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If the system of equations $2 x + 3 k y + (3 k + 4) z = 0,$ $x + (k + 4) y + (4 k + 2) z = 0,$ $x + 2 (k + 4) y + (3 k + 4) z = 0$ has a non-trivial solution, then $k =$

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−8 or 12

8 or −12

−4 or 12

4 or −12

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Correct option is A

The system has non-trivial solution⇒Δ=0⇒23k3k+41k+44k+212(k+4)3k+4=0, Apply R2→2R2-R1, R3→2R3-R1⇒23k3k+40-k+85k0k+163k+4=0⇒2[(-k+8)(3k+4)-5k(k+16)]=0⇒-8k2-60k+32=02k2+15k-8=0∴k=-8 or 12.

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If the system of equations 2x+3ky+(3k+4)z=0, x+(k+4)y+(4k+2)z=0, x+2(k+4)y+(3k+4)z=0 has a non-trivial solution, then k=

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Ready to Test Your Skills?

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If the system of equations $2 x + 3 k y + (3 k + 4) z = 0,$ $x + (k + 4) y + (4 k + 2) z = 0,$ $x + 2 (k + 4) y + (3 k + 4) z = 0$ has a non-trivial solution, then $k =$