array}{cc}1& 2\\ 0& a\end{array}\right|\ne 0$

  $\Rightarrow a\times 1-2\times 0\ne 0$

  $\Rightarrow a\ne 0$

Thus the system associated with the given augmented matrix will have exactly one solution for any real value of a except 0 and for any real values of b .

Thus the required values of a and b are-

  $a\in R-\left\{0\right\}$ and $b\in R$ .

b.

To determine

To find: The values of a andb for which the given augmented matrix has infinitely many solutions.

Answer to Problem 106E

The required values of a and b are $a=0$ and $b=0$ .

Explanation of Solution

Given: The given augmented matrix is $\left[\begin{array}{cc}1& 2\\ 0& a\end{array}|\begin{array}{c}-4\\ b\end{array}\right]$ .

Concept Used: An augmented matrix is a matrix representation of a standard system of simultaneous linear equations. It is a matrix which consists of two parts with a line separating both parts. On the left hand side of the line is the coefficient matrix (generally denoted by A ) of the system of equations and on the right hand side of the line are the constant terms (generally denoted by B ), written in order. Multiplying the coefficient matrix part to the variable column matrix (generally denoted by X) gives the algebraic expressions which are then equated to the constant terms to get the system of equations. Thus, the system of equations can be found using the matrix equation-

  $AX=B$

If a system consists of two linear equations, say-

  $a_{1}x+b_{1}y=c_1$

  $a_{2}x+b_{2}y=c_2$

The associated coefficient matrix and constant matrix are-

  $A=\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$ and $B=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$ respectively.

Then, the above system has-

1. Exactly one solution if $\det (A)\ne 0$ .
2. Infinitely many solutions if $\det (A)=0$ and $(adjA)B=O$ .
3. No solution if $\det (A)=0$ and $(adjA)B\ne O$ .

where $O$ denotes the null matrix.

Also, for any $2\times 2$ matrix, say $X=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ , its adjoint matrix is given by-

  $adjX=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ .

Calculation: The given augmented matrix is $\left[\begin{array}{cc}1& 2\\ 0& a\end{array}|\begin{array}{c}-4\\ b\end{array}\right]$ .

The associated coefficient matrix and constant matrix are-

  $A=\left[\begin{array}{cc}1& 2\\ 0& a\end{array}\right]$ and $B=\left[\begin{array}{c}-4\\ b\end{array}\right]$ respectively.

The given system will have infinite solutions if $\det (A)=0$ and $(adjA)B=O$ .

Now,

  $\det (A)=0$

  $\Rightarrow \left|\begin{array}{cc}1& 2\\ 0& a\end{array}\right|=0$

  $\Rightarrow a\times 1-2\times 0=0$

  $\Rightarrow a=0$

Since, $A=\left[\begin{array}{cc}1& 2\\ 0& a\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right]$ (since a = 0) thus,

  $adjA=\left[\begin{array}{cc}0& -2\\ 0& 1\end{array}\right]$

Now,

  $(adjA)B=O$

  $\Rightarrow \left[\begin{array}{cc}0& -2\\ 0& 1\end{array}\right]\left[\begin{array}{c}-4\\ b\end{array}\right]=O$

  $\Rightarrow \left[\begin{array}{c}0\times (-4)+(-2)\times b\\ 0\times (-4)+1\times b\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

  $\Rightarrow \left[\begin{array}{c}-2b\\ b\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right]$

  $\Rightarrow b=0$ (From equality of matrices)

Thus the system associated with the given augmented matrix will have infinitely many solutions if $a=b=0$ .

Thus the required values of a and b are-

  $a=0$ and $b=0$ .

c.

To determine

To find: The values of a andb for which the given augmented matrix has no solution.

Answer to Problem 106E

The required values of a and b are $a=0$ and $b\in R-\left\{0\right\}$ .

Explanation of Solution

Given: The given augmented matrix is $\left[\begin{array}{cc}1& 2\\ 0& a\end{array}|\begin{array}{c}-4\\ b\end{array}\right]$ .

Concept Used: An augmented matrix is a matrix representation of a standard system of simultaneous linear equations. It is a matrix which consists of two parts with a line separating both parts. On the left hand side of the line is the coefficient matrix (generally denoted by A ) of the system of equations and on the right hand side of the line are the constant terms (generally denoted by B ), written in order. Multiplying the coefficient matrix part to the variable column matrix (generally denoted by X) gives the algebraic expressions which are then equated to the constant terms to get the system of equations. Thus, the system of equations can be found using the matrix equation-

  $AX=B$

If a system consists of two linear equations, say-

  $a_{1}x+b_{1}y=c_1$

  $a_{2}x+b_{2}y=c_2$

The associated coefficient matrix and constant matrix are-

  $A=\left[\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right]$ and $B=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$ respectively.

Then, the above system has-

1. Exactly one solution if $\det (A)\ne 0$ .
2. Infinitely many solutions if $\det (A)=0$ and $(adjA)B=O$ .
3. No solution if $\det (A)=0$ and $(adjA)B\ne O$ .

where $O$ denotes the null matrix.

Also, for any $2\times 2$ matrix, say $X=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ , its adjoint matrix is given by-

  $adjX=\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ .

Calculation: The given augmented matrix is $\left[\begin{array}{cc}1& 2\\ 0& a\end{array}|\begin{array}{c}-4\\ b\end{array}\right]$ .

The associated coefficient matrix and constant matrix are-

  $A=\left[\begin{array}{cc}1& 2\\ 0& a\end{array}\right]$ and $B=\left[\begin{array}{c}-4\\ b\end{array}\right]$ respectively.

The given system will have no solution if $\det (A)=0$ and $(adjA)B\ne O$ .

Now,

  $\det (A)=0$

  $\Rightarrow \left|\begin{array}{cc}1& 2\\ 0& a\end{array}\right|=0$

  $\Rightarrow a\times 1-2\times 0=0$

  $\Rightarrow a=0$

Since, $A=\left[\begin{array}{cc}1& 2\\ 0& a\end{array}\right]=\left[\begin{array}{cc}1& 2\\ 0& 0\end{array}\right]$ (since a = 0) thus,

  $adjA=\left[\begin{array}{cc}0& -2\\ 0& 1\end{array}\right]$

Now,

  $(adjA)B\ne O$

  $\Rightarrow \left[\begin{array}{cc}0& -2\\ 0& 1\end{array}\right]\left[\begin{array}{c}-4\\ b\end{array}\right]\ne O$

  $\Rightarrow \left[\begin{array}{c}0\times (-4)+(-2)\times b\\ 0\times (-4)+1\times b\end{array}\right]\ne \left[\begin{array}{c}0\\ 0\end{array}\right]$

  $\Rightarrow \left[\begin{array}{c}-2b\\ b\end{array}\right]\ne \left[\begin{array}{c}0\\ 0\end{array}\right]$

  $\Rightarrow b\ne 0$ (From equality of matrices)

Thus the system associated with the given augmented matrix will have no solutions if $a=0$ and $b$ has any real value except 0.

Thus the required values of a and b are-

  $a=0$ and $b\in R-\left\{0\right\}$ .