System of equations has no solution












1














enter image description here



I really fail to understand why the answer cannot be option (1).



Here's my method:



For $a = -1$, the coefficient determinant determinant is 0. And if we create $Delta_z$, the determinant in which coefficients of $z$ are replaced by $3,6,b$, (b= 9), that determinant is non zero.



My book says that if coefficient determinant is $0$ and at least one of $Delta_y, Delta_z, Delta_x$ is non $0$ then the system has no solution.



But the answer given is $(2)$. What's wrong with my method? What's the best way to tackle such problems?










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  • For choice (1), isn't (x, y, z) = (4, 1, 0) a solution?
    – Gaffney
    5 hours ago
















1














enter image description here



I really fail to understand why the answer cannot be option (1).



Here's my method:



For $a = -1$, the coefficient determinant determinant is 0. And if we create $Delta_z$, the determinant in which coefficients of $z$ are replaced by $3,6,b$, (b= 9), that determinant is non zero.



My book says that if coefficient determinant is $0$ and at least one of $Delta_y, Delta_z, Delta_x$ is non $0$ then the system has no solution.



But the answer given is $(2)$. What's wrong with my method? What's the best way to tackle such problems?










share|cite|improve this question






















  • For choice (1), isn't (x, y, z) = (4, 1, 0) a solution?
    – Gaffney
    5 hours ago














1












1








1







enter image description here



I really fail to understand why the answer cannot be option (1).



Here's my method:



For $a = -1$, the coefficient determinant determinant is 0. And if we create $Delta_z$, the determinant in which coefficients of $z$ are replaced by $3,6,b$, (b= 9), that determinant is non zero.



My book says that if coefficient determinant is $0$ and at least one of $Delta_y, Delta_z, Delta_x$ is non $0$ then the system has no solution.



But the answer given is $(2)$. What's wrong with my method? What's the best way to tackle such problems?










share|cite|improve this question













enter image description here



I really fail to understand why the answer cannot be option (1).



Here's my method:



For $a = -1$, the coefficient determinant determinant is 0. And if we create $Delta_z$, the determinant in which coefficients of $z$ are replaced by $3,6,b$, (b= 9), that determinant is non zero.



My book says that if coefficient determinant is $0$ and at least one of $Delta_y, Delta_z, Delta_x$ is non $0$ then the system has no solution.



But the answer given is $(2)$. What's wrong with my method? What's the best way to tackle such problems?







determinant






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asked 5 hours ago









Abcd

3,02021134




3,02021134












  • For choice (1), isn't (x, y, z) = (4, 1, 0) a solution?
    – Gaffney
    5 hours ago


















  • For choice (1), isn't (x, y, z) = (4, 1, 0) a solution?
    – Gaffney
    5 hours ago
















For choice (1), isn't (x, y, z) = (4, 1, 0) a solution?
– Gaffney
5 hours ago




For choice (1), isn't (x, y, z) = (4, 1, 0) a solution?
– Gaffney
5 hours ago










4 Answers
4






active

oldest

votes


















4














You miscalculated $Delta_z$. The book solution is correct.
In order to have no solution, it is necessary that the determinant of the matrix of coefficients be zero: this indeed yields $a=-1$. The rank of the matrix is $2$ in this case, and up to a constant multiple, the only nontrivial "dependence" of the equations can be $1cdot (1)-2cdot (2) +1cdot (3) = 0$.
So under the $a=-1$ assumption, there are infinitely many solutions iff this generalizes to the right hand sides, that is, iff $3-2cdot 6 + b =0$, i.e., $b=9$.
Hence, the answer is $a=-1, bneq 9$.






share|cite|improve this answer





























    2














    I suppose the best way to tackle is to row-reduce the matrix and see what happens. You have to eliminate the third and the first columns:
    $$
    begin{pmatrix}
    1 & a & 1 & 3\
    1 & 2 & 2 & 6 \
    1 & 5 & 3 & b
    end{pmatrix}
    to
    begin{pmatrix}
    1 & a & 1 & 3\
    0 & 2-a & 1 & 3 \
    0 & 5-a & 2 & b-3
    end{pmatrix}
    to
    begin{pmatrix}
    1 & 2a-2 & 0 & 0\
    0 & 2-a & 1 & 3 \
    0 & 1+a & 0 & b-9
    end{pmatrix}
    $$

    so if $a=-1$ the last equation reads $0=b-9$, thus having $b=9$ you have $0=0$, which eliminates the equation, and now you have 2 equations in 3 unknowns, hence an infinite number of solutions. Thus (i) cannot be the answer.



    The condition for no solutions would require that $a=-1$ (forcing the last equation to read $0=b-9$) and $b ne 9$ (forcing it to be false).



    UPDATE



    In retrospect, as you say, if $a=-1$ then $Delta = 0$, so there will not be a unique solution. However, if $b = 9$, the last column of $Delta_z$ is $(3,6,9) = 3 times (1,2,3)$, so exactly $3$ times the last column of $Delta$, hence, $Delta_z = 3Delta = 0$, and the idea in your book is correct.






    share|cite|improve this answer























    • Question asks the condition for no solution...
      – Abcd
      5 hours ago










    • @Abcd see update
      – gt6989b
      5 hours ago










    • "you have to eliminate the third and the first columns"- why?
      – Abcd
      5 hours ago










    • @Abcd because they have no variables, much easier to eliminate these than pivoting on column with $a$ in. You can still do it, but it is more arithmetic. BTW method of your book is correct, see another update
      – gt6989b
      5 hours ago










    • Thanks for your time. I understand that you have applied transformations to the determinant. But I am unable to follow after that... What is the new determinant supposed to indicate?
      – Abcd
      5 hours ago



















    0














    I would do it this way, using row reduction. The criterion for a non-homogeneous linear system to have solutions is that the matrix of the linear system (l.h.s.) and the augmented matrix have the same rank.



    From this criterion we deduce readily that if this system has no solution, the matrix of the l.h.s. cannot have rank $3$, hence its determinant, which is equal to $-(a+1)$ is zero, whence $a=-1$. Computing the determinant by row reduction shows this matrix has rank $2$.



    Now, if $a=-1$, the augmented matrix reduces as follows:
    $$begin{bmatrix}1&-1&1&3\1&2&2&6\1&5&3&bend{bmatrix}rightsquigarrow
    begin{bmatrix}1&-1&1&3\0&3&1&3\0&3&1&b-6end{bmatrix}rightsquigarrow
    begin{bmatrix}1&-1&1&3\0&3&1&3\0&0&0&b-9end{bmatrix} $$

    Thus, for $a=-1$, the augmented matrix has rank $3$ if and only if $b-9ne 0$.






    share|cite|improve this answer





























      0














      Rewrite the system as:
      $$
      AX=B
      $$

      where $X=(x,y,z)^t$, $B=(3,6,b)^t$ and
      $$A = begin{bmatrix}
      1 & a & 1 \
      1 & 2 & 2 \
      1 & 5 & 3
      end{bmatrix}$$



      First you must have $det(A)=0$ which after calculations give $a=-1$.
      So the system becomes:



      $$A = begin{bmatrix}
      1 & -1 & 1 \
      1 & 2 & 2 \
      1 & 5 & 3
      end{bmatrix}$$



      Now, row 3 becomes all zero when $r_3rightarrow r_3-2r_2+r_1$. Applying this to the augmented matrix $[A|B]$ gives the $(3,4)$ entry to be:
      $$
      b-9
      $$



      It is required that this entry is nonzero in order to have an inconsistent matrix.
      Thus, $a=-1, bneq 9$.






      share|cite|improve this answer





















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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4














        You miscalculated $Delta_z$. The book solution is correct.
        In order to have no solution, it is necessary that the determinant of the matrix of coefficients be zero: this indeed yields $a=-1$. The rank of the matrix is $2$ in this case, and up to a constant multiple, the only nontrivial "dependence" of the equations can be $1cdot (1)-2cdot (2) +1cdot (3) = 0$.
        So under the $a=-1$ assumption, there are infinitely many solutions iff this generalizes to the right hand sides, that is, iff $3-2cdot 6 + b =0$, i.e., $b=9$.
        Hence, the answer is $a=-1, bneq 9$.






        share|cite|improve this answer


























          4














          You miscalculated $Delta_z$. The book solution is correct.
          In order to have no solution, it is necessary that the determinant of the matrix of coefficients be zero: this indeed yields $a=-1$. The rank of the matrix is $2$ in this case, and up to a constant multiple, the only nontrivial "dependence" of the equations can be $1cdot (1)-2cdot (2) +1cdot (3) = 0$.
          So under the $a=-1$ assumption, there are infinitely many solutions iff this generalizes to the right hand sides, that is, iff $3-2cdot 6 + b =0$, i.e., $b=9$.
          Hence, the answer is $a=-1, bneq 9$.






          share|cite|improve this answer
























            4












            4








            4






            You miscalculated $Delta_z$. The book solution is correct.
            In order to have no solution, it is necessary that the determinant of the matrix of coefficients be zero: this indeed yields $a=-1$. The rank of the matrix is $2$ in this case, and up to a constant multiple, the only nontrivial "dependence" of the equations can be $1cdot (1)-2cdot (2) +1cdot (3) = 0$.
            So under the $a=-1$ assumption, there are infinitely many solutions iff this generalizes to the right hand sides, that is, iff $3-2cdot 6 + b =0$, i.e., $b=9$.
            Hence, the answer is $a=-1, bneq 9$.






            share|cite|improve this answer












            You miscalculated $Delta_z$. The book solution is correct.
            In order to have no solution, it is necessary that the determinant of the matrix of coefficients be zero: this indeed yields $a=-1$. The rank of the matrix is $2$ in this case, and up to a constant multiple, the only nontrivial "dependence" of the equations can be $1cdot (1)-2cdot (2) +1cdot (3) = 0$.
            So under the $a=-1$ assumption, there are infinitely many solutions iff this generalizes to the right hand sides, that is, iff $3-2cdot 6 + b =0$, i.e., $b=9$.
            Hence, the answer is $a=-1, bneq 9$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 5 hours ago









            A. Pongrácz

            5,4051827




            5,4051827























                2














                I suppose the best way to tackle is to row-reduce the matrix and see what happens. You have to eliminate the third and the first columns:
                $$
                begin{pmatrix}
                1 & a & 1 & 3\
                1 & 2 & 2 & 6 \
                1 & 5 & 3 & b
                end{pmatrix}
                to
                begin{pmatrix}
                1 & a & 1 & 3\
                0 & 2-a & 1 & 3 \
                0 & 5-a & 2 & b-3
                end{pmatrix}
                to
                begin{pmatrix}
                1 & 2a-2 & 0 & 0\
                0 & 2-a & 1 & 3 \
                0 & 1+a & 0 & b-9
                end{pmatrix}
                $$

                so if $a=-1$ the last equation reads $0=b-9$, thus having $b=9$ you have $0=0$, which eliminates the equation, and now you have 2 equations in 3 unknowns, hence an infinite number of solutions. Thus (i) cannot be the answer.



                The condition for no solutions would require that $a=-1$ (forcing the last equation to read $0=b-9$) and $b ne 9$ (forcing it to be false).



                UPDATE



                In retrospect, as you say, if $a=-1$ then $Delta = 0$, so there will not be a unique solution. However, if $b = 9$, the last column of $Delta_z$ is $(3,6,9) = 3 times (1,2,3)$, so exactly $3$ times the last column of $Delta$, hence, $Delta_z = 3Delta = 0$, and the idea in your book is correct.






                share|cite|improve this answer























                • Question asks the condition for no solution...
                  – Abcd
                  5 hours ago










                • @Abcd see update
                  – gt6989b
                  5 hours ago










                • "you have to eliminate the third and the first columns"- why?
                  – Abcd
                  5 hours ago










                • @Abcd because they have no variables, much easier to eliminate these than pivoting on column with $a$ in. You can still do it, but it is more arithmetic. BTW method of your book is correct, see another update
                  – gt6989b
                  5 hours ago










                • Thanks for your time. I understand that you have applied transformations to the determinant. But I am unable to follow after that... What is the new determinant supposed to indicate?
                  – Abcd
                  5 hours ago
















                2














                I suppose the best way to tackle is to row-reduce the matrix and see what happens. You have to eliminate the third and the first columns:
                $$
                begin{pmatrix}
                1 & a & 1 & 3\
                1 & 2 & 2 & 6 \
                1 & 5 & 3 & b
                end{pmatrix}
                to
                begin{pmatrix}
                1 & a & 1 & 3\
                0 & 2-a & 1 & 3 \
                0 & 5-a & 2 & b-3
                end{pmatrix}
                to
                begin{pmatrix}
                1 & 2a-2 & 0 & 0\
                0 & 2-a & 1 & 3 \
                0 & 1+a & 0 & b-9
                end{pmatrix}
                $$

                so if $a=-1$ the last equation reads $0=b-9$, thus having $b=9$ you have $0=0$, which eliminates the equation, and now you have 2 equations in 3 unknowns, hence an infinite number of solutions. Thus (i) cannot be the answer.



                The condition for no solutions would require that $a=-1$ (forcing the last equation to read $0=b-9$) and $b ne 9$ (forcing it to be false).



                UPDATE



                In retrospect, as you say, if $a=-1$ then $Delta = 0$, so there will not be a unique solution. However, if $b = 9$, the last column of $Delta_z$ is $(3,6,9) = 3 times (1,2,3)$, so exactly $3$ times the last column of $Delta$, hence, $Delta_z = 3Delta = 0$, and the idea in your book is correct.






                share|cite|improve this answer























                • Question asks the condition for no solution...
                  – Abcd
                  5 hours ago










                • @Abcd see update
                  – gt6989b
                  5 hours ago










                • "you have to eliminate the third and the first columns"- why?
                  – Abcd
                  5 hours ago










                • @Abcd because they have no variables, much easier to eliminate these than pivoting on column with $a$ in. You can still do it, but it is more arithmetic. BTW method of your book is correct, see another update
                  – gt6989b
                  5 hours ago










                • Thanks for your time. I understand that you have applied transformations to the determinant. But I am unable to follow after that... What is the new determinant supposed to indicate?
                  – Abcd
                  5 hours ago














                2












                2








                2






                I suppose the best way to tackle is to row-reduce the matrix and see what happens. You have to eliminate the third and the first columns:
                $$
                begin{pmatrix}
                1 & a & 1 & 3\
                1 & 2 & 2 & 6 \
                1 & 5 & 3 & b
                end{pmatrix}
                to
                begin{pmatrix}
                1 & a & 1 & 3\
                0 & 2-a & 1 & 3 \
                0 & 5-a & 2 & b-3
                end{pmatrix}
                to
                begin{pmatrix}
                1 & 2a-2 & 0 & 0\
                0 & 2-a & 1 & 3 \
                0 & 1+a & 0 & b-9
                end{pmatrix}
                $$

                so if $a=-1$ the last equation reads $0=b-9$, thus having $b=9$ you have $0=0$, which eliminates the equation, and now you have 2 equations in 3 unknowns, hence an infinite number of solutions. Thus (i) cannot be the answer.



                The condition for no solutions would require that $a=-1$ (forcing the last equation to read $0=b-9$) and $b ne 9$ (forcing it to be false).



                UPDATE



                In retrospect, as you say, if $a=-1$ then $Delta = 0$, so there will not be a unique solution. However, if $b = 9$, the last column of $Delta_z$ is $(3,6,9) = 3 times (1,2,3)$, so exactly $3$ times the last column of $Delta$, hence, $Delta_z = 3Delta = 0$, and the idea in your book is correct.






                share|cite|improve this answer














                I suppose the best way to tackle is to row-reduce the matrix and see what happens. You have to eliminate the third and the first columns:
                $$
                begin{pmatrix}
                1 & a & 1 & 3\
                1 & 2 & 2 & 6 \
                1 & 5 & 3 & b
                end{pmatrix}
                to
                begin{pmatrix}
                1 & a & 1 & 3\
                0 & 2-a & 1 & 3 \
                0 & 5-a & 2 & b-3
                end{pmatrix}
                to
                begin{pmatrix}
                1 & 2a-2 & 0 & 0\
                0 & 2-a & 1 & 3 \
                0 & 1+a & 0 & b-9
                end{pmatrix}
                $$

                so if $a=-1$ the last equation reads $0=b-9$, thus having $b=9$ you have $0=0$, which eliminates the equation, and now you have 2 equations in 3 unknowns, hence an infinite number of solutions. Thus (i) cannot be the answer.



                The condition for no solutions would require that $a=-1$ (forcing the last equation to read $0=b-9$) and $b ne 9$ (forcing it to be false).



                UPDATE



                In retrospect, as you say, if $a=-1$ then $Delta = 0$, so there will not be a unique solution. However, if $b = 9$, the last column of $Delta_z$ is $(3,6,9) = 3 times (1,2,3)$, so exactly $3$ times the last column of $Delta$, hence, $Delta_z = 3Delta = 0$, and the idea in your book is correct.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 5 hours ago

























                answered 5 hours ago









                gt6989b

                33k22452




                33k22452












                • Question asks the condition for no solution...
                  – Abcd
                  5 hours ago










                • @Abcd see update
                  – gt6989b
                  5 hours ago










                • "you have to eliminate the third and the first columns"- why?
                  – Abcd
                  5 hours ago










                • @Abcd because they have no variables, much easier to eliminate these than pivoting on column with $a$ in. You can still do it, but it is more arithmetic. BTW method of your book is correct, see another update
                  – gt6989b
                  5 hours ago










                • Thanks for your time. I understand that you have applied transformations to the determinant. But I am unable to follow after that... What is the new determinant supposed to indicate?
                  – Abcd
                  5 hours ago


















                • Question asks the condition for no solution...
                  – Abcd
                  5 hours ago










                • @Abcd see update
                  – gt6989b
                  5 hours ago










                • "you have to eliminate the third and the first columns"- why?
                  – Abcd
                  5 hours ago










                • @Abcd because they have no variables, much easier to eliminate these than pivoting on column with $a$ in. You can still do it, but it is more arithmetic. BTW method of your book is correct, see another update
                  – gt6989b
                  5 hours ago










                • Thanks for your time. I understand that you have applied transformations to the determinant. But I am unable to follow after that... What is the new determinant supposed to indicate?
                  – Abcd
                  5 hours ago
















                Question asks the condition for no solution...
                – Abcd
                5 hours ago




                Question asks the condition for no solution...
                – Abcd
                5 hours ago












                @Abcd see update
                – gt6989b
                5 hours ago




                @Abcd see update
                – gt6989b
                5 hours ago












                "you have to eliminate the third and the first columns"- why?
                – Abcd
                5 hours ago




                "you have to eliminate the third and the first columns"- why?
                – Abcd
                5 hours ago












                @Abcd because they have no variables, much easier to eliminate these than pivoting on column with $a$ in. You can still do it, but it is more arithmetic. BTW method of your book is correct, see another update
                – gt6989b
                5 hours ago




                @Abcd because they have no variables, much easier to eliminate these than pivoting on column with $a$ in. You can still do it, but it is more arithmetic. BTW method of your book is correct, see another update
                – gt6989b
                5 hours ago












                Thanks for your time. I understand that you have applied transformations to the determinant. But I am unable to follow after that... What is the new determinant supposed to indicate?
                – Abcd
                5 hours ago




                Thanks for your time. I understand that you have applied transformations to the determinant. But I am unable to follow after that... What is the new determinant supposed to indicate?
                – Abcd
                5 hours ago











                0














                I would do it this way, using row reduction. The criterion for a non-homogeneous linear system to have solutions is that the matrix of the linear system (l.h.s.) and the augmented matrix have the same rank.



                From this criterion we deduce readily that if this system has no solution, the matrix of the l.h.s. cannot have rank $3$, hence its determinant, which is equal to $-(a+1)$ is zero, whence $a=-1$. Computing the determinant by row reduction shows this matrix has rank $2$.



                Now, if $a=-1$, the augmented matrix reduces as follows:
                $$begin{bmatrix}1&-1&1&3\1&2&2&6\1&5&3&bend{bmatrix}rightsquigarrow
                begin{bmatrix}1&-1&1&3\0&3&1&3\0&3&1&b-6end{bmatrix}rightsquigarrow
                begin{bmatrix}1&-1&1&3\0&3&1&3\0&0&0&b-9end{bmatrix} $$

                Thus, for $a=-1$, the augmented matrix has rank $3$ if and only if $b-9ne 0$.






                share|cite|improve this answer


























                  0














                  I would do it this way, using row reduction. The criterion for a non-homogeneous linear system to have solutions is that the matrix of the linear system (l.h.s.) and the augmented matrix have the same rank.



                  From this criterion we deduce readily that if this system has no solution, the matrix of the l.h.s. cannot have rank $3$, hence its determinant, which is equal to $-(a+1)$ is zero, whence $a=-1$. Computing the determinant by row reduction shows this matrix has rank $2$.



                  Now, if $a=-1$, the augmented matrix reduces as follows:
                  $$begin{bmatrix}1&-1&1&3\1&2&2&6\1&5&3&bend{bmatrix}rightsquigarrow
                  begin{bmatrix}1&-1&1&3\0&3&1&3\0&3&1&b-6end{bmatrix}rightsquigarrow
                  begin{bmatrix}1&-1&1&3\0&3&1&3\0&0&0&b-9end{bmatrix} $$

                  Thus, for $a=-1$, the augmented matrix has rank $3$ if and only if $b-9ne 0$.






                  share|cite|improve this answer
























                    0












                    0








                    0






                    I would do it this way, using row reduction. The criterion for a non-homogeneous linear system to have solutions is that the matrix of the linear system (l.h.s.) and the augmented matrix have the same rank.



                    From this criterion we deduce readily that if this system has no solution, the matrix of the l.h.s. cannot have rank $3$, hence its determinant, which is equal to $-(a+1)$ is zero, whence $a=-1$. Computing the determinant by row reduction shows this matrix has rank $2$.



                    Now, if $a=-1$, the augmented matrix reduces as follows:
                    $$begin{bmatrix}1&-1&1&3\1&2&2&6\1&5&3&bend{bmatrix}rightsquigarrow
                    begin{bmatrix}1&-1&1&3\0&3&1&3\0&3&1&b-6end{bmatrix}rightsquigarrow
                    begin{bmatrix}1&-1&1&3\0&3&1&3\0&0&0&b-9end{bmatrix} $$

                    Thus, for $a=-1$, the augmented matrix has rank $3$ if and only if $b-9ne 0$.






                    share|cite|improve this answer












                    I would do it this way, using row reduction. The criterion for a non-homogeneous linear system to have solutions is that the matrix of the linear system (l.h.s.) and the augmented matrix have the same rank.



                    From this criterion we deduce readily that if this system has no solution, the matrix of the l.h.s. cannot have rank $3$, hence its determinant, which is equal to $-(a+1)$ is zero, whence $a=-1$. Computing the determinant by row reduction shows this matrix has rank $2$.



                    Now, if $a=-1$, the augmented matrix reduces as follows:
                    $$begin{bmatrix}1&-1&1&3\1&2&2&6\1&5&3&bend{bmatrix}rightsquigarrow
                    begin{bmatrix}1&-1&1&3\0&3&1&3\0&3&1&b-6end{bmatrix}rightsquigarrow
                    begin{bmatrix}1&-1&1&3\0&3&1&3\0&0&0&b-9end{bmatrix} $$

                    Thus, for $a=-1$, the augmented matrix has rank $3$ if and only if $b-9ne 0$.







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                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 5 hours ago









                    Bernard

                    118k639112




                    118k639112























                        0














                        Rewrite the system as:
                        $$
                        AX=B
                        $$

                        where $X=(x,y,z)^t$, $B=(3,6,b)^t$ and
                        $$A = begin{bmatrix}
                        1 & a & 1 \
                        1 & 2 & 2 \
                        1 & 5 & 3
                        end{bmatrix}$$



                        First you must have $det(A)=0$ which after calculations give $a=-1$.
                        So the system becomes:



                        $$A = begin{bmatrix}
                        1 & -1 & 1 \
                        1 & 2 & 2 \
                        1 & 5 & 3
                        end{bmatrix}$$



                        Now, row 3 becomes all zero when $r_3rightarrow r_3-2r_2+r_1$. Applying this to the augmented matrix $[A|B]$ gives the $(3,4)$ entry to be:
                        $$
                        b-9
                        $$



                        It is required that this entry is nonzero in order to have an inconsistent matrix.
                        Thus, $a=-1, bneq 9$.






                        share|cite|improve this answer


























                          0














                          Rewrite the system as:
                          $$
                          AX=B
                          $$

                          where $X=(x,y,z)^t$, $B=(3,6,b)^t$ and
                          $$A = begin{bmatrix}
                          1 & a & 1 \
                          1 & 2 & 2 \
                          1 & 5 & 3
                          end{bmatrix}$$



                          First you must have $det(A)=0$ which after calculations give $a=-1$.
                          So the system becomes:



                          $$A = begin{bmatrix}
                          1 & -1 & 1 \
                          1 & 2 & 2 \
                          1 & 5 & 3
                          end{bmatrix}$$



                          Now, row 3 becomes all zero when $r_3rightarrow r_3-2r_2+r_1$. Applying this to the augmented matrix $[A|B]$ gives the $(3,4)$ entry to be:
                          $$
                          b-9
                          $$



                          It is required that this entry is nonzero in order to have an inconsistent matrix.
                          Thus, $a=-1, bneq 9$.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            Rewrite the system as:
                            $$
                            AX=B
                            $$

                            where $X=(x,y,z)^t$, $B=(3,6,b)^t$ and
                            $$A = begin{bmatrix}
                            1 & a & 1 \
                            1 & 2 & 2 \
                            1 & 5 & 3
                            end{bmatrix}$$



                            First you must have $det(A)=0$ which after calculations give $a=-1$.
                            So the system becomes:



                            $$A = begin{bmatrix}
                            1 & -1 & 1 \
                            1 & 2 & 2 \
                            1 & 5 & 3
                            end{bmatrix}$$



                            Now, row 3 becomes all zero when $r_3rightarrow r_3-2r_2+r_1$. Applying this to the augmented matrix $[A|B]$ gives the $(3,4)$ entry to be:
                            $$
                            b-9
                            $$



                            It is required that this entry is nonzero in order to have an inconsistent matrix.
                            Thus, $a=-1, bneq 9$.






                            share|cite|improve this answer












                            Rewrite the system as:
                            $$
                            AX=B
                            $$

                            where $X=(x,y,z)^t$, $B=(3,6,b)^t$ and
                            $$A = begin{bmatrix}
                            1 & a & 1 \
                            1 & 2 & 2 \
                            1 & 5 & 3
                            end{bmatrix}$$



                            First you must have $det(A)=0$ which after calculations give $a=-1$.
                            So the system becomes:



                            $$A = begin{bmatrix}
                            1 & -1 & 1 \
                            1 & 2 & 2 \
                            1 & 5 & 3
                            end{bmatrix}$$



                            Now, row 3 becomes all zero when $r_3rightarrow r_3-2r_2+r_1$. Applying this to the augmented matrix $[A|B]$ gives the $(3,4)$ entry to be:
                            $$
                            b-9
                            $$



                            It is required that this entry is nonzero in order to have an inconsistent matrix.
                            Thus, $a=-1, bneq 9$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 5 hours ago









                            Test123

                            2,855828




                            2,855828






























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