Question Set A

cbse 12th Standard Maths Determinants Important Questions And Answers Study Material - QB365

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Questions:

Generic 2

4 x 2 = 8

If A and B are matrix of order 3 and IAI = 5, IBI = 3, then find the value of |3AB|.

For what value of \(\lambda\),the matrix \(\left[\begin{array}{rrr} 1 & \lambda & 0 \\ 3 & -1 & 2 \\ 4 & 1 & 5 \end{array}\right]\) is singular?

If \(A=\left| \begin{matrix} 1 \\ 3x \end{matrix}\begin{matrix} 2 \\ -1 \end{matrix} \right| \) and \(B=\left| \begin{matrix} 1 \\ -1 \end{matrix}\begin{matrix} 3 \\ 1 \end{matrix} \right| \), write the value of |AB|.

If A=\(\begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}\) write A^-1 in terms of A .

Generic 5

8 x 5 = 40

Determine the product \(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] ,\) and use it to solve the system of equations: x - y + z = 4, x - 2y - 2z = 9, 2x + y + 3z = 1.

A(adj A) = (adj A)A = |A| I for matrix \(A=\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| .\)

If \(A=\left\lceil \begin{matrix} 2 & 3 \\ 1 & -4 \end{matrix} \right\rceil \), \(B=\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \end{matrix} \right] \), verify that (AB)^-1= B^-1A^-1

\(A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] and\quad B=\left[ \begin{matrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{matrix} \right] \), find the product AB and use this result to solve the following system of linear equations:
2x - y + z = -1
-x + 2y - z = 4
x - y + 2z = -3

The management committee of a residential colony decided to award some of its members (say x) for Honesty, some (say y) for Helping others and some others (say z) for Supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for Cooperation and Supervision added to two times the number of awardees for Honesty is 33. If the sum of the number of awardees for Honest and Supervision is twice the number of awardees for helping others, Using matrix method find the number of awardees of each category. Apart from these values, namely, Honesty, Cooperation and supervision, suggest one more value which the management of the colony must include for awards.

10.

A school wants to award its student or the values of Honesty, Regularity and Hard Work with a total cash award of Rs.6,000. Three times the award money for Hard work added to that given for Honesty amounts to Rs.11,000. The award money given for Honesty and Hard work together is double the one given for regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, namely, Honesty, Regularity and Hard work, suggest one more value which the school must include for awards.

11.

Using matrices solve the following system of linear equations:
x - y + 2z = 7;
3x + 4y - 5z = -5;
2x - y + 3z = 12

12.

If A = \(\begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}\) find A^-1. Using A^-1 solve the following system of equations:

2x - 3y + 5z = 16;
3x + 2y - 4z = -4;
x + y - 2z = -3

Multiple Choice Question

8 x 1 = 8

13.

Let A be a non-angular square matrix of order 3 x 3, then |A . adj A| is equal to

(a)

|A|³

(b)

|A|²

(c)

|A|

(d)

3|A|

14.

If A and B are invertible matrices then which of the following is not correct

(a)

\(A d j A=|A| \cdot A^{-1}\)

(b)

\(\operatorname{det}\left(A^{-1}\right)=(\operatorname{det} A)^{-1}\)

(c)

\((A B)^{-1}=B^{-1} A^{-1}\)

(d)

\((A+B)^{-1}=A^{-1}+B^{-1}\)

15.

Asquare matrix A is said to be non-singular, if

(a)

\(|A|=0\)

(b)

\(|A| \neq 0\)

(c)

\(|A|=-1\)

(d)

\(|A|=1\)

16.

If area of a triangle is 35 sq. units with vertices (2, - 6), (5, 4) and (k, 4),then k is

(a)

(b)

-2

(c)

-12, -2

(d)

12, -2

17.

Area of the triangle whose vertices are (a, b + c), (b, c + a) and (c, a + b), is

(a)

2 sq units

(b)

3 sq unit

(c)

0 sq unit

(d)

None of the above

18.

A and B are invertible matrices of the same order such that |(AB)^-1| = 8, If |A| = 2, then |B| is

(a)

(b)

(c)

(d)

\(\frac{1}{16}\)

19.

Let A be a square matrix of order 2 × 2, then |KA| is equal to

(a)

K|A|

(b)

K²|A|

(c)

K³|A|

(d)

2K|A|

20.

If \(\begin{vmatrix} 2x & -1 \\ 4 & 2 \end{vmatrix}=\begin{vmatrix} 3 & 0 \\ 2 & 1 \end{vmatrix}\) then x is

(a)

(b)

\(\frac { 2 }{ 3 } \)

(c)

\(\frac { 3 }{ 2 } \)

(d)

\(-\frac { 1 }{ 4 } \)

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Answers Key:

Generic 2

4 x 2 = 8

Clearly, |3AB| = 3³|A| |B| = 27 × 5 × 3 = 405
[if matrix A is of order n \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\times\) n, then |kA| = kⁿ|A| and |AB| = |A||B|

Let \(A=\left[\begin{array}{rrr} 1 & \lambda & 0 \\ 3 & -1 & 2 \\ 4 & 1 & 5 \end{array}\right] \)Since the matrix is singular.
\(\therefore |A|=0 \)
\(\Rightarrow\left|\begin{array}{rrr} 1 & \lambda & 0 \\ 3 & -1 & 2 \\ 4 & 1 & 5 \end{array}\right|=0 \)
\(\Rightarrow 1(-5-2)-\lambda(15-8)+0=0 \)
\(\Rightarrow -7-7 \lambda=0 \)
\(\Rightarrow -7 \lambda=7 \)
\(\therefore\lambda=-1\)

\(AB=\left| \begin{matrix} 1 \\ 3 \end{matrix}\begin{matrix} 2 \\ -1 \end{matrix} \right| \left| \begin{matrix} 1 \\ -1 \end{matrix}\begin{matrix} 3 \\ 1 \end{matrix} \right| \)
\(=\left| \begin{matrix} -1 \\ 4 \end{matrix}\begin{matrix} 5 \\ 8 \end{matrix} \right| \)
\(\left| AB \right| =-1\times 8-5\times 4 =-8-20=-28\)

\(A^{-1}=\frac{1}{|A|} \text { adj } A\)
\(|A|=\left[\begin{array}{lr} 2 & 3 \\ 5 & -2 \end{array}\right]=-4-15=-19 \)
\(\Rightarrow A^{-1}=-\frac{1}{19}\left[\begin{array}{rr} -2 & -3 \\ -5 & 2 \end{array}\right]=\frac{1}{19}\left[\begin{array}{cc} 2 & 3 \\ 5 & -2 \end{array}\right]=\frac{1}{19} A \)

Generic 5

8 x 5 = 40

\(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} -4+4+8 & 4-8+4 & -4-8+12 \\ -7+1+6 & 7-2+3 & -7-2+9 \\ 5-3-2 & -5+6-1 & 5+6-3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] =8I\)
where I is the identity matrix
Let \(A=\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \)
and \(B=\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \)
\(\Rightarrow AB=8I\)
Post Multiplyingbith sides by B ^-1, we get
\(AB{ B }^{ -1 }=8I{ B }^{ -1 }\)
\(\Rightarrow A=8{ B }^{ -1 }\)
\(\Rightarrow { B }^{ -1 }=\frac { A }{ 8 } \)
Given Equations are:
x - y + z = 4
x - 2y - 2z = 9
and 2x + y + 3z = 1
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow BX=C\)
where \(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \ and\ C=\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(\Rightarrow X={ B }^{ -1 }C\)
Using \({ B }^{ -1 }=\frac { A }{ 8 } \)
\(\Rightarrow X=\frac { A }{ 8 } C\)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{matrix} \right] \)
\(=\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ -16 \\ -8 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right] \)
(x, y, z) = (3, -2, -1)

Here, \(\left| A \right| =\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \)
= 1(0+0) + 1(9+2) + 2(0-0)
= 11
\(\Rightarrow \left| A \right| =\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| ...(i)\)
\(adj\quad A=\left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \quad \)
\(A(adj\quad A)=\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \)
\(=\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| \)
\((adj\quad A)A=\left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \)
\(=\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| ....(i)\)
Thus, it is verified that A(adj A) = (adj A)A = |A|I

Given \(A=\left\lceil \begin{matrix} 2 & 3 \\ 1 & -4 \end{matrix} \right\rceil \) and \(B=\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \end{matrix} \right] \)
then \(AB=\left[ \begin{matrix} 2 & 3 \\ 1 & -4 \end{matrix} \right] \left[ \begin{matrix} 1 & -2 \\ -1 & 3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 5 \\ 5 & -14 \end{matrix} \right] \)
Taking L.H.S = \((AB)^{ -1 }=\frac { adj(AB) }{ \left| AB \right| } \)
Here adj(AB) \(=\left[ \begin{matrix} -14 & -5 \\ -5 & -1 \end{matrix} \right] \) and |AB| = 14 - 25 = 11
\(\left| AB \right| =14-25=-11\)
\((AB)^{ -1 }=-\frac { 1 }{ 11 } \left[ \begin{matrix} -14 & -5 \\ -5 & -1 \end{matrix} \right] =\frac { 1 }{ 11 } \left[ \begin{matrix} 14 & 5 \\ 5 & 1 \end{matrix} \right] \)
\(adjB=\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right] \)
\(|B|=3-2=1\)
\(adjA=\left[ \begin{matrix} -4 & -3 \\ -1 & 2 \end{matrix} \right] \)
\(\left| A \right| =-8-3=-11\)
\(B^{ -1 }=\frac { 1 }{ \left| B \right| } (adjB)=\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right] \)
\(A^{ -1 }=\frac { 1 }{ 11 } \left[ \begin{matrix} -4 & -3 \\ -1 & 2 \end{matrix} \right] \)
Taking R.H.S = B^-1A^-1
\(=\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \end{matrix} \right] ^{ -1 }\left[ \begin{matrix} 2 & 3 \\ 1 & -4 \end{matrix} \right] ^{ -1 }\)
\(=1\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right] \times \frac { 1 }{ 11 } \left[ \begin{matrix} -4 & -3 \\ -1 & 2 \end{matrix} \right] \)
\(=\frac { 1 }{ 11 } \left[ \begin{matrix} 14 & 5 \\ 5 & 1 \end{matrix} \right] \)
\(L.H.S=R.H.S\)

\((I)AB=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right] \left[ \begin{matrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{matrix} \right] \)
\(=\begin{bmatrix}6-1-1&2-3+1&-2-1+3\\-3+2+1&-1+6-1&1+2-3\\3-1-2&1-3+2&-1-1+6 \end{bmatrix}\)
\(=\begin{bmatrix} 4&0&0\\0&4&0\\0&0&4\end{bmatrix}=4\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix}4I_3\)
\(\Rightarrow A\left({1\over4}B\right)=I_3\Rightarrow A^{-1}={1\over2}B\)
\(\Rightarrow A^{-1}={1\over4}\begin{bmatrix}3&1&-1\\1&3&1\\-1&1&3 \end{bmatrix}\)....(1)
(II) the given system of linear equations is:
2x - y + z = -1
-x + 2y - z = 4
x - y + 2z = -3
These equations can be written as AX = B'
where \(A=\begin{bmatrix} 2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}X=\begin{bmatrix} x\\y\\z\end{bmatrix}\ B'\begin{bmatrix} -1\\4\\-3\end{bmatrix}\)
Now \(|A|=\begin{bmatrix} 2&-1&1\\-1&2&-1\\1&-1&2\end{bmatrix}\)
= 2(4-1) + (1)(-2+1) + (1)(1-2)
= 6-1-1 = 4 ≠ 0
⇒ A is non-singular.
∴ the given system has a unique solution given by:
X = A^-1B'
[∵ AX = B' ⇒ A^-1(AX) = A¹B'
⇒ (A^-1A)X = A^-1B'
⇒ IX = A^-1B' ⇒ X = A^-1B' ]
i.e \(\begin{bmatrix}x\\y\\z \end{bmatrix}={1\over4}\begin{bmatrix} 3&1&-1\\1&3&1\\-1&1&3\end{bmatrix}\begin{bmatrix}1\\4\\3 \end{bmatrix}\)
\(={1\over 4}\begin{bmatrix}-3+4+3\\-1+123\\1+4-9 \end{bmatrix}={1\over4}\begin{bmatrix} 4\\8\\-4\end{bmatrix}=\begin{bmatrix}1\\2\\-1 \end{bmatrix}\)
Hence x = 1, y = 2 and z = -1

Let x,y and z be the number of awardees for honesty, co-operation and supervision respectively Acc to the question.
x + y + z = 12
2x + 3y + 3z = 33
x - 2y + z = 0
Matrix equation is
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 12 \\ 33 \\ 0 \end{matrix} \right] \)
AX = B
|A| = \(\left[ \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 3 \\ 1 & -2 & 1 \end{matrix} \right] \)
= 1(3+6)-1(2-3)+1(-4-3)
\(\therefore\) A-1 exists.
adj A=\(\left[ \begin{matrix} 9 & 1 & -7 \\ -3 & 0 & 3 \\ 0 & -1 & 1 \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{matrix} \right] \)
X=A-1B
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ 3 } \left[ \begin{matrix} 9 & -3 & 0 \\ 1 & 0 & -1 \\ -7 & 3 & 1 \end{matrix} \right] \left[ \begin{matrix} 12 \\ 33 \\ 0 \end{matrix} \right] \)
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 4 \\ 5 \end{matrix} \right] \)
By equality of matrices, we have : x = 3, y = 4, z = 5
i.e., the number of awardees for honesty co-operation and supervision respectively are 3,4 and 4.
Value: Another value which the management can include may be regularity and sincerity.

10.

x=500, y=2000, z=3500

11.

Given equation can be written as
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
=\(\left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] \) or AX = B
|A| = 1(7)+1(19)+2(-11)
= 4\(\neq \)0
\(\therefore \)A^-1 exists
Co-factors
a₁₁= 7, a₁₂ = -19 a₁₃ = -11
a₂₁ = 1, a₂₂ = -1 a₂₃ = -1
a₃₁ = -3, a₃₂ = 11 a₃₃ = 7
\(\Rightarrow { A }^{ -1 }=\frac { 1 }{ 4 } \left[ \begin{matrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{matrix} \right] \)
\(\therefore \) X = A^-1B
\(\therefore \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{matrix} \right] \left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 2 \\ 1 \\ 3 \end{matrix} \right] \)
\(\therefore \) x = 2, y = 1 and z = 3

12.

\(|A| =\left|\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right| \)
\(=2(0)+3(-2)+5(1)=-1 \neq 0 \)
\(\operatorname{Adj} A =\left[\begin{array}{rrr} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{array}\right]^{\top} \)
\(=\left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \)
\(A^{-1} =\frac{1}{|A|} \operatorname{adj} A=-\frac{1}{1}\left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \)
\(=\left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]\)
Consider equations,
\(2 x-3 y+5 z=16 \)
\(3 x+2 y-4 z=-4 \)
\(x+y-2 z=-3\)
Corresponding matrix equation is
\(\left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 16 \\ -4 \\ -3 \end{array}\right]\)
i.e., AX = B, Its solution is X = A ^-1B
\(\text { [from {i} } X=\left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & 13 \end{array}\right]\left[\begin{array}{r} 16 \\ -4 \\ -3 \end{array}\right] \)
\(\text { or } \left.\quad \begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 0-4+6 \\ -32-36+69 \\ -16-20+39 \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right] \)
\(\therefore \quad x=2, y=1, z=3\)

Multiple Choice Question

8 x 1 = 8

13.

(a)
|A|³

14.

(d)
\((A+B)^{-1}=A^{-1}+B^{-1}\)

15.

(b)
\(|A| \neq 0\)

16.

\(\frac{1}{2}\left|\begin{array}{ccc} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{array}\right|=\pm 35\)

17.

Area of triangle, \(\Delta=\frac{1}{2}\left|\begin{array}{lll} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{array}\right|\)

18.

As \(\left| { (AB) }^{ -1 } \right| =\frac { 1 }{ |AB| } =\frac { 1 }{ |A||B| } \)
\(\Rightarrow 8=\frac { 1 }{ 2|B| } \Rightarrow B=\frac { 1 }{ 16 } \)

19.

As if A = \(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\) then \(\left| A \right| =\begin{bmatrix} a & b \\ c & d \end{bmatrix}\)
\(KA=\begin{bmatrix} Ka & Kb \\ Kc & Kd \end{bmatrix}\) and \(\left| KA \right| =\begin{bmatrix} Ka & Kb \\ Kc & Kd \end{bmatrix}\)
\(={ K }^{ 2 }\begin{vmatrix} a & b \\ c & d \end{vmatrix}={ K }^{ 2 }|A|\)

20.

As \(\begin{vmatrix} 2x & -1 \\ 4 & 2 \end{vmatrix}=\begin{vmatrix} 3 & 0 \\ 2 & 1 \end{vmatrix}\)
⇒ 4x + 4 = 3 - 0
⇒ x = \(-\frac { 1 }{ 4 } \)