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Questions:

Multiple Choice Question

5 x 1 = 5

The inverse matrix of \(\begin{pmatrix} 3 & 1 \\ 5 & 2\end{pmatrix}\) is ________.

(a)

\(\begin{pmatrix} 2 & -1 \\-5 & 3 \end{pmatrix}\)

(b)

\(\begin{pmatrix} -2 & 5 \\1 & -3 \end{pmatrix}\)

(c)

\(\begin{pmatrix} 3 & -1 \\-5 & -3 \end{pmatrix}\)

(d)

\(\begin{pmatrix} -3 & 5 \\1 & -2 \end{pmatrix}\)

Which of the following matrix has no inverse.

(a)

\(\begin{pmatrix} -1 & 1 \\ 1 &-4 \end{pmatrix}\)

(b)

\(\begin{pmatrix} 2 & -1 \\ -4 &2 \end{pmatrix}\)

(c)

\(\begin{pmatrix} cos\ a & sin\ a \\ -sin\ a & cos\ a \end{pmatrix}\)

(d)

\(\begin{pmatrix} sin\ a & cos\ a \\ -cos\ a & sin\ a \end{pmatrix}\)

The value of the determinant \({\begin{vmatrix} a & 0 & 0 \\ 0 & a & 0 \\ 0 & 0 & c \end{vmatrix}}^{2}\)is ________.

(a)

abc

(b)

(c)

a²b²c²

(d)

-abc

If \(\triangle=\begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix}\) then \(\begin{vmatrix} 3 & 1 & 2 \\ 1 & 2 & 3 \\ 2 & 3 & 1 \end{vmatrix}\) is ________.

(a)

\(\triangle\)

(b)

-\(\triangle\)

(c)

3\(\triangle\)

(d)

-3\(\triangle\)

The value of x if \(\begin{vmatrix} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{vmatrix}=0\) is_________.

(a)

0, - 1

(b)

0, 1

(c)

- 1, 1

(d)

- 1, - 1

Generic 2

6 x 2 = 12

Using the property of determinant, evaluate \(\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}.\)

Find the values of x if \(\begin{vmatrix} 2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4\\6 & x \end{vmatrix}.\)

If \(A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}\) then show that |2A| = 4 |A|.

The technology matrix of an economic system of two industries is \(\begin{bmatrix} 0.50 & 0.25 \\ 0.40 & 0.67 \end{bmatrix}\). Test whether the system is viable as per Hawkins-Simon conditions.

10.

The technology matrix of an economic system of two industries is\(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\) .Test whether the system is viable as per Hawkins-Simon conditions.

11.

The technology matrix of an economic system of two industries is\(\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}\). Test whether the system is viable as per Hawkins Simon conditions.

Generic 5

3 x 5 = 15

12.

Let a, b and c denote the sides BC, CA and AB respectively of \(\Delta\) ABC. If \(\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{matrix} \right| =0\), then find the value of sin² A + sin²B + sin²C.

13.

If A = \(\begin{bmatrix}3 & -1 & 1 \\ -15 & 6 & -5\\5 & -2 & 2 \end{bmatrix}\) then, find the Inverse of A.

14.

Evaluate:\(\begin{vmatrix} 1&a&a^2-bc\\1&b&b^2-ca\\1&c&c^2-ab \end{vmatrix}\)

Generic 3

6 x 3 = 18

15.

Show that \(\begin{vmatrix} a & a+b&a+b+c \\2a &3a+2b &4a+3b+2c\\3a&6a+3b&10a+6b+3c \end{vmatrix}=a^3.\)

16.

if A =\(\left[ \begin{matrix} cos\ \alpha & sin\ \alpha \\ -sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right] \) is such that A^T = A^-1, find \(\alpha\)

17.

Using matrix method, solve x + 2y + z = 7, x + 3z = 11 and 2x - 3y =1.

18.

Show that \(\begin{vmatrix}0 &ab^2 &ac^2 \\a^2b & 0 & bc^2\\a^2c&b^2c&0\end{vmatrix}=2a^3b^3c^3.\)

19.

Without actual expansion show that the value of the determinant \(\begin{vmatrix}5 &5^2 &5^3 \\5^2 & 5^3 & 5^4\\5^4&5^5&5^6 \end{vmatrix}\)is zero.

20.

Solve: \(\begin{vmatrix} x & 2 & -1 \\ 2 & 5 & x \\ -1 & 2 & x \end{vmatrix}=0.\)

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Answers Key:

Multiple Choice Question

5 x 1 = 5

\(|A|=6-5=1\)
\(A^{-1}=\left(\begin{array}{cc} 2 & -1 \\ -5 & 3 \end{array}\right)\)

\(\text {Since }|A|=4-4=0\)

(b)
-\(\triangle\)

\(\left|\begin{array}{lll} 0 & 1 & 0 \\ x & 2 & x \\ 1 & 3 & x \end{array}\right|=0 \Rightarrow-1\left[x^2-x\right]=0\)
\(\Rightarrow x(x-1)=0 \Rightarrow x=0,1\)

Generic 2

6 x 2 = 12

Let |A| = \(\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}\)
Taking 2 common from C₁ and 4 common from C₃, we get,
\(|A|=2\times4\begin{vmatrix} 3 & 5&3 \\ 1 & 4 & 1\\1 &1 &1\end{vmatrix}=8\times 0\ [\because C_1\equiv C_3]=0\)

Given \(\begin{vmatrix}2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\Rightarrow\) 2 - 20 = 2x² - 24
\(\Rightarrow\) -18 = 2x² - 24
\(\Rightarrow\) -18 + 24 = 2x²
\(\Rightarrow\) 6 = 2x²
\(\Rightarrow\) x² = 3
\(\Rightarrow\) x = \(\pm\sqrt{3}\)

Given = \(\begin{bmatrix} 1 & 2 \\ 4 & 2\end{bmatrix},\) then 2A = \(\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}\)
\(\therefore\) |2A| = \(\begin{vmatrix}2 & 4 \\8 & 4 \end{vmatrix}\) = 8 - 32 = -24 ...(1)
Also, |A| = \(\begin{vmatrix} 1&2 \\4 & 2 \end{vmatrix}\) = 2 - 8 = -6
\(\therefore\) 4|A| = 4(-6) = -24 ....(2)
From (1) and (2), |2A| = 4.|A|

The technology matrix is
\(B=\begin{bmatrix} 0.50 & 0.25 \\ 0.40 & 0.67 \end{bmatrix}\)
I - B =\(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.50 & 0.25 \\ 0.40 & 0.67 \end{bmatrix}=\begin{bmatrix} 0.50 & -0.25 \\ -0.40 & 0.33 \end{bmatrix}\)
\(|\mathrm{I}-\mathrm{B}|\) = 0.165 - 0.1 = 0.065 > 0
Since the diagonal elements of (I - B) are positive and |I - B| is positive, Hawkins - Simon conditions are satisfied.
Therefore the given system is viable.

10.

B = \(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\)
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}=\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
|I - B| =\(\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
= 0.08 - 0.18 = - 0.1 < 0
Since |I - B| is negative, Hawkins - Simon conditions are not satisfied.
Therefore the given system is not viable.

11.

B \(=\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}\)
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.50 & 0.30 \\ 0.41 & 0.33 \end{bmatrix}=\begin{bmatrix} 0.50 & -0.30 \\ -0.41 & 0.67 \end{bmatrix}\)
= (0.50) (0.67) - (0.30) (0.41)
\(|I-B|\) = 0.335 - 0.123 = 0.212 > 0
Since the main diagonal elements of I - B are positive and |I-B| is positive. Hawkins Simon conditions are satisfied. Therefore given system is viable

Generic 5

3 x 5 = 15

12.

Give A = \(\left| \begin{matrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{matrix} \right| =0\)
Applying R₂ \(\rightarrow\) R₂-R₁ and R₃ \(\rightarrow\) R₃ - R₁
we get A = \(\left| \begin{matrix} 1 & a & b \\ 0 & c-a & a-b \\ 0 & b-a & c-b \end{matrix} \right| =0\)
Expanding along C₁ we get
\(1\left| \begin{matrix} c-a & a-b \\ b-a & c-b \end{matrix} \right| =0\)
\(\Rightarrow\) (c - a) (c - b) - (b - a) (a - b) = 0
\(\Rightarrow\) c² - bc - ac + ab - (ab - b²- a²+ ab) = 0
\(\Rightarrow\) a² + b²+ c²- ab - bc - ca = 0
Multiplying both sides by 2 we get, 2a²+ 2b²+ 2c²- 2ab - 2bc - 2ca = 0
\(\Rightarrow\) (a - b)²+ (b - c)²+ (c - a)²= 0
\(\Rightarrow\) a = b = 0, b - c = 0,c - a = 0
\(\Rightarrow\) a = b = c
\(\Rightarrow\) \(\Delta\) ABC is equilateral.
\(\therefore A=B=C=\frac { \pi }{ 3 } \)
\(\therefore { sin }^{ 2 }A+{ sin }^{ 2 }B+{ sin }^{ 2 }C=3{ sin }^{ 2 }\frac { \pi }{ 3 } =3{ \left( sin\frac { \pi }{ 3 } \right) }^{ 2 }=3{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }^{ 2 }=3\times \frac { 3 }{ 4 } =\frac { 9 }{ 4 } \)

13.

\(A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right)\)
\(|A|=3(12-10)+1(-30+25)+1(30-30)\)
\(=6-5=1 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right)\)

14.

Let A \(=\begin{vmatrix} 1 & a&a^2&-bc \\1 &b&{b}^{2}&-ca\\1&c&c^2&-ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|+\left|\begin{array}{ccc} 1 & a & -b c \\ 1 & b & -c a \\ 1 & c & -a b \end{array}\right|\)
\(A=\begin{vmatrix} 1 & a&{a}^{2} \\ 1 &b&b^2\\1&c&c^2 \end{vmatrix}-\begin{vmatrix} 1 & a&bc \\1 &b&ca\\1&c&ab \end{vmatrix}\)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{1}{a b c}\left|\begin{array}{ccc} a & a^2 & a b c \\ b & b^2 & a b c \\ c & c^2 & a b c \end{array}\right|\)
(Multiplying R₁,R₂ and R₃ of II det by a, b, c respectively)
\(=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\frac{a b c}{a b c}\left|\begin{array}{lll} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{array}\right|\)
\(\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|-\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|=0\)

Generic 3

6 x 3 = 18

15.

LHS = \(\begin{vmatrix} a & a+b&a+b+c \\2a &3a+2b &4a+3b+2c\\3a&6a+3b&10a+6b+3c \end{vmatrix}\)
Applying R2 \(\rightarrow\) R2 - 2R1 and R3 \(\rightarrow\) 3R1 we get,
\(=\begin{vmatrix} a & a+b&a+b+c \\0 &0 &2a+b\\0&3a&7a+3b\end{vmatrix}\)

Expanding along C₁ we get
\(=a\begin{vmatrix}a&2a+b\\3a&7a+b\end{vmatrix}-0+0\)
= a(7a² + 3ab - 6a² - 3ab)
= a(7a² - 6a²) = a(a²) = a³ = RHS
Hence proved.

16.

Given that A^T= A^-1
\(\Rightarrow\) AA^T = AA^-1
\(\Rightarrow\) AA^T^{= I}
Now, AA^T = \(\left[ \begin{matrix} cos\ \alpha & sin\ \alpha \\ -sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right] \)\(\left[ \begin{matrix} cos\ \alpha & -sin\ \alpha \\ sin\ \alpha & \ cos\ \alpha \ \end{matrix} \right] \)
\(=\left[ \begin{matrix} { cos }^{ 2 }\alpha +{ sin }^{ 2 }\alpha & -sin\alpha cos\alpha +sin\alpha cos\alpha \\ -sin\alpha cos\alpha +sin\alpha cos\alpha & +{ sin }^{ 2 }\alpha +{ cos }^{ 2 }\alpha \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)[\(\because\) sin²\(\alpha\) + cos²\(\alpha\) = 1]
Thus, AA^T = I is true for all \(\alpha\).
Hence \(\alpha\) can take any real value.

17.

The system of equations can be written in the form AX = B where,
\(A=\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 7 \\ 11 \\ 1 \end{matrix} \right] \)
Now, |A| = \(\left| \begin{matrix} 1 & 2 & 1 \\ 1 & 0 & 3 \\ 2 & -3 & 0 \end{matrix} \right| =1\left| \begin{matrix} 0 & 3 \\ -3 & 0 \end{matrix} \right| -2\left| \begin{matrix} 1 & 3 \\ 2 & 0 \end{matrix} \right| +1 \left| \begin{matrix} 1 & 0 \\ 2 & -3 \end{matrix} \right| \)
= 1(0 + 9) - 2(0 - 6) + 1(-3 - 0) = 9 + 12 - 3 = 18 \(\neq \) 0
\(\Rightarrow\) A^-1 exists.
A₁₁ = 0 + 9 = 9, A₁₂= -(0 - 6) = 6, A₁₃ = -3 - 0 = -3
A₂₁ = -(0 + 3) = -3, A₂₂= 0 - 2 = -2, A₂₃= -(-3 - 4) = 7
A₃₁ = 6 - 0 = 6, A₃₂= -(3 - 1) = -2, A₃₃= 0 - 2 = -2
\(\therefore adj\quad A={ \left[ \begin{matrix} 9 & 6 & -3 \\ -3 & -2 & 7 \\ 6 & -2 & -2 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ |A| } adj\quad A=\frac { 1 }{ 18 } \left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \)
\(\therefore \ X={ A }^{ -1 }B=\frac { 1 }{ 18 } \left[ \begin{matrix} 9 & -3 & 6 \\ 6 & -2 & -2 \\ -3 & 7 & -2 \end{matrix} \right] \left[ \begin{matrix} 7 \\ 11 \\ 1 \end{matrix} \right] \)
\(=\frac { 1 }{ 18 } \left[ \begin{matrix} 63 & -33 & +6 \\ 42 & -22 & -2 \\ -21 & +77 & -2 \end{matrix} \right] =\frac { 1 }{ 18 } \left[ \begin{matrix} 36 \\ 18 \\ 54 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ 3 \end{matrix} \right] \)
\(\therefore\) x = 2, y = 1, and z = 3.

18.

LHS = \(\begin{vmatrix}0 &ab^2 &ac^2 \\a^2b & 0 & bc^2\\a^2c&b^2c&0\end{vmatrix}\)
Taking a, b and c common from R₁, R₂, R₃
\(=\operatorname{abc}\left|\begin{array}{ccc} 0 & b^2 & c^2 \\ a^2 & 0 & c^2 \\ a^2 & b^2 & 0 \end{array}\right|\)
Taking a², b², c² common from C₁, C₂, C₃
= \(a^2b^2c^2\begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1\\ 1 & 1 & 0 \end{vmatrix}\)
= a³b³c³ [0 - 1 (0 - 1)+ 1(1 - 0)]
= a³b³c³ (1 + 1) = 2 a³b³c³ = RHS

19.

\(=\left|\begin{array}{ccc} 5 & 5^2 & 5^3 \\ 5^2 & 5^3 & 5^4 \\ 5^4 & 5^5 & 5^6 \end{array}\right|\)
Taking 5 and 5² common from R₁and R₂
\(5 \times 5^2\left|\begin{array}{ccc} 1 & 5 & 5^2 \\ 1 & 5 & 5^2 \\ 5^4 & 5^5 & 5^6 \end{array}\right|=0\left(\text {Since } R_1=R_2\right)\)

20.

\(\left|\begin{array}{ccc} x & 2 & -1 \\ 2 & 5 & x \\ -1 & 2 & x \end{array}\right|=0\)
x(5x - 2x)-2(2x + x)-1 (4 + 5) = 0
3x² - 6x - 9 = 0
x² - 2x - 3 = 0 (Divided by 3)
(x - 3)(x + 1) = 0
\(x=+3,-1\)