Exercise-4.1, Class 12th, Maths, Chapter 4, NCERT

DETERMINANTS

Question 1

Evaluate ${\displaystyle \mid \begin{array}{cc}2& 4\\ -5& -1\end{array}\mid }$

Solution.
$\det =2(-1)-4(-5)=-2+20=\boxed{{\displaystyle 18}}\mathrm{}$$\mathrm{}$

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Question 2

(i) ${\displaystyle \mid \begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\mid }$

Solution 2 (i):

$$\det =\cos \theta \cdot \cos \theta -(-\sin \theta )\cdot \sin \theta ={\cos }^2\theta +{\sin }^2\theta =\boxed{{\displaystyle 1}}\mathrm{.}$$

Question 2 (ii):

$$\mid \begin{array}{cc}{x}^2-x+1& x-1\\ x+1& x+1\end{array}\mid \mathrm{}$$Solution:

We know that for a $2\times 2$ determinant.$$\mathrm{}$$

Now compute:

$$\det =(x^2-x+1)(x+1)-(x-1)(x+1)\mathrm{.}$$

Step 1: Expand both terms

$$(x^2-x+1)(x+1)=x^3+x^2-x^2-x+x+1=x^3+1$$
$$(x-1)(x+1)=x^2-1$$

Step 2: Substitute back

$$\det =(x^3+1)-(x^2-1)=x^3-x^2+2$$

Question 3

If $A=\left(\begin{array}{cc}1& 2\\ 4& 2\end{array}\right)$, show $\mathrm{\mid }2A\mathrm{\mid }=4\mathrm{\mid }A\mathrm{\mid }$

Solution.
For an $n\times n$ matrix scaling by scalar $k$ multiplies the determinant by $k^n$. Here $n=2$ and $k=2$. So $\mathrm{\mid }2A\mathrm{\mid }=2^2\mathrm{\mid }A\mathrm{\mid }=4\mathrm{\mid }A\mathrm{\mid }$$\boxed{{\displaystyle }}$

(Direct check: $\mathrm{\mid }A\mathrm{\mid }=1\cdot 2-2\cdot 4=2-8=-6$. Then $\mathrm{\mid }2A\mathrm{\mid }=\det \left(\begin{array}{cc}2& 4\\ 8& 4\end{array}\right)=2\cdot 4-4\cdot 8=8-32=-24=4(-6)$

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Question 4

If $A=\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 4\end{array}\right)$, show $\mathrm{\mid }3A\mathrm{\mid }=27\mathrm{\mid }A\mathrm{\mid }$

Solution.
$A$ is $3\times 3$. Scaling by 3 multiplies determinant by $3^3=27$. Hence $\mathrm{\mid }3A\mathrm{\mid }=27\mathrm{\mid }A\mathrm{\mid }$. $\boxed{{\displaystyle }}$​

(One can check: $\mathrm{\mid }A\mathrm{\mid }$ is product of diagonal entries $1\cdot 1\cdot 4=4$.

$\mathrm{\mid }3A\mathrm{\mid }$ has diagonal $3,3,12$ so determinant $3\cdot 3\cdot 12=108=27\cdot 4$

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Question 5 — Evaluate the following determinants

(i) ${\displaystyle \mid \begin{array}{ccc}3& -1& -2\\ 0& 0& -1\\ 3& -5& 0\end{array}\mid }$

Solution (5.i).
Expand along second row (only one nonzero entry $a_{23}=-1$):

Minor for $a_{23}$ is $\mid \begin{array}{cc}3& -1\\ 3& -5\end{array}\mid =3(-5)-(-1)3=-15+3=-12$

Cofactor $C_{23}=(-1{)}^{2+3}\cdot (-12)=-1\cdot (-12)=12$. Then determinant = $a_{23}{C}_{23}=(-1)\cdot 12=\boxed{{\displaystyle -12}}\mathrm{}$

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(ii) ${\displaystyle \mid \begin{array}{ccc}3& -4& 5\\ 1& 1& -2\\ 2& 3& 1\end{array}\mid }$

Solution (5.ii).
Compute by expansion / rule of Sarrus:

$$\det =3\mid \begin{array}{cc}1& -2\\ 3& 1\end{array}\mid -(-4)\mid \begin{array}{cc}1& -2\\ 2& 1\end{array}\mid +5\mid \begin{array}{cc}1& 1\\ 2& 3\end{array}\mid \mathrm{.}$$

Evaluate minors: $\mathrm{}$$\mathrm{}\cdot \mathrm{}\mathrm{}$
So $\det =3\cdot 7+4\cdot 5+5\cdot 1=21+20+5=\boxed{{\displaystyle 46}}\mathrm{}$

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(iii) ${\displaystyle \mid \begin{array}{ccc}0& 1& 2\\ -1& 0& -3\\ -2& 3& 0\end{array}\mid }$

Solution (5.iii).
Compute (expanding first row):

$$\det =0\cdot (\cdots )-1\cdot \mid \begin{array}{cc}-1& -3\\ -2& 0\end{array}\mid +2\cdot \mid \begin{array}{cc}-1& 0\\ -2& 3\end{array}\mid \mathrm{}$$

First minor: $(-1)\cdot 0-(-3)(-2)=0-6=-6$; the corresponding contribution is $-1\cdot (-6)=6$
Second minor: $(-1)\cdot 3-0\cdot (-2)=-3$; contribution $2\cdot (-3)=-6$. Sum $6-6=\boxed{{\displaystyle 0}}\mathrm{}$

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(iv) ${\displaystyle \mid \begin{array}{ccc}2& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\mid }$

Solution (5.iv).
Expand along first row:

$$\det =2\mid \begin{array}{cc}2& -1\\ -5& 0\end{array}\mid -(-1)\mid \begin{array}{cc}0& -1\\ 3& 0\end{array}\mid +(-2)\mid \begin{array}{cc}0& 2\\ 3& -5\end{array}\mid \mathrm{}$$

Compute minors: first $=2\cdot 0-(-1)(-5)=0-5=-5$ ⇒ $2(-5)=-10$
second minor $=0\cdot 0-(-1)\cdot 3=3$ ⇒ sign gives $+1\cdot 3=+3$
third minor $=0\cdot (-5)-2\cdot 3=-6$ ⇒ multiplied by $-2$ gives $+12$.
Sum $-10+3+12=\boxed{{\displaystyle 5}}\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$

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Question 6

If $A=\left(\begin{array}{ccc}1& 1& -2\\ 2& 1& -3\\ 5& 4& -9\end{array}\right)$, find $\mathrm{\mid }A\mathrm{\mid }$

Solution.
Expand along first row:

$$\mathrm{\mid }A\mathrm{\mid }=1\mid \begin{array}{cc}1& -3\\ 4& -9\end{array}\mid -1\mid \begin{array}{cc}2& -3\\ 5& -9\end{array}\mid +(-2)\mid \begin{array}{cc}2& 1\\ 5& 4\end{array}\mid \mathrm{}$$

Compute minors: first = $1(-9)-(-3)4=-9+12=3$
second = $2(-9)-(-3)5=-18+15=-3$. With the minus sign yields $-1\cdot (-3)=+3$.
third minor = $2\cdot 4-1\cdot 5=8-5=3$, times $-2$ gives $-6$. Sum $3+3-6=\boxed{{\displaystyle 0}}\mathrm{}$

So $\mathrm{\mid }A\mathrm{\mid }=0$

Question 7 (i):

$$\text{Find the value of }x\text{ if }\mid \begin{array}{cc}2& 4\\ 5& 1\end{array}\mid =\mid \begin{array}{cc}2x& 4\\ 6& x\end{array}\mid \mathrm{}$$

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Solution:

For any $2\times 2$ matrix
$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,

$$\det =ad-bc\mathrm{}$$Step 1: Compute the determinant on the left-hand side (LHS)

$$\mid \begin{array}{cc}2& 4\\ 5& 1\end{array}\mid =(2)(1)-(4)(5)=2-20=-18$$Step 2: Compute the determinant on the right-hand side (RHS)

$$\mid \begin{array}{cc}2x& 4\\ 6& x\end{array}\mid =(2x)(x)-(4)(6)=2x^2-24$$

Step 3: Set them equal

$$-18=2x^2-24.$$

Simplify:

$$2x^2=-18+24=6.$$
$$x^2=3.$$
$$\boxed{{\displaystyle x=\pm \sqrt{3}}}\mathrm{}$$

Question 7 (ii):

$$\text{Find the value of }x\text{ if }\mid \begin{array}{cc}2& 3\\ 4& 5\end{array}\mid =\mid \begin{array}{cc}x& 3\\ 2x& 5\end{array}\mid \mathrm{}$$

Solution:

For any $2\times 2$ matrix
$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,

$$\det =ad-bc\mathrm{.}$$

Step 1: Compute LHS

$$\mid \begin{array}{cc}2& 3\\ 4& 5\end{array}\mid =(2)(5)-(3)(4)=10-12=-2$$

Step 2: Compute RHS

$$\mid \begin{array}{cc}x& 3\\ 2x& 5\end{array}\mid =(x)(5)-(3)(2x)=5x-6x=-x\mathrm{}$$

Step 3: Equate LHS and RHS

$$-2=-x\Rightarrow x=2$$

 

Question 8:

$$\text{If }\mid \begin{array}{cc}x& 2\\ 18& x\end{array}\mid =\mid \begin{array}{cc}6& 2\\ 18& 6\end{array}\mid ,\text{ then find }x\mathrm{}$$

Options:
(A) $6$ (B) $\pm 6$ (C) $-6$ (D) $0$

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Solution:

For any $2\times 2$ determinant
$\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$,

$$\det =ad-bc\mathrm{.}$$

Step 1: Compute LHS

$$\mid \begin{array}{cc}x& 2\\ 18& x\end{array}\mid =x\cdot x-2\cdot 18=x^2-36$$Step 2: Compute RHS

$$\mid \begin{array}{cc}6& 2\\ 18& 6\end{array}\mid =6\cdot 6-2\cdot 18=36-36=0$$

Step 3: Equate LHS and RHS

$$x^2-36=0$$

$$x^2=36$$

$$x=\pm 6$$

Final Answer:

$$\boxed{{\displaystyle (B)x=\pm 6}}$$