
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)
= cos^{2} θ+ sin^{2} θ
= 1
(ii) \begin{vmatrix} \mathbf{x^2 − x + 1} & \mathbf{x − 1} \\ \mathbf{x + 1} & \mathbf{x + 1} \end{vmatrix}
= (x^{2} − x + 1)(x + 1) − (x − 1)(x + 1)
= x^{3} − x^{2} + x + x^{2} − x + 1 − (x^{2} − 1)
= x^{3} + 1 − x^{2} + 1
= x^{3} − x^{2} + 2
Q3 If A=\(\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\), then show that 2A = 4A 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= 6So R.H.S. = 4 A = 4 x (6) = 24So 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 = 27A. 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 = 27A
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(32)
= 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 = 2x^{2} – 24
⇒2x^{2} = 6
⇒ x^{2} = 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