1,733 views

1 Answer

Best answer
4 votes
4 votes

$A = \begin{bmatrix}
2 & b\\ 
4 & 8
\end{bmatrix}$

and let the augmented matrix be 

$A|B = \begin{bmatrix}
2 & b & 16\\ 
4 & 8 & g
\end{bmatrix}$


For singularity, $|A| = 0$ but this says that the system of linear equations can have either no solution or infinite solution .

We get $b=4$

Now, since to make it solvable i.e. having infinite solutions, we need to make the last row of augmented matrix = $0$.

Now, this is only possible if we take $g=32$, then last row completely gets eliminated and hence, we have infinite solutions .

selected by

Related questions

1 votes
1 votes
1 answer
1
sarika asked Jul 5, 2017
1,159 views
If B has eigenvalues 1, 2, 3, C has eigenvalues 4, 5, 6, and D has eigenvalues 7, 8,9, what are the eigenvalues of the 6 by 6 matrix A =[ B C , 0 D] ?
0 votes
0 votes
0 answers
2
5 votes
5 votes
1 answer
3
Tesla! asked Oct 23, 2017
456 views
How many n*n (0,1) matrix are invertable ?PS: original question in book is for 2*2