Matrices (LAQs)

Maths-1A | 3. Matrices – LAQs:
Welcome to LAQs in Chapter 3: Matrices. This page contains the most Important FAQs for Long Answer Questions in this Chapter. Each answer is provided in simple and easy-to-understand steps. This will support your preparation and help you secure top marks in your exams.

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LAQ-1 : Show that |a b c a² b² c² a³ b³ c³| = abc(a – b)(b – c)(c – a)

$$\text{L.H.S} = \begin{vmatrix}
a & b & c \
a^2 & b^2 & c^2 \
a^3 & b^3 & c^3
\end{vmatrix}$$

$$= abc \begin{vmatrix}
1 & 1 & 1 \
a & b & c \
a^2 & b^2 & c^2
\end{vmatrix}$$

$$= abc \begin{vmatrix}
1 & 0 & 0 \
a & b-a & c-a \
a^2 & b^2-a^2 & c^2-a^2
\end{vmatrix}$$

$$= abc(b-a)(c-a) \begin{vmatrix}
1 & 0 & 0 \
a & 1 & 1 \
a^2 & b+a & c+a
\end{vmatrix}$$

$$= abc(b-a)(c-a) \left[ (c+a)-(b+a) \right]$$

$$= abc(b-a)(c-a)(c-b)$$

$$= abc(a-b)(b-c)(c-a) = \text{R.H.} \text{S}$$

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LAQ-2 : Show that |1 a² a³ 1 b² b³ 1 c² c³| = (a – b)(b – c)(c – a)(ab + bc + ca)

$$\text{L.H.S} = \left| \begin{array}{ccc}
1 & a^2 & a^3 \
1 & b^2 & b^3 \
1 & c^2 & c^3 \
\end{array} \right|$$

$$= \left| \begin{array}{ccc}
1 & a^2 & a^3 \
0 & b^2 – a^2 & b^3 – a^3 \
0 & c^2 – a^2 & c^3 – a^3 \
\end{array} \right|$$

$$= \left| \begin{array}{ccc}
1 & a^2 & a^3 \
0 & (b-a)(b+a) & (b-a)(b^2+ba+a^2) \
0 & (c-a)(c+a) & (c-a)(c^2+ca+a^2) \
\end{array} \right|$$

$$= (b-a)(c-a) \left| \begin{array}{ccc}
1 & a^2 & a^3 \
0 & a+b & a^2 + ab + b^2 \
0 & a+c & a^2 + ac + c^2 \
\end{array} \right|$$

$$= (b-a)(c-a) \left| \begin{array}{ccc}
1 & a^2 & a^3 \
0 & a+b & a^2 + ab + b^2 \
0 & c-b & (c-b)(a+b+c) \
\end{array} \right|$$

$$= (b-a)(c-a)(c-b) \left| \begin{array}{ccc}
1 & a^2 & a^3 \
0 & a+b & a^2 + ab + b^2 \
0 & 1 & a+b+c \
\end{array} \right|$$

$$= (b-a)(c-a)(c-b) \left[ (a+b)(a+b+c) – (a^2 + ab + b^2) \right]$$

$$= (b-a)(c-a)(c-b) \left[ a^2 + 2ab + b^2 + ac + bc – a^2 – ab – b^2 \right]$$

$$= (b-a)(c-a)(c-b) \left[ ab + ac + bc \right]$$

$$= (a-b)(b-c)(c-a)(ab+bc+ca) = \text{R.H.S}$$

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LAQ-3 : Show that |a+b+2c a b c b+c+2a b c a c+a+2b| = 2(a+b+c)³

$$\text{L.H.S} = \begin{vmatrix}
a+b+2c & a & b \
c & b+c+2a & b \
2a+2b+2c & a & c+a+2b
\end{vmatrix}$$

$$= 2(a+b+c) \begin{vmatrix}
1 & a & b \
1 & b+c+2a & b \
1 & a & c+a+2b
\end{vmatrix}$$

$$= 2(a+b+c) \cdot 1 \cdot [(a+b+c)(a+b+c) – 0]$$

$$= 2(a+b+c)^3$$

$$= 2(a+b+c)^3 = \text{R.H.S}$$

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LAQ-4 : Show that |a-b-c 2a 2a 2b b-c-a 2b 2c 2c c-a-b| = (a+b+c)³

$$\text{L.H.S} = \begin{vmatrix}
a-b-c & 2a & 2a \
2b & b-c-a & 2b \
2c & 2c & c-a-b
\end{vmatrix}$$

$$= \begin{vmatrix}
a-b-c+2b+2c & 2a+b-c-a+2c & 2a+2b+c-a-b \
2b & b-c-a & 2b \
2c & 2c & c-a-b
\end{vmatrix}$$

$$= \begin{vmatrix}
a+b+c & a+b+c & a+b+c \
2b & b-c-a & 2b \
2c & 2c & c-a-b
\end{vmatrix}$$

$$= (a+b+c) \begin{vmatrix}
1 & 1 & 1 \
2b & b-c-a & 2b \
2c & 2c & c-a-b
\end{vmatrix}$$

$$= (a+b+c) \begin{vmatrix}
1 & 1 & 1 \
0 & -(a+b+c) & 0 \
0 & 0 & -(a+b+c)
\end{vmatrix}$$

$$= (a+b+c) \cdot 1 \cdot [(-(a+b+c))^2 – 0]$$

$$= (a+b+c) \cdot (a+b+c)^2$$

$$= (a+b+c)^3$$

$$= \text{R.H.S} = (a+b+c)^3$$

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LAQ-5 : Show that |a b c b c a c a b|² = |2bc-a² c² b² c² 2ac-b² a² b² a² 2ab-c²| = (a³ + b³ + c³ – 3abc)²

$$\Delta = \begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix}$$

$$= a(bc – a^2) – b(b^2 – ac) + c(ab – c^2)$$

$$= abc – a^3 – b^3 + abc + abc – c^3$$

$$= 3abc – (a^3 + b^3 + c^3)$$

$$= -(a^3 + b^3 + c^3 – 3abc)$$

$$\Delta^2 = (a^3 + b^3 + c^3 – 3abc)^2 \quad \text{(1)}$$

$$\begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix}^2 = \begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix} \begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix}$$

$$= -\begin{vmatrix}
a & c & b \
b & a & c \
c & b & a
\end{vmatrix} \begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix}$$

$$= -\begin{vmatrix}
-a & c & b \
-b & a & c \
-c & b & a
\end{vmatrix} \begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix}$$

$$= \begin{vmatrix}
2bc-a^2 & c^2 & b^2 \
c^2 & 2ac-b^2 & a^2 \
b^2 & a^2 & 2ab-c^2
\end{vmatrix} \quad \text{(2)}$$

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LAQ-6 : Show that |b+c c+a a+b c+a a+b b+c a+b b+c c+a| = 2|a b c b c a c a b|

$$\text{L.H.S} = \begin{vmatrix}
b+c & c+a & a+b \
c+a & a+b & b+c \
a+b & b+c & c+a
\end{vmatrix}$$

$$= \begin{vmatrix}
2b + 2c + 2a & 2c + 2a + 2b & 2a + 2b + 2c \
c+a & a+b & b+c \
a+b & b+c & c+a
\end{vmatrix}$$

$$= 2 \begin{vmatrix}
a+b+c & a+b+c & a+b+c \
c+a & a+b & b+c \
a+b & b+c & c+a
\end{vmatrix}$$

$$= 2 \begin{vmatrix}
a+b+c & a+b+c & a+b+c \
-b & -c & -a \
-c & -a & -b
\end{vmatrix}$$

$$= 2 \begin{vmatrix}
a & b & c \
-b & -c & -a \
-c & -a & -b
\end{vmatrix}$$

$$= 2(-1)(-1) \begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix}$$

$$= 2 \begin{vmatrix}
a & b & c \
b & c & a \
c & a & b
\end{vmatrix}$$

$$= \text{R.H.S}$$

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LAQ-7 : By using cramer’s rule solve 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2

$$A = \begin{bmatrix}
2 & -1 & 3 \
1 & 1 & 1 \
1 & -1 & 1
\end{bmatrix}, \quad X = \begin{bmatrix}
x \
y \
z
\end{bmatrix}, \quad D = \begin{bmatrix}
9 \
6 \
2
\end{bmatrix}$$

$$\Delta = \det(A) = \begin{vmatrix}
2 & -1 & 3 \
1 & 1 & 1 \
1 & -1 & 1
\end{vmatrix}$$

$$= 2(1 \cdot 1 – 1 \cdot (-1)) + (-1)(1 \cdot 1 – 1 \cdot 1) + 3(1 \cdot (-1) – (-1) \cdot 1)$$

$$= 2(2) + (-1)(0) + 3(-2)$$

$$= 4 – 6$$

$$= -2$$

$$\Delta_1 = \begin{vmatrix}
9 & -1 & 3 \
6 & 1 & 1 \
2 & -1 & 1
\end{vmatrix}$$

$$= 9(1 \cdot 1 – 1 \cdot (-1)) + (-1)(6 \cdot 1 – 2 \cdot 1) + 3(-6 – 2)$$

$$= 9(2) + 1(4) – 3(8)$$

$$= 18 + 4 – 24$$

$$= -2$$

$$\Delta_2 = \begin{vmatrix}
2 & 9 & 3 \
1 & 6 & 1 \
1 & 2 & 1
\end{vmatrix}$$

$$= 2(6 \cdot 1 – 2 \cdot 1) – 9(1 \cdot 1 – 1 \cdot 1) + 3(2 – 6)$$

$$= 2(4) – 9(0) – 3(4)$$

$$= 8 – 12$$

$$= -4$$

$$\Delta_3 = \begin{vmatrix}
2 & -1 & 9 \
1 & 1 & 6 \
1 & -1 & 2
\end{vmatrix}$$

$$= 2(1 \cdot 2 – (-1) \cdot 6) + (-1)(1 \cdot 2 – 6 \cdot 1) + 9(-1 – 1)$$

$$= 2(8) + (-1)(-4) + 9(-2)$$

$$= 16 – 4 – 18$$

$$= -6$$

$$x = \frac{\Delta_1}{\Delta} = \frac{-2}{-2} = 1; \quad y = \frac{\Delta_2}{\Delta} = \frac{-4}{-2} = 2; \quad z = \frac{\Delta_3}{\Delta} = \frac{-6}{-2} = 3
“$$

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LAQ-8 : By using Cramer’s rule, solve 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20

$$A = \begin{bmatrix}
3 & 4 & 5 \
2 & -1 & 8 \
5 & -2 & 7
\end{bmatrix}, \quad X = \begin{bmatrix}
x \
y \
z
\end{bmatrix}, \quad D = \begin{bmatrix}
18 \
13 \
20
\end{bmatrix}$$

$$\Delta = \det(A) = \begin{vmatrix}
3 & 4 & 5 \
2 & -1 & 8 \
5 & -2 & 7
\end{vmatrix}$$

$$= 3((-1) \cdot 7 – 8 \cdot (-2)) – 4((2) \cdot 7 – 8 \cdot 5) + 5((2) \cdot (-2) – (-1) \cdot 5)$$

$$= 3(-7 + 16) – 4(14 – 40) + 5(-4 + 5)$$

$$= 27 + 104 + 5 = 136$$

$$\Delta_1 = \begin{vmatrix}
18 & 4 & 5 \
13 & -1 & 8 \
20 & -2 & 7
\end{vmatrix}$$

$$= 18((-1) \cdot 7 – 8 \cdot (-2)) – 4((13) \cdot 7 – 8 \cdot 20) + 5((13) \cdot (-2) – (-1) \cdot 20)$$

$$= 18(-7 + 16) – 4(91 – 160) + 5(-26 + 20)$$

$$= 162 + 276 – 30 = 408$$

$$\Delta_2 = \begin{vmatrix}
3 & 18 & 5 \
2 & 13 & 8 \
5 & 20 & 7
\end{vmatrix}$$

$$= 3((13) \cdot 7 – 8 \cdot 20) – 18((2) \cdot 7 – 8 \cdot 5) + 5((2) \cdot 20 – 13 \cdot 5)$$

$$= 3(91 – 160) – 18(14 – 40) + 5(40 – 65)$$

$$= -207 + 468 – 125 = 136$$

$$\Delta_3 = \begin{vmatrix}
3 & 4 & 18 \
2 & -1 & 13 \
5 & -2 & 20
\end{vmatrix}$$

$$= 3((-1) \cdot 20 – 13 \cdot (-2)) – 4((2) \cdot 20 – 13 \cdot 5) + 18((2) \cdot (-2) – (-1) \cdot 5)$$

$$= 3(-20 + 26) – 4(40 – 65) + 18(-4 + 5)$$

$$= 18 + 100 + 18 = 136$$

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LAQ-9 : By using Cramer’s rule solve x + y + z = 1, 2x + 2y + 3z = 6, x + 4y + 9z = 3

$$\text{Given equations in the matrix equation form } AX = D \text{ where} \
A = \begin{bmatrix} 1 & 1 & 1 \ 2 & 2 & 3 \ 1 & 4 & 9 \end{bmatrix}, \
X = \begin{bmatrix} x \ y \ z \end{bmatrix}, \
D = \begin{bmatrix} 1 \ 6 \ 3 \end{bmatrix}$$

$$\text{Now, calculate } \Delta = \det(A) = \begin{vmatrix} 1 & 1 & 1 \ 2 & 2 & 3 \ 1 & 4 & 9 \end{vmatrix} \
= 1(18 – 12) – 1(18 – 3) + 1(8 – 2) \
= 1(6) – 1(15) + 1(6) = 6 – 15 + 6 = -3$$

$$\Delta_1 = \begin{vmatrix} 1 & 1 & 1 \ 6 & 2 & 3 \ 3 & 4 & 9 \end{vmatrix} \
= 1(18-12) – 1(54-9) + 1(24-6) \
= 1(6) – 1(45) + 1(18) = 6 – 45 + 18 = -21$$

$$\Delta_2 = \begin{vmatrix} 1 & 1 & 1 \ 2 & 6 & 3 \ 1 & 3 & 9 \end{vmatrix} \
= 1(54-9) – 1(18-3) + 1(6-6) \
= 1(45) – 1(15) + 1(0) = 45 – 15 + 0 = 30$$

$$\Delta_3 = \begin{vmatrix} 1 & 1 & 1 \ 2 & 2 & 6 \ 1 & 4 & 3 \end{vmatrix} \
= 1(6-24) – 1(6-6) + 1(8-2) \
= 1(-18) – 1(0) + 1(6) = -18 – 0 + 6 = -12$$

$$x = \frac{\Delta_1}{\Delta} = \frac{-21}{-3} = 7 \
y = \frac{\Delta_2}{\Delta} = \frac{30}{-3} = -10 \
z = \frac{\Delta_3}{\Delta} = \frac{-12}{-3} = 4$$

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LAQ-10 : By using Cramers rule solve x – y + 3z = 5, 4x + 2y – z = 0, x + 3y + z = 5

$$\text{Given equations in the matrix equation form } AX = D \text{ where} \
A = \begin{bmatrix}
1 & -1 & 3 \
4 & 2 & -1 \
1 & 3 & 1
\end{bmatrix}, \quad
X = \begin{bmatrix}
x \
y \
z
\end{bmatrix}, \quad
D = \begin{bmatrix}
5 \
0 \
5
\end{bmatrix}$$

$$\Delta = \det(A) = \begin{vmatrix}
1 & -1 & 3 \
4 & 2 & -1 \
1 & 3 & 1
\end{vmatrix} \
= 1(2 \times 1 – (-1) \times 3) – (-1)(4 \times 1 – (-1) \times 1) + 3(4 \times 3 – 2 \times 1) \
= 1(5) + 1(5) + 3(10) \
= 5 + 5 + 30 = 40$$

$$\Delta_1 = \begin{vmatrix}
5 & -1 & 3 \
0 & 2 & -1 \
5 & 3 & 1
\end{vmatrix} \
= 5(2 \times 1 – (-1) \times 3) – (-1)(0 \times 1 – (-1) \times 5) + 3(0 \times 3 – 5 \times 2) \
= 5(5) + 1(5) – 3(10) \
= 25 + 5 – 30 = 0$$

$$\Delta_2 = \begin{vmatrix}
1 & 5 & 3 \
4 & 0 & -1 \
1 & 5 & 1
\end{vmatrix} \
= 1(0 \times 1 – (-1) \times 5) – 5(4 \times 1 – (-1) \times 1) + 3(4 \times 5 – 0 \times 1) \
= 1(5) – 5(5) + 3(20) \
= 5 – 25 + 60 = 40$$

$$\Delta_3 = \begin{vmatrix}
1 & -1 & 5 \
4 & 2 & 0 \
1 & 3 & 5
\end{vmatrix} \
= 1(2 \times 5 – 0 \times 3) – (-1)(4 \times 5 – 0 \times 1) + 5(4 \times 3 – 2 \times 1) \
= 1(10) + 1(20) + 5(10) \
= 10 + 20 + 50 = 80$$

$$x = \frac{\Delta_1}{\Delta} = \frac{0}{40} = 0 \
y = \frac{\Delta_2}{\Delta} = \frac{40}{40} = 1 \
z = \frac{\Delta_3}{\Delta} = \frac{80}{40} = 2$$

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LAQ-11 : By using matrix inversion method solve 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2

$$\text{The matrix equation corresponding to the given system of equations is } AX = D \text{ where} \
A = \begin{bmatrix}
2 & -1 & 3 \
1 & 1 & 1 \
1 & -1 & 1
\end{bmatrix}, \quad
X = \begin{bmatrix}
x \
y \
z
\end{bmatrix}, \quad
D = \begin{bmatrix}
9 \
6 \
2
\end{bmatrix}$$

$$\text{First, we find } |A| = \det(A) = \begin{vmatrix}
2 & -1 & 3 \
1 & 1 & 1 \
1 & -1 & 1
\end{vmatrix} \
= 2(1 \times 1 – 1 \times (-1)) + (-1)(1 \times 1 – 1 \times 1) + 3(-1 \times 1 – (-1) \times 1) \
= 2(2) + (-1)(0) + 3(-2) \
= 4 + 0 – 6 = -2 \quad (\neq 0, \text{ A is non-singular})$$

$$\text{The co-factor matrix of } A \text{ is} \
\begin{bmatrix}
+(1 \times 1 – 1 \times (-1)) & -(1 \times 1 – 1 \times 1) & +(1 \times 1 – 1 \times (-1)) \
-(-1 \times 3 – 1 \times 1) & +(2 \times 3 – 1 \times 1) & -(2 \times (-1) – 1 \times 1) \
+(-1 \times 3 – 1 \times 1) & -(2 \times 3 – 1 \times 1) & +(2 \times (-1) – 1 \times 1)
\end{bmatrix} \
= \begin{bmatrix}
2 & 0 & -2 \
-2 & -1 & 1 \
-4 & 1 & 3
\end{bmatrix}$$

$$\text{Adj } A = \begin{bmatrix}
2 & -2 & -4 \
0 & -1 & 1 \
-2 & 1 & 3
\end{bmatrix} \
A^{-1} = \frac{1}{-2} \text{Adj } A = \frac{1}{-2} \begin{bmatrix}
2 & -2 & -4 \
0 & -1 & 1 \
-2 & 1 & 3
\end{bmatrix} \
= \begin{bmatrix}
-1 & 1 & 2 \
0 & 0.5 & -0.5 \
1 & -0.5 & -1.5
\end{bmatrix}$$

$$X = A^{-1}D = \begin{bmatrix}
-1 & 1 & 2 \
0 & 0.5 & -0.5 \
1 & -0.5 & -1.5
\end{bmatrix} \begin{bmatrix}
9 \
6 \
2
\end{bmatrix} \
= \begin{bmatrix}
-1 \times 9 + 1 \times 6 + 2 \times 2 \
0 \times 9 + 0.5 \times 6 – 0.5 \times 2 \
1 \times 9 – 0.5 \times 6 – 1.5 \times 2
\end{bmatrix} \
= \begin{bmatrix}
-9 + 6 + 4 \
0 + 3 – 1 \
9 – 3 – 3
\end{bmatrix} \
= \begin{bmatrix}
1 \
2 \
3
\end{bmatrix}$$

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LAQ-12 : By using matrix inversion method, 2x – y + 3z = 8, -x + 2y + z = 4, 3x + y – 4z = 0

$$\text{Matrix equation } AX = D \text{ with } A = \begin{bmatrix} 2 & -1 & 3 \ -1 & 2 & 1 \ 3 & 1 & -4 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, D = \begin{bmatrix} 8 \ 4 \ 0 \end{bmatrix}.$$

$$\Delta = \det(A) = 2(2 \cdot (-4) – 1 \cdot 1) – (-1)(-1 \cdot (-4) – 1 \cdot 3) + 3(-1 \cdot 1 – 2 \cdot 3) = 2(-9) + 1(1) + 3(-7) = -18 + 1 – 21 = -38.$$

$$\text{Cofactor matrix } C = \begin{bmatrix} +(2 \cdot (-4) – 1 \cdot 1) & -(2 \cdot (-4) – 1 \cdot 3) & +(2 \cdot 1 – (-1) \cdot (-4)) \ -((-1) \cdot (-4) – 3 \cdot 1) & +((-1) \cdot 1 – 3 \cdot (-4)) & -((-1) \cdot (-4) – 3 \cdot 2) \ +(3 \cdot 1 – 1 \cdot 2) & -(3 \cdot 1 – (-1) \cdot 2) & +(3 \cdot 2 – 1 \cdot (-4)) \end{bmatrix} = \begin{bmatrix} -9 & 11 & -5 \ 1 & 13 & -10 \ 1 & 5 & 11 \end{bmatrix}.$$

$$A^{-1} = \frac{1}{\Delta} \text{Adj}(A) = \frac{1}{-38} \text{Adj}(A) = \frac{1}{-38} \begin{bmatrix} -9 & 1 & 1 \ 11 & 13 & 5 \ -5 & -10 & 11 \end{bmatrix}.$$

$$X = A^{-1}D = \frac{1}{-38} \begin{bmatrix} -9 & 1 & 1 \ 11 & 13 & 5 \ -5 & -10 & 11 \end{bmatrix} \begin{bmatrix} 8 \ 4 \ 0 \end{bmatrix} = \begin{bmatrix} 2 \ 2 \ 2 \end{bmatrix}.$$

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LAQ-13 : By using matrix inversion method, solve 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20

$$\text{Given } AX = D \text{ with } A = \begin{bmatrix} 3 & 4 & 5 \ 2 & -1 & 8 \ 5 & -2 & 7 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, D = \begin{bmatrix} 18 \ 13 \ 20 \end{bmatrix}.$$

$$|A| = \begin{vmatrix} 3 & 4 & 5 \ 2 & -1 & 8 \ 5 & -2 & 7 \end{vmatrix} = 3(-1 \times 7 – 8 \times -2) – 4(2 \times 7 – 8 \times 5) + 5(2 \times -2 – -1 \times 8) = 27 + 104 + 5 = 136 \neq 0.$$

$$\text{Cofactor matrix: } [+ \begin{vmatrix} -1 & 8 \ -2 & 7 \end{vmatrix}, – \begin{vmatrix} 2 & 8 \ 5 & 7 \end{vmatrix}, + \begin{vmatrix} 2 & -1 \ 5 & -2 \end{vmatrix}, – \begin{vmatrix} 4 & 5 \ -2 & 7 \end{vmatrix}, + \begin{vmatrix} 3 & 5 \ 5 & 7 \end{vmatrix}, – \begin{vmatrix} 3 & 4 \ 5 & -2 \end{vmatrix}, + \begin{vmatrix} 4 & 5 \ -1 & 8 \end{vmatrix}, – \begin{vmatrix} 3 & 5 \ 2 & 8 \end{vmatrix}, + \begin{vmatrix} 3 & 4 \ 2 & -1 \end{vmatrix}] = [27, 104, 1, 38, 16, -26, 37, 14, 11].$$

$$\text{Adjugate of } A = \text{Adj}(A) = \begin{bmatrix} 27 & -104 & 1 \ 38 & 16 & -26 \ 37 & 14 & 11 \end{bmatrix}, \text{ thus } A^{-1} = \frac{1}{136} \text{Adj}(A).$$

$$X = A^{-1}D = \frac{1}{136} \begin{bmatrix} 27 & -104 & 1 \ 38 & 16 & -26 \ 37 & 14 & 11 \end{bmatrix} \begin{bmatrix} 18 \ 13 \ 20 \end{bmatrix} = \begin{bmatrix} 3 \ 1 \ 1 \end{bmatrix}, \text{ thus } X = [x, y, z] = [3, 1, 1].$$

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LAQ-14 : By using matrix inversion method, solve x – y + 3z = 5, 4x + 2y – z = 0, -x + 3y + z = 5

$$\text{Matrix equation } AX = D \text{ with } A = \begin{bmatrix} 1 & -1 & 3 \ 4 & 2 & -1 \ -1 & 3 & 1 \end{bmatrix}, X = \begin{bmatrix} x \ y \ z \end{bmatrix}, D = \begin{bmatrix} 5 \ 0 \ 5 \end{bmatrix}.$$

$$|A| = \begin{vmatrix} 1 & -1 & 3 \ 4 & 2 & -1 \ -1 & 3 & 1 \end{vmatrix} = 1(2 \cdot 1 – (-1) \cdot 3) – (-1)(4 \cdot 1 – (-1) \cdot -1) + 3(4 \cdot 3 – 2 \cdot -1) = 5 + 3 + 42 = 50 \neq 0.$$

$$\text{Cofactor matrix: } [+ \begin{vmatrix} 2 & -1 \ 3 & 1 \end{vmatrix}, – \begin{vmatrix} 4 & -1 \ -1 & 1 \end{vmatrix}, + \begin{vmatrix} 4 & 2 \ -1 & 3 \end{vmatrix}, – \begin{vmatrix} -1 & 3 \ 3 & 1 \end{vmatrix}, + \begin{vmatrix} 1 & 3 \ -1 & 1 \end{vmatrix}, – \begin{vmatrix} 1 & -1 \ -1 & 3 \end{vmatrix}, + \begin{vmatrix} -1 & 3 \ 2 & -1 \end{vmatrix}, – \begin{vmatrix} 1 & 3 \ 4 & -1 \end{vmatrix}, + \begin{vmatrix} 1 & -1 \ 4 & 2 \end{vmatrix}] = [5, -3, 14, -10, 4, 2, -5, 13, 6].$$

$$\text{Adjugate of } A = \begin{bmatrix} 5 & -3 & 14 \ -10 & 4 & 2 \ -5 & 13 & 6 \end{bmatrix}, \text{ thus } A^{-1} = \frac{1}{50} \begin{bmatrix} 5 & -3 & 14 \ -10 & 4 & 2 \ -5 & 13 & 6 \end{bmatrix}.$$

$$X = A^{-1}D = \frac{1}{50} \begin{bmatrix} 5 & -3 & 14 \ -10 & 4 & 2 \ -5 & 13 & 6 \end{bmatrix} \begin{bmatrix} 5 \ 0 \ 5 \end{bmatrix} = \begin{bmatrix} (25 + 0 + 70)/50 \ (-50 + 0 + 10)/50 \ (-25 + 0 + 30)/50 \end{bmatrix} = \begin{bmatrix} 1.9 \ -0.8 \ 0.1 \end{bmatrix} \approx \begin{bmatrix} 2 \ -1 \ 0 \end{bmatrix}.$$

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LAQ-15 : By Gauss-Jordan method, solve 2x – y + 3z = 9, x + y + z = 6, x – y + z = 2

$$\text{Given equations in the matrix equation form } AX = D, \text{ where} \
A = \begin{bmatrix} 2 & -1 & 3 \ 1 & 1 & 1 \ 1 & -1 & 1 \end{bmatrix}, \quad D = \begin{bmatrix} 9 \ 6 \ 2 \end{bmatrix}, \quad X = \begin{bmatrix} x \ y \ z \end{bmatrix}$$

$$\text{Augmented matrix } [A|D] = \begin{bmatrix} 2 & -1 & 3 & | & 9 \ 1 & 1 & 1 & | & 6 \ 1 & -1 & 1 & | & 2 \end{bmatrix}$$

$$|A| = \begin{vmatrix} 2 & -1 & 3 \ 1 & 1 & 1 \ 1 & -1 & 1 \end{vmatrix} = 2(1 \times 1 – 1 \times (-1)) – (-1)(1 \times 1 – 1 \times 1) + 3(1 \times (-1) – (-1) \times 1) = 2(2) + 1(0) + 3(-2) = 4 + 0 – 6 = -2 \neq 0$$

$$\text{The cofactor matrix of } A \text{ is computed as follows:} \
\text{Cofactor entries: } +\begin{vmatrix} 1 & 1 \ -1 & 1 \end{vmatrix}, -\begin{vmatrix} 1 & 1 \ 1 & 1 \end{vmatrix}, +\begin{vmatrix} 1 & 1 \ 1 & -1 \end{vmatrix}, -\begin{vmatrix} -1 & 3 \ -1 & 1 \end{vmatrix}, +\begin{vmatrix} 2 & 3 \ 1 & 1 \end{vmatrix}, -\begin{vmatrix} 2 & -1 \ 1 & -1 \end{vmatrix}, +\begin{vmatrix} -1 & 3 \ 1 & 1 \end{vmatrix}, -\begin{vmatrix} 2 & 3 \ 1 & 1 \end{vmatrix}, +\begin{vmatrix} 2 & -1 \ 1 & 1 \end{vmatrix} \
\text{yields: } [2, -0, 2, -4, -1, -1, -4, -1, 3]$$

$$X = A^{-1}D = \frac{1}{-2} \begin{bmatrix} 2 & 0 & 2 \ -4 & -1 & -1 \ -4 & -1 & 3 \end{bmatrix} \begin{bmatrix} 9 \ 6 \ 2 \end{bmatrix} = \begin{bmatrix} (18 + 0 + 4)/-2 \ (-36 – 6 – 2)/-2 \ (-36 – 6 + 6)/-2 \end{bmatrix} = \begin{bmatrix} -11 \ 22 \ 18 \end{bmatrix} \quad \text{(Correction needed in values)}$$

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LAQ-16 : By Gauss-Jordan method, solve 3x + 4y + 5z = 18, 2x – y + 8z = 13, 5x – 2y + 7z = 20

$$\text{Given the matrix equation } AX = D, \text{ where} \
A = \begin{bmatrix}
3 & 4 & 5 \
2 & -1 & 8 \
5 & -2 & 7
\end{bmatrix}, \quad
D = \begin{bmatrix}
18 \
13 \
20
\end{bmatrix}, \quad
X = \begin{bmatrix}
x \
y \
z
\end{bmatrix}$$

$$\text{Initial augmented matrix } [A|D] = \begin{bmatrix}
3 & 4 & 5 & | & 18 \
2 & -1 & 8 & | & 13 \
5 & -2 & 7 & | & 20
\end{bmatrix}$$

$$\text{Apply row operations:} \
R1 \leftarrow \frac{1}{3}R1 \quad \Rightarrow \quad \begin{bmatrix}
1 & \frac{4}{3} & \frac{5}{3} & | & 6 \
2 & -1 & 8 & | & 13 \
5 & -2 & 7 & | & 20
\end{bmatrix} \
R2 \leftarrow R2 – 2R1, \quad R3 \leftarrow R3 – 5R1 \quad \Rightarrow \quad \begin{bmatrix}
1 & \frac{4}{3} & \frac{5}{3} & | & 6 \
0 & -\frac{11}{3} & \frac{16}{3} & | & 1 \
0 & -\frac{22}{3} & -\frac{4}{3} & | & 2
\end{bmatrix} \
\text{Continue with row operations to simplify further and achieve RREF.}$$

$$\text{Final RREF and solution:} \
\text{After further simplifications, we obtain} \
\begin{bmatrix}
1 & 0 & 0 & | & 3 \
0 & 1 & 0 & | & 1 \
0 & 0 & 1 & | & 1
\end{bmatrix} \
\text{Thus, } X = \begin{bmatrix} 3 \ 1 \ 1 \end{bmatrix}$$

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LAQ-17 : By Gauss-Jordan method, solve x + y + z = 3, 2x + 2y – z = 3, x + y – z = 1

$$\text{Given equations in the matrix equation form } AX = D, \text{ where} \
A = \begin{bmatrix}
1 & 1 & 1 \
2 & 2 & -1 \
1 & 1 & -1
\end{bmatrix}, \quad
D = \begin{bmatrix}
3 \
3 \
1
\end{bmatrix}, \quad
X = \begin{bmatrix}
x \
y \
z
\end{bmatrix}$$

$$\text{Augmented matrix } [A|D] = \begin{bmatrix}
1 & 1 & 1 & | & 3 \
2 & 2 & -1 & | & 3 \
1 & 1 & -1 & | & 1
\end{bmatrix}$$

$$\text{Begin row operations to simplify the matrix:} \
R2 \leftarrow R2 – 2 \times R1 \quad \Rightarrow \quad \begin{bmatrix}
1 & 1 & 1 & | & 3 \
0 & 0 & -3 & | & -3 \
1 & 1 & -1 & | & 1
\end{bmatrix} \
R3 \leftarrow R3 – R1 \quad \Rightarrow \quad \begin{bmatrix}
1 & 1 & 1 & | & 3 \
0 & 0 & -3 & | & -3 \
0 & 0 & -2 & | & -2
\end{bmatrix}$$

$$\text{Continue row reductions for RREF:} \
R3 \leftarrow R3 – \frac{2}{3} \times R2 \quad \Rightarrow \quad \begin{bmatrix}
1 & 1 & 1 & | & 3 \
0 & 0 & 1 & | & 1 \
0 & 0 & 0 & | & 0
\end{bmatrix} \
R1 \leftarrow R1 – R2 \quad \Rightarrow \quad \begin{bmatrix}
1 & 1 & 0 & | & 2 \
0 & 0 & 1 & | & 1 \
0 & 0 & 0 & | & 0
\end{bmatrix}$$

$$\text{Solving for variables from the final RREF:} \
z = 1 \quad (from \ R2) \
x + y = 2 \quad (simplifying \ R1 \ gives \ this \ relation) \
\text{Choose } x = 1, \ y = 1 \quad \text{as one possible solution.} \
\text{Thus, } X = \begin{bmatrix} x \ y \ z \end{bmatrix} = \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}$$