Question
Evaluate the determinant $\left|\begin{array}{cc}x^2-x+1 & x-1 \\ x+1 & x+1\end{array}\right|$
✓
Answer
Applying $C_1\rightarrow C_1-C_2,$
we get, .$\left| {\begin{array}{*{20}{c}} {{x^2} - x + 1}&{x - 1} \\ {x + 1}&{x + 1} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{x^2} - 2x + 2}&{x - 1} \\ 0&{x + 1} \end{array}} \right|$
$= \left( {{x^2} - 2x + 2} \right).\left( {x + 1} \right) - \left( {x - 1} \right).0$
$= x^3 - 2x^2 + 2x + x^2 - 2x + 2$
$= x^3 - x^2 + 2$
we get, .$\left| {\begin{array}{*{20}{c}} {{x^2} - x + 1}&{x - 1} \\ {x + 1}&{x + 1} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {{x^2} - 2x + 2}&{x - 1} \\ 0&{x + 1} \end{array}} \right|$
$= \left( {{x^2} - 2x + 2} \right).\left( {x + 1} \right) - \left( {x - 1} \right).0$
$= x^3 - 2x^2 + 2x + x^2 - 2x + 2$
$= x^3 - x^2 + 2$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Evaluate the definite integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
→If $A_{ij}$ is the cofactor of the element $a_{ij}$ of the determinant$\begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix},$ then write the value of $a_{32}. A_{32}.$
→In the adjacent figure, find the value of $\overrightarrow{ OB }$.
→Prove that $\left|\begin{array}{cc}x^2+x+1 & x+1 \\ x & x-1\end{array}\right| =x^3-x^2-x-1$
→Let $A = \left[\begin{array}{ccc} {1} & {\sin \theta} & {1} \\ {-\sin \theta} & {1} & {\sin \theta} \\ {-1} & {-\sin \theta} & {1} \end{array}\right]$ where $0 \leq \theta \leq 2 \pi$. Then
→Evaluate the following:
$\sin^{-1}(\sin4)$
→$\sin^{-1}(\sin4)$
Write a vector of magnitude 9 units in the direction of vector$-2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
→Find the value of $\tan^{-1}\Big(\tan\frac{9\pi}{8}\Big)$
→Evaluate: $\int\text{(1-x)}\sqrt{x}\text{dx}.$
→Evaluate the determinant $\left|\begin{array}{ccc}3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0\end{array}\right|$
→