GATE CSE 2003 | Question: 41

Question

Consider the following system of linear equations $$\left( \begin{array}{ccc} 2 & 1 & -4 \\ 4 & 3 & -12 \\ 1 & 2 & -8 \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} \alpha \\ 5 \\ 7 \end{array} \right)$$ Notice that the second and the third columns of the coefficient matrix are linearly dependent. For how many values of $\alpha$, does this system of equations have infinitely many solutions?

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Comments

Procedure to solve this question when solution is given but some element value isn't given then

step1: make the augmented matrix into echelon form of a matrix.

step2: If it's given as an unique solution then try to make last row totally not equals to zero.

          Infinite solution exists then make the last row totally 0 then we can get linear independent solution.

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Text me if i am wrong...

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Ans

Answer 1

Since the second and third columns of the coefficient matrix are linearly deoendent, determis $0.$ So, the system of equations either has infinitely many solutions (if they are consistent) or no solution. To check for consistency, we apply reduction method on $(A\mid B)$

$R_{2}\leftarrow R_{2}-2R_{1},R_{3}\leftarrow R_{3}-0.5R_{1},R_{3}\leftarrow R_{3}-1.5R_{2}, R_1 \leftarrow R_1/2$

obtain the resultant matrix

$\large\begin{pmatrix}2&1&{-4}&\alpha\\4&3&{-12}&5\\1&2&{-8}&7\end{pmatrix}\rightarrow\begin{pmatrix}1&0.5&{-2}&0.5\alpha\\0&1&{-4}&5-2\alpha\\0&0&{-0}& -0.5+2.5\alpha\end{pmatrix}$

or infinitely many solutions, we must have $-0.5+2.5\alpha =0\; i.e., \alpha=\dfrac{1}{5}.$ So, for only $1$ value of $\alpha,$ this system has infinitely many solutions. So, option (B) is correct.

Comments

$Ax=b $$\implies x = A^{-1}b$

$A^{-1}$ would only exist if $|A| \neq 0$

so this means the system would have no solution or infinite solutions, if $|A|=0$

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@Parimal Paritosh You said that there are just 4 3x3 subsquare matrices by eliminating column 1, column 2, column 3, column 4 once respectively. But according to this source we need to consider only adjacent rows and columns for square sub matrices.

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@Satbir This is true only for square matrix. False for rectangular matrix.

Answer 2

Given system of equations is in form AX=B.

For infinitely many solutions here, effectively we have only two variables i.e. w=(y-4z). but they should be consistent i.e. rank(A)=rank(AB)<3(no. of variables)

 Since column2 and column3 are dependent so make column3 all zero. So we got rank(A)=2

Now for the augmented matrix, AB put all zero column at last because it's of no use it will give 0 determinant value of all 3*3 submatrices.

After excluding that dependent column, we got a 3*3 matrix, solving it gives alpha= .2 this is single value.

Or see the image below

Option B ...

Note: The row rank and the column rank of a matrix A are equal. Here column rank is easy to find for A. And then for AB we must get rank 2 for having infinite solutions. 

P.S. -- Ignore that "3 equation and 2 variable" in the image.

Comments

Sir, i am also using writometer pen.

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What is the use of crammers rule here btw?

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True @ritiksri8. Why to follow the complex method.

 

Answer 3

For infinitely many solution we've to check this condition, rank(A) = rank(A|B) < n, where n is the unknown variable.

Now make the given matrix(augmented) into echleon form by Gauss-elimination method.

$\begin{bmatrix} 2 &1 &-4 &\alpha \\ 4 &3 &-12 &5 \\ 1 &2 &-8 &7 \end{bmatrix}$      R2 = r2 - 2r1

$\begin{bmatrix} 2 &1 &-4 &\alpha \\ 0 &1 &-4 &5-2\alpha \\ 1 &2 &-8 &7 \end{bmatrix}$        r3 = 2r3 -r1

$\begin{bmatrix} 2 &1 &-4 &\alpha \\ 0 &1 &-4 &5-2\alpha \\ 0 &3 &-12 &14 - \alpha \end{bmatrix}$     r3 = 3r2 - r3

$\begin{bmatrix} 2 &1 &-4 &\alpha \\ 0 &1 &-4 &5-2\alpha \\ 0 &0 &0 &1 - 5\alpha \end{bmatrix}$

Now to satisfy the above written condition for infinite solution 1 - 5$\alpha$ = 0 i.e  $\alpha$ = $\frac{1}{5}$

Comments

Forgot to mention answer B  😀

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But if we use det then the elementary row operations that are used in the above question are invalid right..??

Answer 4

​​​​​​(b) ​1 value of alpha is only there . 

 

Answer 5

When you reduce the augmented matrix in row echelon form row 1 will be 2 1 -4 a , row 2 will be R2=R2-2R1 which will produce 0 1 -4 | 5-2a , row 3 will be R3= R3-1/2 R1 which will produce 0 3/2 -6 | 7- a/2 Now just make second element of R3 0 . Which will be R3= R3 - 3/2R2 which will give R3 as 0 0 0 | 7-a/2 -3/2*7 +3/4a . Then rank of a has to be equal to rank (a|b) for the system to have consistent solution . Then 7-a/2-3/2*7 +3/4a has to be 0  then a will have one solution . Also you can check others like 5-2a=0 ,7-a/2 =0 , in each case you have check whether a satisfies last row . So finally you will have only one value of a 1/5

Answer 6

Ans