Matrices (SAQs)
Maths-1A | 3. Matrices – SAQs:
Welcome to SAQs in Chapter 3: Matrices. This page contains the most Important FAQs for Short 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.
SAQ-1 : If A is a non-singular matrix then prove that A-1 = Adj A/det A
$$We take ( A = \begin{bmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{bmatrix} )$$
$$We take cofactors of ( a_1, b_1, c_1, \ldots ) as ( A_1, B_1, C_1, \ldots )$$
$$\text{Adj } A = \begin{bmatrix} A_1 & B_1 & C_1 \ A_2 & B_2 & C_2 \ A_3 & B_3 & C_3 \end{bmatrix}^T$$
$$\text{Adj } A = \begin{bmatrix} A_1 & A_2 & A_3 \ B_1 & B_2 & B_3 \ C_1 & C_2 & C_3 \end{bmatrix}$$
$$A \cdot (\text{Adj } A) = \begin{bmatrix} a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \ a_3 & b_3 & c_3 \end{bmatrix} \begin{bmatrix} A_1 & A_2 & A_3 \ B_1 & B_2 & B_3 \ C_1 & C_2 & C_3 \end{bmatrix}$$
$$= \begin{bmatrix} a_1A_1 + b_1B_1 + c_1C_1 & a_1A_2 + b_1B_2 + c_1C_2 & a_1A_3 + b_1B_3 + c_1C_3 \ a_2A_1 + b_2B_1 + c_2C_1 & a_2A_2 + b_2B_2 + c_2C_2 & a_2A_3 + b_2B_3 + c_2C_3 \ a_3A_1 + b_3B_1 + c_3C_1 & a_3A_2 + b_3B_2 + c_3C_2 & a_3A_3 + b_3B_3 + c_3C_3 \end{bmatrix}$$
$$= \begin{bmatrix} \det A & 0 & 0 \ 0 & \det A & 0 \ 0 & 0 & \det A \end{bmatrix}$$
$$= (\det A) \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix} = (\det A) I$$
$$A (\text{Adj } A) = (\det A) I$$
$$A \left( \frac{\text{Adj } A}{\det A} \right) = I$$
$$A^{-1} = \frac{\text{Adj } A}{\det A}$$
SAQ-2 : Find the adjoint and the inverse of the matrix A = [1 3 3 1 4 3 1 3 4]
$$Given ( A = \begin{bmatrix} 1 & 3 & 3 \ 1 & 4 & 3 \ 1 & 3 & 4 \end{bmatrix} )$$
$$\det A = 1(16 – 9) – 3(4 – 3) + 3(3 – 4)$$
$$= 7 – 3 – 3 = 1 \neq 0$$
$$A \text{ is non-singular}$$
$$\text{Now, } \text{Adj } A = \begin{bmatrix} |4 & 3 \ 3 & 4| & -|1 & 3 \ 1 & 4| & |1 & 4 \ 1 & 3| \ -|3 & 3 \ 3 & 4| & |1 & 3 \ 1 & 4| & -|1 & 3 \ 1 & 3| \ |3 & 3 \ 4 & 3| & -|1 & 3 \ 1 & 3| & |1 & 3 \ 1 & 4| \end{bmatrix}^T$$
$$= \begin{bmatrix} 7 & -1 & -1 \ -3 & 1 & 0 \ -3 & 0 & 1 \end{bmatrix}^T$$
$$\text{Adj } A = \begin{bmatrix} 7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$$
$$A^{-1} = \frac{1}{\det A} (\text{Adj } A)$$
$$= \frac{1}{1} \begin{bmatrix} 7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} 7 & -3 & -3 \ -1 & 1 & 0 \ -1 & 0 & 1 \end{bmatrix}$$
SAQ-3 : Find the adjoint and the inverse of the matrix A = [1 0 2 2 1 0 3 2 1]
$$Given ( A = \begin{bmatrix} 1 & 0 & 2 \ 2 & 1 & 0 \ 3 & 2 & 1 \end{bmatrix} )$$
$$\det A = 1(1 – 0) – 0 + 2(4 – 3)$$
$$= 1 + 2 = 3 \neq 0$$
$$A \text{ is non-singular}$$
$$Now, \text{Adj } A = \begin{bmatrix} |1 & 0 \ 2 & 1| & -|2 & 0 \ 3 & 1| & |2 & 1 \ 3 & 2| \ -|0 & 2 \ 2 & 1| & |1 & 2 \ 3 & 1| & -|1 & 0 \ 3 & 2| & |0 & 2 \ 1 & 0| & -|1 & 2 \ 2 & 0| & |1 & 0 \ 2 & 1| \end{bmatrix}$$
$$= \begin{bmatrix} 1 & -2 & 1 \ 4 & -5 & -2 \ -2 & 4 & 1 \end{bmatrix}$$
$$\text{Adj } A = \begin{bmatrix} 1 & 4 & -2 \ -2 & -5 & 4 \ 1 & -2 & 1 \end{bmatrix}$$
$$A^{-1} = \frac{1}{\det A} (\text{Adj } A)$$
$$= \frac{1}{3} \begin{bmatrix} 1 & 4 & -2 \ -2 & -5 & 4 \ 1 & -2 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} \frac{1}{3} & \frac{4}{3} & -\frac{2}{3} \ -\frac{2}{3} & -\frac{-5}{3} & \frac{4}{3} \ \frac{1}{3} & -\frac{2}{3} & \frac{1}{3} \end{bmatrix}$$
SAQ-4 : Show that the matrix A = [1 2 1 3 2 3 1 1 2] is non-singular and find A-1
$$Given ( A = \begin{bmatrix} 1 & 2 & 1 \ 3 & 2 & 3 \ 1 & 1 & 2 \end{bmatrix} )$$
$$\det A = 1(4 – 3) – 2(6 – 3) + 1(3 – 2)$$
$$= 1 – 6 + 1 = -4 \neq 0$$
$$A \text{ is non-singular}$$
$$Now, \text{Adj } A = \begin{bmatrix} |2 & 3 \ 1 & 2| & -|3 & 3 \ 1 & 2| & |3 & 2 \ 1 & 1| \ -|2 & 1 \ 1 & 2| & |1 & 1 \ 1 & 2| & -|1 & 2 \ 1 & 1| \ |2 & 1 \ 2 & 3| & -|1 & 1 \ 3 & 3| & |1 & 2 \ 3 & 2| \end{bmatrix}^T$$
$$= \begin{bmatrix} 1 & -3 & 1 \ -3 & 1 & 1 \ 4 & 0 & -4 \end{bmatrix}^T$$
$$\text{Adj } A = \begin{bmatrix} 1 & -3 & 4 \ -3 & 1 & 0 \ 1 & 1 & -4 \end{bmatrix}$$
$$A^{-1} = \frac{1}{\det A} (\text{Adj } A)$$
$$= \frac{1}{-4} \begin{bmatrix} 1 & -3 & 4 \ -3 & 1 & 0 \ 1 & 1 & -4 \end{bmatrix}$$
$$= \begin{bmatrix} -\frac{1}{4} & \frac{3}{4} & -1 \ \frac{3}{4} & -\frac{1}{4} & 0 \ -\frac{1}{4} & -\frac{1}{4} & 1 \end{bmatrix}$$
SAQ-5 : Find the inverse of A = [2 1 2 1 0 1 2 2 1]
$$Given ( A = \begin{bmatrix} 2 & 1 & 2 \ 1 & 0 & 1 \ 2 & 2 & 1 \end{bmatrix} )$$
$$\det A = 2(0 – 2) – 1(1 – 2) + 2(2 – 0)$$
$$= -4 + 1 + 4 = 1 \neq 0$$
$$A \text{ is non-singular}$$
$$Now, \text{Adj } A = \begin{bmatrix} |0 & 1 \ 2 & 1| & -|1 & 1 \ 2 & 1| & |1 & 0 \ 2 & 2| \ -|1 & 2 \ 2 & 1| & |2 & 2 \ 2 & 1| & -|2 & 1 \ 2 & 2| \ |1 & 2 \ 0 & 1| & -|2 & 2 \ 1 & 1| & |2 & 1 \ 1 & 0| \end{bmatrix}^T$$
$$= \begin{bmatrix} -2 & 1 & 2 \ 3 & -2 & -2 \ 1 & 0 & -1 \end{bmatrix}^T$$
$$\text{Adj } A = \begin{bmatrix} -2 & 3 & 1 \ 1 & -2 & 0 \ 2 & -2 & -1 \end{bmatrix}$$
$$A^{-1} = \frac{1}{\det A} (\text{Adj } A)$$
$$= \frac{1}{1} \begin{bmatrix} -2 & 3 & 1 \ 1 & -2 & 0 \ 2 & -2 & -1 \end{bmatrix}$$
$$= \begin{bmatrix} -2 & 3 & 1 \ 1 & -2 & 0 \ 2 & -2 & -1 \end{bmatrix}$$
SAQ-6 : If A = [1 2 2 2 1 2 2 2 1] then show that A2 – 4A – 5I = 0
$$A^2 = A \cdot A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} 1+4+4 & 2+2+4 & 2+4+2 \ 2+2+4 & 4+1+4 & 4+2+2 \ 2+4+2 & 4+2+2 & 4+4+1 \end{bmatrix}$$
$$= \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix}$$
$$4A = 4 \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & 2 \ 2 & 2 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix}$$
$$5I = 5 \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix}$$
$$\text{Hence } A^2 – 4A – 5I = \begin{bmatrix} 9 & 8 & 8 \ 8 & 9 & 8 \ 8 & 8 & 9 \end{bmatrix} – \begin{bmatrix} 4 & 8 & 8 \ 8 & 4 & 8 \ 8 & 8 & 4 \end{bmatrix} – \begin{bmatrix} 5 & 0 & 0 \ 0 & 5 & 0 \ 0 & 0 & 5 \end{bmatrix}$$
$$= \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$$
$$= 0$$
SAQ-7 : If A = [1 -2 1 0 1 -1 3 -1 1] then show that A3 – 3A2 – A – 3I = 0
$$A^2 = A \cdot A = \begin{bmatrix} 1 & -2 & 1 \ 0 & 1 & -1 \ 3 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -2 & 1 \ 0 & 1 & -1 \ 3 & -1 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} 1+0+3 & -2-2-1 & 1+2+1 \ 0+0-3 & 0+1+1 & 0-1-1 \ 3-0+3 & -6-1-1 & 3+1+1 \end{bmatrix}$$
$$= \begin{bmatrix} 4 & -5 & 4 \ -3 & 2 & -2 \ 6 & -8 & 5 \end{bmatrix}$$
$$A^3 = A^2 \cdot A = \begin{bmatrix} 4 & -5 & 4 \ -3 & 2 & -2 \ 6 & -8 & 5 \end{bmatrix} \begin{bmatrix} 1 & -2 & 1 \ 0 & 1 & -1 \ 3 & -1 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} 4+0+12 & -8-5-4 & 4+5+4 \ -3+0-6 & 6+2+2 & -3-2-2 \ 6+0+15 & -12-8-5 & 6+8+5 \end{bmatrix}$$
$$= \begin{bmatrix} 16 & -17 & 13 \ -9 & 10 & -7 \ 21 & -25 & 16 \end{bmatrix}$$
$$A^3 – 3A^2 – A – 3I = \begin{bmatrix} 16 & -17 & 13 \ -9 & 10 & -7 \ 21 & -25 & 16 \end{bmatrix} – 3\begin{bmatrix} 4 & -5 & 4 \ -3 & 2 & -2 \ 6 & -8 & 5 \end{bmatrix} – \begin{bmatrix} 1 & -2 & 1 \ 0 & 1 & -1 \ 3 & -1 & 1 \end{bmatrix} – 3\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} 16 & -17 & 13 \ -9 & 10 & -7 \ 21 & -25 & 16 \end{bmatrix} – \begin{bmatrix} 12 & -15 & 12 \ -9 & 6 & -6 \ 18 & -24 & 15 \end{bmatrix} – \begin{bmatrix} 1 & -2 & 1 \ 0 & 1 & -1 \ 3 & -1 & 1 \end{bmatrix} – \begin{bmatrix} 3 & 0 & 0 \ 0 & 3 & 0 \ 0 & 0 & 3 \end{bmatrix}$$
$$= \begin{bmatrix} 16 – 12 – 1 – 3 & -17 + 15 + 2 – 0 & 13 – 12 – 1 – 0 \ -9 + 9 + 0 – 0 & 10 – 6 – 1 – 3 & -7 + 6 + 1 + 0 \ 21 – 18 – 3 + 0 & -25 + 24 + 1 + 0 & 16 – 15 – 1 – 3 \end{bmatrix}$$
$$= \begin{bmatrix} 0 & 0 & 0 \ 0 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$$
$$= 0$$
SAQ-8 : If I = [1 0 0 1], E = [0 1 0 0] then S.T (aI + bE)3 = a3I + 3a2bE, where I is unit matrix of order 2
$$Given I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}, E = \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$$
$$aI + bE = a \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + b \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix} = \begin{bmatrix} a & 0 \ 0 & a \end{bmatrix} + \begin{bmatrix} 0 & b \ 0 & 0 \end{bmatrix} = \begin{bmatrix} a & b \ 0 & a \end{bmatrix}$$
$$\text{L.H.S} = (aI + bE)^3 = \left( \begin{bmatrix} a & b \ 0 & a \end{bmatrix} \right)^3 = \begin{bmatrix} a & b \ 0 & a \end{bmatrix} \begin{bmatrix} a & b \ 0 & a \end{bmatrix}$$
$$= \begin{bmatrix} a^2 + 0 & ab + ab \ 0 + 0 & 0 + a^2 \end{bmatrix} \begin{bmatrix} a & b \ 0 & a \end{bmatrix}$$
$$= \begin{bmatrix} a^2 & 2ab \ 0 & a^2 \end{bmatrix} \begin{bmatrix} a & b \ 0 & a \end{bmatrix}$$
$$= \begin{bmatrix} a^3 & 3a^2b \ 0 & a^3 \end{bmatrix} \quad \cdots (1)$$
$$\text{R.H.S} = a^3I + 3a^2bE$$
$$= a^3 \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} + 3a^2b \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}$$
$$= \begin{bmatrix} a^3 & 0 \ 0 & a^3 \end{bmatrix} + \begin{bmatrix} 0 & 3a^2b \ 0 & 0 \end{bmatrix}$$
$$= \begin{bmatrix} a^3 & 3a^2b \ 0 & a^3 \end{bmatrix} \quad \cdots (2)$$
$$\text{L.H.S} = \text{R.H.S}$$
SAQ-9 : If θ – ∅ = π/2, then S.T [cos2θ cosθsinθ cosθsinθ sin2θ] [cos2∅ cos∅sin∅ cos∅sin∅ sin2∅] = 0
$$Given that \theta – \phi = \frac{\pi}{2}$$
$$\theta = \frac{\pi}{2} + \phi$$
$$\text{Hence, } \cos\theta = \cos\left(\frac{\pi}{2} + \phi\right) = -\sin\phi$$
$$\sin\theta = \sin\left(\frac{\pi}{2} + \phi\right) = \cos\phi$$
$$\begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta & \cos\theta\sin\theta & \sin^2\theta \end{bmatrix} \begin{bmatrix} \cos^2\phi & \cos\phi\sin\phi & \cos\phi\sin\phi & \sin^2\phi \end{bmatrix}$$
$$= \begin{bmatrix} \sin^2\phi & -\sin\phi\cos\phi & -\sin\phi\cos\phi & \cos^2\phi \end{bmatrix} \begin{bmatrix} \cos^2\phi & \cos\phi\sin\phi & \cos\phi\sin\phi & \sin^2\phi \end{bmatrix}$$
$$= \begin{bmatrix} \sin^2\phi\cos^2\phi – \sin^2\phi\cos^2\phi & \sin^3\phi\cos\phi – \sin^3\phi\cos\phi & -\sin\phi\cos^3\phi + \sin\phi\cos^3\phi & -\sin^2\phi\cos^2\phi + \sin^2\phi\cos^2\phi \end{bmatrix}$$
$$= \begin{bmatrix} 0 & 0 & 0 & 0 \end{bmatrix}$$
$$= 0$$
SAQ-10 : If A = [3 -3 4 2 -3 4 0 -1 1] then show that A-1 = A3
$$A^{-1} = A^3 \Leftrightarrow A A^{-1} = A A^3 \Leftrightarrow I = A^4 \Leftrightarrow A^4 = I$$
$$\text{Given } A = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix}, \text{ then}$$
$$A^2 = A \cdot A = \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} 3 & -3 & 4 \ 2 & -3 & 4 \ 0 & -1 & 1 \end{bmatrix}$$
$$= \begin{bmatrix} 9-6+0 & -9+9-4 & 12-12+4 \ 6-6+0 & -6+9-4 & 0+0+0 \ 0-2+0 & 0+3-1 & 0-4+1 \end{bmatrix}$$
$$= \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix}$$
$$A^4 = (A^2)(A^2) = \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix} \begin{bmatrix} 3 & -4 & 4 \ 0 & -1 & 0 \ -2 & 2 & -3 \end{bmatrix}$$
$$= \begin{bmatrix} 9+0-8 & -12+4+8 & 12+0-12 \ 0+0+0 & 0+1+0 & 0+0+0 \ -6+0+6 & 8-2-6 & -8+0+9 \end{bmatrix}$$
$$= \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$$
$$= I$$
SAQ-11 : If 3A = [1 2 2 2 1 -2 -2 2 -1] then show that A-1 = AT
$$A^{-1} = A^T \Leftrightarrow A A^{-1} = A A^T \Leftrightarrow I = A A^T \Leftrightarrow A A^T = I$$
$$\text{Given } 3A = \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ -2 & 2 & -1 \end{bmatrix}, \text{ then}$$
$$A = \frac{1}{3} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ -2 & 2 & -1 \end{bmatrix}$$
$$A^T = \frac{1}{3} \begin{bmatrix} 1 & 2 & -2 \ 2 & 1 & 2 \ 2 & -2 & -1 \end{bmatrix}$$
$$\text{Now, } A A^T = \frac{1}{3} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ -2 & 2 & -1 \end{bmatrix} \cdot \frac{1}{3} \begin{bmatrix} 1 & 2 & -2 \ 2 & 1 & 2 \ 2 & -2 & -1 \end{bmatrix}$$
$$= \frac{1}{9} \begin{bmatrix} 1 & 2 & 2 \ 2 & 1 & -2 \ -2 & 2 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 & -2 \ 2 & 1 & 2 \ 2 & -2 & -1 \end{bmatrix}$$
$$= \frac{1}{9} \begin{bmatrix} 1 \times 1 + 2 \times 2 + 2 \times 2 & 1 \times 2 + 2 \times 1 – 2 \times 2 & 1 \times -2 + 2 \times 2 – 2 \times 1 \ 2 \times 1 + 1 \times 2 – 2 \times 2 & 2 \times 2 + 1 \times 1 + -2 \times -2 & 2 \times -2 + 1 \times 2 – 2 \times -1 \ -2 \times 1 + 2 \times 2 – 1 \times 2 & -2 \times 2 + 2 \times 1 + 2 \times -1 & -2 \times -2 + 2 \times -2 + 1 \times -1 \end{bmatrix}$$
$$= \frac{1}{9} \begin{bmatrix} 1 + 4 + 4 & 2 + 2 – 4 & -2 + 4 – 2 \ 2 + 2 – 4 & 4 + 1 + 4 & -4 + 2 + 2 \ -2 + 4 – 2 & -4 + 2 – 2 & 4 – 4 – 1 \end{bmatrix}$$
$$= \frac{1}{9} \begin{bmatrix} 9 & 0 & 0 \ 0 & 9 & 0 \ 0 & 0 & 9 \end{bmatrix}$$
$$= \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}$$
$$= I$$