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# Determinants

Determinant is a continuation of the previous chapter matrix. Matrix and determinant are the core topics of algebra. These will help us to solve many algebraic linear equations very easily. Matrix and determinants both are interlinked. Topics which are covered in this chapter - determinant of a square matrix (upto 3x3), minors. cofactors, finding the area of triangle, adjoint and inverse of square matrix, consistency & inconsistency and number of solutions of system of linear equation, solving system of linear equation in two or three variables using inverse of a matrix.

•  Q1 Evaluate the determinants \begin{vmatrix} \mathbf{2} & \mathbf{4} \\ \mathbf{-5} & \mathbf{-1} \end{vmatrix} Ans: $\begin{vmatrix} \mathbf{2} & \mathbf{4} \\ \mathbf{-5} & \mathbf{-1} \end{vmatrix}$ = 2(−1) − 4(−5) = − 2 + 20 = 18 Q2 Evaluate the determinants (i) \begin{vmatrix} \mathbf{Cosθ} & \mathbf{−sin θ} \\ \mathbf{sin θ} & \mathbf{cos θ} \end{vmatrix}(ii) \begin{vmatrix} \mathbf{x^2 − x + 1} & \mathbf{x − 1} \\ \mathbf{x + 1} & \mathbf{x + 1} \end{vmatrix} Ans: (i) \begin{vmatrix} \mathbf{Cosθ} & \mathbf{−sin θ} \\ \mathbf{sin θ} &  \mathbf{cos θ} \end{vmatrix} = (cos θ)(cos θ) − (−sin θ)(sin θ) = cos2 θ+ sin2 θ = 1   (ii) \begin{vmatrix} \mathbf{x^2 − x + 1} & \mathbf{x − 1} \\ \mathbf{x + 1} &  \mathbf{x + 1} \end{vmatrix} = (x2 − x + 1)(x + 1) − (x − 1)(x + 1) = x3 − x2 + x + x2 − x + 1 − (x2 − 1) = x3 + 1 − x2 + 1 = x3 − x2 + 2 Q3 If A=$$\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}$$, then show that |2A| = 4|A| Ans: The given matrix is $$u=\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}$$    So 2A = 2$$\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}$$             $$= \begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}$$   so L.H.S. = |2A| $$= \begin{bmatrix}2 & 4\\8 & 4\end{bmatrix}$$                  = 2 x 4 - 4 x 8                 = 8 - 32                  = -24   Now, |A| $$= \begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}$$   = 1 x 2 - 2 x 4 = 2 - 8 = -6   So R.H.S. = 4 |A| = 4 x (-6) = -24   So L.H.S. = R.H.S. Q4 If A=$$\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}$$, then show that |3A| = 27|A|. Ans: The given matrix is   A=$$\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}$$   It can be observed that in the first column, two entries are zero. Thus, we expand along the first column (C1) for easier calculation.   | A| = 1$$\begin{vmatrix}1 & 2\\0 & 4\end{vmatrix}$$ - 0$$\begin{vmatrix}0 & 1\\0 & 4\end{vmatrix}$$ + 0$$\begin{vmatrix}0 &1\\1 & 2\end{vmatrix}$$ = 1(4 – 0) – 0 + 0 = 4   So 27 |A| = 27 (4) = 108 ……. (i)      Now 3A = 3$$\begin{bmatrix}1 & 0 & 1\\0 & 1 & 2\\0 & 0 & 4\end{bmatrix}$$=$$\begin{bmatrix}3 & 0 & 3\\0 & 3 & 6\\0 & 0 & 12\end{bmatrix}$$    So |3A| = 3$$\begin{vmatrix}3 & 6\\0 & 12\end{vmatrix}$$ - 0$$\begin{vmatrix}0 & 3\\0 & 12\end{vmatrix}$$ + 0$$\begin{vmatrix}0 &3\\0 & 6\end{vmatrix}$$               =  3 (36 – 0) = 3(36) 108 ……….. (ii)   From equations (i) and (ii), we have: |3A| = 27|A| Q5 Evaluate the determinants (i) $$\begin{vmatrix}3 & -1 & -2\\0 & 1 & 2\\0 & 0 & 4\end{vmatrix}$$ (iii) $$\begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}$$ (ii) $$\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}$$(iv) $$\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}$$ Ans: (i) Let A = $$\begin{vmatrix}3 & -1 & -2\\0 & 1 & 2\\0 & 0 & 4\end{vmatrix}$$   It can be observed that in the second row, two entries are zero. Thus, we expand along the second row for easier calculation.   |A|  = -0$$\begin{vmatrix}-1 & -2\\-5 & 0\end{vmatrix}$$ + 0$$\begin{vmatrix}3 & -2\\3 & 0\end{vmatrix}$$ – (-1)$$\begin{vmatrix}3 &-1\\3 &-5\end{vmatrix}$$ = (-15 + 3) = -12   (ii) Let A = $$\begin{vmatrix}0 & 1 & 2\\-1 & 0 & -3\\-2 & 3 & 0\end{vmatrix}$$   By expanding along the first row, we have:   |A|  = 3$$\begin{vmatrix}1 & -2\\3 & 1\end{vmatrix}$$ + 4$$\begin{vmatrix}1 & -2\\2 & 1\end{vmatrix}$$  + 5$$\begin{vmatrix}1 &1\\2 &3\end{vmatrix}$$ = 3 (1+6) + 4(1+4) + 5(3-2) = 3 (7) + 4 (5) + 5 (1) = 21 + 20 + 5 = 46   (iii) Let  A = $$\begin{vmatrix}3 & -4 & 5\\1 & 1 & -2\\2 & 3 & 1\end{vmatrix}$$   By expanding along the first row, we have:   |A|  = 0$$\begin{vmatrix}0 & -3\\3 & 0\end{vmatrix}$$ - 1$$\begin{vmatrix}-1 & -3\\-2 & 0\end{vmatrix}$$  + 2$$\begin{vmatrix}-1 & 0\\-2 &3\end{vmatrix}$$ = 0 – 1(0 – 6) + 2 (-3 - 0) = -1 (-6) + 2(-3) = 6 – 6 = 0   (iv) Let  A = $$\begin{vmatrix}2 & -1 & -2\\0 & 2 & -1\\3 & -5 & 0\end{vmatrix}$$   By expanding along the first column, we have:   |A|  = 2$$\begin{vmatrix}2 & -1\\-5 & 0\end{vmatrix}$$ - 0$$\begin{vmatrix}-1 & -2\\-5 & 0\end{vmatrix}$$  + 3$$\begin{vmatrix}-1 & -2\\2 & -1\end{vmatrix}$$   = 2(0 – 5) – 0 + 3(1 + 4) = -10 + 15 = 5 Q6 If A = $$\begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}$$, Find |A| Ans: Let  A = $$\begin{bmatrix}1 & 1 & -2\\2 & 1 & -3\\5 & 4 & -9\end{bmatrix}$$   By expanding along the first row, we have:   |A|  = 1$$\begin{vmatrix}1 & -3\\4 & -9\end{vmatrix}$$ - 1$$\begin{vmatrix}2 & -3\\5 & -9\end{vmatrix}$$  -  2$$\begin{vmatrix}2 & 1\\5 & 4\end{vmatrix}$$ = 1(-9 + 12) – 1(-18 + 15) -2(8 – 5) = 1(3) – 1 (-3) – 2(3) = 3 + 3 – 6 = 6 – 6 = 0 Q7 Find values of x, if (i) $$\begin{vmatrix}2 & 4\\2 & 1\end{vmatrix}$$ = $$\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}$$ (ii) $$\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}$$ = $$\begin{vmatrix}x & 3\\2x & 5\end{vmatrix}$$ Ans: (i) $$\begin{vmatrix}2 & 4\\2 & 1\end{vmatrix}$$ = $$\begin{vmatrix}2x & 4\\6 & x\end{vmatrix}$$ ⇒2 x 1 – 5 x 4 = 2x x x – 6 x 4) ⇒ 2- 20 = 2x2 – 24 ⇒2x2 = 6 ⇒ x2 = 3 ⇒ x = ±√3   (ii) $$\begin{vmatrix}2 & 3\\4 & 5\end{vmatrix}$$ = $$\begin{vmatrix}x & 3\\2x & 5\end{vmatrix}$$ ⇒ 2 x 5 – 3 x 4 = x x 5 – 3 x 2x ⇒10  – 12  = 5x – 6x ⇒  -2  = -x ⇒ x = 2