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If the matrix [−1x−y4205x+yz6]\begin{bmatrix} -1 & x-y & 4 \\ 2 & 0 & 5 \\ x+y & z & 6 \end{bmatrix} is symmetric, then x+3y+2zx + 3y + 2z is equal to

Solution

✅ Correct Option: 1

A symmetric matrix satisfies the condition that element at position (i,j) equals element at position (j,i).

For the matrix [−1x−y4205x+yz6]\begin{bmatrix} -1 & x-y & 4 \\ 2 & 0 & 5 \\ x+y & z & 6 \end{bmatrix} to be symmetric:

Element (1,2) = Element (2,1): x−y=2x - y = 2

Element (1,3) = Element (3,1): 4=x+y4 = x + y

Element (2,3) = Element (3,2): 5=z5 = z


From the symmetry conditions:

x−y=2x - y = 2 ... (1)

x+y=4x + y = 4 ... (2)

z=5z = 5 ... (3)


Adding equations (1) and (2):

(x−y)+(x+y)=2+4(x - y) + (x + y) = 2 + 4

2x=62x = 6

x=3x = 3


Subtracting equation (1) from equation (2):

(x+y)−(x−y)=4−2(x + y) - (x - y) = 4 - 2

2y=22y = 2

y=1y = 1


The expression x+3y+2zx + 3y + 2z becomes:

x+3y+2zx + 3y + 2z

=3+3(1)+2(5)= 3 + 3(1) + 2(5)

=3+3+10= 3 + 3 + 10

=16= 16

Therefore, x+3y+2z=16x + 3y + 2z = 16

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