1.

For which value of x, matrix \(\begin{bmatrix}1&-2&3\\[0.3em]1&2&1\\[0.3em]x&2&-3\end{bmatrix}\) is singular ?

Answer»

Then   \(\begin{bmatrix}1&-2&3\\[0.3em]1&2&1\\[0.3em]x&2&-3\end{bmatrix}\) = 0

⇒ 1 (- 6 – 2) + 2 (-3 – x) + 3(2 – 2x) = 0

⇒ -8 – 6 – 2x + 6 – 6x = 0

⇒ -8 = 8

⇒ x = 8/-8 = -1

Hence, x = -1



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