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Question 102 of 110

Q.If A=[1−23121x2−3]A = \begin{bmatrix}1 & -2 & 3\\1 & 2 & 1\\x & 2 & -3\end{bmatrix} is singular, find the value of xx.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2023Subjective· 2mImportance★★★★★
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Expanding det⁡(A)\det(A) along the first row and setting it to 0 gives x=−1x=-1.

A=[1−23121x2−3]A=\begin{bmatrix}1&-2&3\\1&2&1\\x&2&-3\end{bmatrix}

Expand along the first row:

det⁡(A)=1(2(−3)−1(2))−(−2)(1(−3)−1(x))+3(1(2)−2(x))\det(A) = 1\big(2(-3)-1(2)\big) - (-2)\big(1(-3)-1(x)\big) + 3\big(1(2)-2(x)\big)

=1(−6−2)+2(−3−x)+3(2−2x)= 1(-6-2) + 2(-3-x) + 3(2-2x)

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