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GATE20011.1
+11
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Consider the following statements:
S1: The sum of two singular $n \times n$ matrices may be nonsingular
S2: The sum of two $n \times n$ nonsingular matrices may be singular
Which one of the following statements is correct?
S1 and S2 both are true
S1 is true, S2 is false
S1 is false, S2 is true
S1 and S2 both are false
gate2001
linearalgebra
normal
matrices
asked
Sep 14, 2014
in
Linear Algebra
by
Kathleen
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Jun 20, 2017
by
Silpa

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4
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+14
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Best answer
Yes A is correct option!, Both statements are True
$S_{1}:\text{True}$
$A=\begin{matrix} 2&10 \\ 1&5 \end{matrix}$$\qquad B=\begin{matrix} 3&6 \\ 2&4 \end{matrix}$
$A=0 \quad B=0$
$A+B={3}$
$S_{2}:\text{True}$
$A=\begin{matrix} 1&0 \\ 0&1 \end{matrix}$$\qquad B=\begin{matrix} 0&1 \\ 1&0 \end{matrix}$
$A=1 \quad B={1}$
$A+B=0$
answered
Dec 28, 2016
by
Vijay Thakur
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edited
Nov 29, 2017
by
pavan singh
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+6
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ans should be A.
for S1 : singular matrices matrix[0,0,0,1] + matrix[1,0,0,0] = matrix[1,0,0,1], which is non singular
for S2 : nonsingular matrices matrix[1,0,0,1] + matrix[0,1,1,0] = matrix[1,1,1,1], which is singular
answered
Dec 30, 2014
by
jayendra
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but matrices should be square ..?
yes, that is true
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+2
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I think there is a contradiction Let me explain with example
If A = [ 2 10
1 5]
B = [ 3 6
2 4]
Sum [ 5 16
3 9]
 sum = 9×5_16×3= 3 which is not equal to zero . So it is non singular
answered
Nov 18, 2016
by
signals8692
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41
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edited
Dec 28, 2016
by
Vijay Thakur
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+2
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A singular matrix is a square matrix whose determinant is 0. Whenever you solve question related to determinant, think of triangular matrix, like upper triangular matrix, because its determinant is just product of diagonal entries, and so easy to visualize. So definitely, in a singular matrix, one of the entries in diagonal must be zero. Similarly, in nonsingular, none of the entries should be zero.
Now take any matrices and just check .
So we see that both S1 and S2 are true. So option (A) is correct.
answered
Apr 15, 2017
by
ashwina
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Apr 15, 2017
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ashwina
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