Find the value of x, if [1x1][1322511532][12x]=0
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We have, [1x1]1×3[1322511532]3×3[12x]3×1=0 ⇒ [1+2x+153+5x+32+x+2]1×3[12x]3×1=0 ⇒ [16 + 2x + (5x + 6)⋅2 + (x + 4)x]1×1 = 0
⇒ [16 + 2x + 10x + 12 + x2 + 4x] = 0
⇒ [x2 + 16x + 28] = 0
⇒ [x2 + 2x + 14x + 28] = 0
⇒ (x + 2)(x + 14) = 0
∴ x = 2, –14