Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 4 Ex 4.1 page covers all 8 questions, solved step-by-step — evaluating determinants of order 2 and order 3, the effect of scalar multiplication |kA|=k^n|A| on a determinant, and solving determinant equations for an unknown, exactly the way CBSE Determinants answers are marked.
For a 2\times2 determinant \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad-bc:
\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix} = 2(-1)-4(-5) = -2+20
(i) \cos\theta(\cos\theta)-(-\sin\theta)(\sin\theta) = \cos^2\theta+\sin^2\theta
(ii) (x^2-x+1)(x+1)-(x-1)(x+1)
Expanding the first product: (x^2-x+1)(x+1) = x^3+x^2-x^2-x+x+1 = x^3+1
And the second product: (x-1)(x+1) = x^2-1
So the determinant = (x^3+1)-(x^2-1) = x^3-x^2+2
|A| = 1(2)-2(4) = 2-8 = -6
2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix} \Rightarrow |2A| = 2(4)-4(8) = 8-32 = -24
Also 4|A| = 4(-6) = -24
Since A is upper triangular, |A| is the product of its diagonal entries: |A| = 1\times1\times4 = 4.
3A = \begin{bmatrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{bmatrix}, which is also upper triangular, so |3A| = 3\times3\times12 = 108.
Also 27|A| = 27(4) = 108
(i) Expanding along R_1:
= 3\begin{vmatrix} 0 & -1 \\ -5 & 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} +(-2)\begin{vmatrix} 0 & 0 \\ 3 & -5 \end{vmatrix}
= 3(0-5)+1(0+3)-2(0-0) = -15+3-0
(ii) Expanding along R_1:
= 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) = 21+20+5
(iii) Expanding along R_1:
= 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) = 0+6-6
(iv) Expanding along R_1:
= 2\begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} -(-1)\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} +(-2)\begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}
= 2(0-5)+1(0+3)-2(0-6) = -10+3+12
Expanding along R_1:
|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
(i) LHS = 2(1)-4(5) = 2-20 = -18. RHS = 2x(x)-4(6) = 2x^2-24.
Setting them equal: -18 = 2x^2-24 \Rightarrow 2x^2 = 6 \Rightarrow x^2 = 3 \Rightarrow x = \pm\sqrt{3}
(ii) LHS = 2(5)-3(4) = 10-12 = -2. RHS = x(5)-3(2x) = 5x-6x = -x.
Setting them equal: -2 = -x \Rightarrow x = 2
LHS = x(x)-2(18) = x^2-36. RHS = 6(6)-2(18) = 36-36 = 0.
Setting them equal: x^2-36 = 0 \Rightarrow x^2 = 36 \Rightarrow x = \pm6
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