Matrices (VSAQs)

Maths-1A | 3. Matrices – VSAQs:
Welcome to VSAQs in Chapter 3: Matrices. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.

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VSAQ-1 : If [x-3 2y-8 z+2 6] = [5 2 -2 a-4] then find the values of x, y, z and a

$$\text{Given } \begin{bmatrix} x-3 & 2y-8 & z+2 & 6 \end{bmatrix} = \begin{bmatrix} 5 & 2 & -2 & a-4 \end{bmatrix}$$

$$x – 3 = 5$$

$$x = 5 + 3 = 8$$

$$2y – 8 = 2$$

$$2y = 2 + 8$$

$$2y = 10$$

$$y = 5$$

$$z + 2 = -2$$

$$z = -2 – 2 = -4$$

$$a – 4 = 6$$

$$a = 6 + 4$$

$$a = 10$$

$$x = 8, \quad y = 5, \quad z = -4, \quad a = 10$$

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VSAQ-2 : If [x-1 2 5-y 0 z-1 7 1 0 a-5] = [1 2 3 0 4 7 1 0 0] then find the values of x, y, z and a

$$\text{Given } \begin{bmatrix} x-1 & 2 & 5-y & 0 & z-1 & 7 & 1 & 0 & a-5 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 & 0 & 4 & 7 & 1 & 0 & 0 \end{bmatrix}$$

$$x – 1 = 1$$

$$x = 1 + 1 = 2$$

$$5 – y = 3$$

$$y = 5 – 3 = 2$$

$$z – 1 = 4$$

$$z = 4 + 1 = 5$$

$$a – 5 = 0$$

$$a = 5$$

$$x = 2, \quad y = 2, \quad z = 5, \quad a = 5$$

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VSAQ-3 : If A = [1 2 3 3 2 1] and B = [3 2 1 1 2 3], find 3B – 2A

$$\text{Given } B = \begin{bmatrix} 3 & 2 & 1 \ 1 & 2 & 3 \end{bmatrix}, \quad A = \begin{bmatrix} 1 & 2 & 3 \ 3 & 2 & 1 \end{bmatrix}$$

$$3B – 2A = 3 \begin{bmatrix} 3 & 2 & 1 \ 1 & 2 & 3 \end{bmatrix} – 2 \begin{bmatrix} 1 & 2 & 3 \ 3 & 2 & 1 \end{bmatrix}$$

$$= \begin{bmatrix} 9 & 6 & 3 \ 3 & 6 & 9 \end{bmatrix} – \begin{bmatrix} 2 & 4 & 6 \ 6 & 4 & 2 \end{bmatrix}$$

$$= \begin{bmatrix} 9-2 & 6-4 & 3-6 \ 3-6 & 6-4 & 9-2 \end{bmatrix}$$

$$= \begin{bmatrix} 7 & 2 & -3 \ -3 & 2 & 7 \end{bmatrix}$$

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VSAQ-4 : If A = [1 2 3 4], B = [3 8 7 2] and 2X + A = B then find X

$$\text{Given } A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}, \quad B = \begin{bmatrix} 3 & 8 \ 7 & 2 \end{bmatrix}$$

$$2X + A = B$$

$$2X = B – A = \begin{bmatrix} 3 & 8 \ 7 & 2 \end{bmatrix} – \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$$

$$= \begin{bmatrix} 3-1 & 8-2 \ 7-3 & 2-4 \end{bmatrix}$$

$$= \begin{bmatrix} 2 & 6 \ 4 & -2 \end{bmatrix}$$

$$= 2 \begin{bmatrix} 1 & 3 \ 2 & -1 \end{bmatrix}$$

$$2X = 2 \begin{bmatrix} 1 & 3 \ 2 & -1 \end{bmatrix}$$

$$X = \begin{bmatrix} 1 & 3 \ 2 & -1 \end{bmatrix}$$

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VSAQ-5 : If A = [2 3 -1 7 8 5] and B = [1 0 1 2 -4 -1] then find A + B

$$A + B = \begin{bmatrix} 2 & 3 & -1 \ 7 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 1 & 0 & 1 \ 2 & -4 & -1 \end{bmatrix}$$

$$= \begin{bmatrix} 2+1 & 3+0 & -1+1 \ 7+2 & 8-4 & 5-1 \end{bmatrix}$$

$$= \begin{bmatrix} 3 & 3 & 0 \ 9 & 4 & 4 \end{bmatrix}$$

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VSAQ- 6 : Define Trace of a matrix, find the trace of [1 3 -5 2 -1 5 2 0 1]

$$\text{The trace of a square matrix is the sum of elements in the principal diagonal.}$$

$$\text{Given:}$$

$$A = \begin{bmatrix} 1 & 3 & -5 \ 2 & -1 & 5 \ 2 & 0 & 1 \end{bmatrix}$$

$$\text{The trace, } \text{Tr}(A), \text{ is:}$$

$$\text{Tr}(A) = 1 + (-1) + 1 = 1$$

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VSAQ-7 : If A = [i 0 0 i] then find A²

$$A^2 = A \times A = \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix} \begin{bmatrix} i & 0 \ 0 & i \end{bmatrix}$$

$$= \begin{bmatrix} i \cdot i + 0 \cdot 0 & i \cdot 0 + 0 \cdot i \ 0 \cdot i + i \cdot 0 & 0 \cdot 0 + i \cdot i \end{bmatrix}$$

$$= \begin{bmatrix} i^2 & 0 \ 0 & i^2 \end{bmatrix}$$

$$= \begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}$$

$$= (-1) \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$$

$$= -I$$

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VSAQ-8 : If A = [2 4 -1 k] and A² = [0 0 0 0] then find the value of k

$$\text{Given } A = \begin{bmatrix} 2 & 4 \ -1 & k \end{bmatrix}$$

$$A^2 = A \times A = \begin{bmatrix} 2 & 4 \ -1 & k \end{bmatrix} \begin{bmatrix} 2 & 4 \ -1 & k \end{bmatrix}$$

$$= \begin{bmatrix} 2(2) + 4(-1) & 2(4) + 4(k) \ -1(2) + k(-1) & -1(4) + k(k) \end{bmatrix}$$

$$= \begin{bmatrix} 4 – 4 & 8 + 4k \ -2 – k & -4 + k^2 \end{bmatrix}$$

$$= \begin{bmatrix} 0 & 8 + 4k \ -2 – k & -4 + k^2 \end{bmatrix}$$

$$8 + 4k = 0$$

$$4k = -8$$

$$k = -2$$

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VSAQ-9 : If A = [2 -4 -5 3] then find A + A’ and AA’

$$A = \begin{bmatrix} 2 & -4 \ -5 & 3 \end{bmatrix}, \quad A’ = \begin{bmatrix} 2 & -5 \ -4 & 3 \end{bmatrix}$$

(i) Calculation of A+A′:

$$A + A’ = \begin{bmatrix} 2 & -4 \ -5 & 3 \end{bmatrix} + \begin{bmatrix} 2 & -5 \ -4 & 3 \end{bmatrix}$$

$$= \begin{bmatrix} 2+2 & -4-5 \ -5-4 & 3+3 \end{bmatrix}$$

$$= \begin{bmatrix} 4 & -9 \ -9 & 6 \end{bmatrix}$$

(ii) Calculation of AA′:

$$AA’ = \begin{bmatrix} 2 & -4 \ -5 & 3 \end{bmatrix} \begin{bmatrix} 2 & -5 \ -4 & 3 \end{bmatrix}$$

$$= \begin{bmatrix} 2 \cdot 2 + (-4) \cdot (-4) & 2 \cdot (-5) + (-4) \cdot 3 \ -5 \cdot 2 + 3 \cdot (-4) & -5 \cdot (-5) + 3 \cdot 3 \end{bmatrix}$$

$$= \begin{bmatrix} 4 + 16 & -10 – 12 \ -10 – 12 & 25 + 9 \end{bmatrix}$$

$$= \begin{bmatrix} 20 & -22 \ -22 & 34 \end{bmatrix}$$

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VSAQ-10 : If A = [2 -1 2 1 3 -4], B = [1 -2 -3 0 5 4] then verify that (AB)’ = B’A’

$$\text{Given:}$$

$$A = \begin{bmatrix} 2 & -1 \ 1 & 3 \ 3 & -4 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & -2 \ -3 & 0 \ 5 & 4 \end{bmatrix}$$

(i) Calculation of AB:

$$AB = \begin{bmatrix} 2 & -1 \ 1 & 3 \ 3 & -4 \end{bmatrix} \begin{bmatrix} 1 & -2 \ -3 & 0 \ 5 & 4 \end{bmatrix}$$

$$= \begin{bmatrix} (2 \cdot 1) + (-1 \cdot -3) + (2 \cdot 5) & (2 \cdot -2) + (-1 \cdot 0) + (2 \cdot 4) \ (1 \cdot 1) + (3 \cdot -3) + (-4 \cdot 5) & (1 \cdot -2) + (3 \cdot 0) + (-4 \cdot 4) \end{bmatrix}$$

$$= \begin{bmatrix} 2 + 3 + 10 & -4 + 0 + 8 \ 1 – 9 – 20 & -2 + 0 – 16 \end{bmatrix}$$

$$= \begin{bmatrix} 15 & 4 \ -28 & -18 \end{bmatrix}$$

(ii) Calculation of B′A′:

$$B’A’ = \begin{bmatrix} 1 & -2 \ -3 & 0 \ 5 & 4 \end{bmatrix} \begin{bmatrix} 2 & 1 \ -1 & 3 \ 2 & -4 \end{bmatrix}$$

$$= \begin{bmatrix} (1 \cdot 2) + (-3 \cdot -1) + (5 \cdot 2) & (1 \cdot 1) + (-3 \cdot 3) + (5 \cdot -4) \ (-2 \cdot 2) + (0 \cdot -1) + (4 \cdot 2) & (-2 \cdot 1) + (0 \cdot 3) + (4 \cdot -4) \end{bmatrix}$$

$$= \begin{bmatrix} 2 + 3 + 10 & 1 – 9 – 20 \ -4 + 0 + 8 & -2 + 0 – 16 \end{bmatrix}$$

$$= \begin{bmatrix} 15 & 4 \ -28 & -18 \end{bmatrix}$$

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

$$\begin{bmatrix} 15 & 4 \ -28 & -18 \end{bmatrix} = \begin{bmatrix} 15 & 4 \ -28 & -18 \end{bmatrix}$$

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VSAQ-11 : Define a symmetric matrix. If A = [-1 2 3 2 5 6 3 x 7] is symmetric find the value of x

$$\text{A square matrix } A \text{ is said to be a symmetric matrix if } A = A^T$$

$$\text{Given:}$$

$$A = \begin{bmatrix} -1 & 2 & 3 \ 2 & 5 & 6 \ 3 & x & 7 \end{bmatrix}$$

$$A^T = \begin{bmatrix} -1 & 2 & 3 \ 2 & 5 & x \ 3 & 6 & 7 \end{bmatrix}$$

$$\text{Now, since } A = A^T$$

$$x = 6$$

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VSAQ-12 : Define a skew symmetric matrix. If [0 2 1 -2 0 -2 -1 x 0] is a skew symmetric matrix then find the value of x

$$\text{A square matrix } A \text{ is said to be a skew symmetric matrix if } A = -A^T$$

$$\text{Given:}$$

$$A = \begin{bmatrix} 0 & 2 & 1 \ -2 & 0 & -2 \ -1 & x & 0 \end{bmatrix}$$

$$-A^T = -\begin{bmatrix} 0 & -2 & -1 \ 2 & 0 & x \ 1 & -2 & 0 \end{bmatrix}$$

$$= \begin{bmatrix} 0 & 2 & 1 \ -2 & 0 & -x \ -1 & 2 & 0 \end{bmatrix}$$

$$\text{Now, since } A = -A^T$$

$$x = 2$$

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VSAQ-13 : If A = [1 0 0 2 3 4 5 -6 x] and det A = 45 then find x

$$\text{Given: } \left| \begin{matrix} 1 & 0 & 0 \ 2 & 3 & 4 \ 5 & -6 & x \end{matrix} \right| = 45$$

$$1 \cdot \left| \begin{matrix} 3 & 4 \ -6 & x \end{matrix} \right| + 0 + 0 = 45$$

$$1 \cdot (3x + 24) = 45$$

$$3x + 24 = 45$$

$$3x = 45 – 24$$

$$3x = 21$$

$$x = \frac{21}{3} = 7$$

$$x = 7$$

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VSAQ-14 : Find the determinant of the matrix [1² 2² 3² 2² 3² 4² 3² 4² 5²]

$$\text{Given:} \left| \begin{matrix} 12 & 22 & 32 \ 22 & 32 & 42 \ 32 & 42 & 52 \end{matrix} \right| = \left| \begin{matrix} 1 & 4 & 9 \ 4 & 9 & 16 \ 9 & 16 & 25 \end{matrix} \right|$$

$$= 1 \cdot \left( 9 \times 25 – 16 \times 16 \right) – 4 \cdot \left( 4 \times 25 – 9 \times 16 \right) + 9 \cdot \left( 16 \times 4 – 9 \times 9 \right)$$

$$= 1 \cdot (225 – 256) – 4 \cdot (100 – 144) + 9 \cdot (64 – 81)$$

$$= 1 \cdot (-31) – 4 \cdot (-44) + 9 \cdot (-17)$$

$$= -31 + 176 – 153$$

$$= 176 – 184$$

$$= -8$$

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VSAQ-15 : Find the Adjoint and inverse of A = [2 -3 4 6]

$$|A| = 2 \cdot 6 – 4 \cdot (-3)$$

$$= 12 + 12$$

$$= 24$$

$$\text{Adj } A = \begin{bmatrix} 6 & 3 \ -4 & 2 \end{bmatrix}$$

$$A^{-1} = \frac{1}{|A|} (\text{Adj } A)$$

$$= \frac{1}{24} \begin{bmatrix} 6 & 3 \ -4 & 2 \end{bmatrix}$$

$$= \begin{bmatrix} \frac{6}{24} & \frac{3}{24} \ \frac{-4}{24} & \frac{2}{24} \end{bmatrix}$$

$$= \begin{bmatrix} \frac{1}{4} & \frac{1}{8} \ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}$$

$$A^{-1} = \begin{bmatrix} \frac{1}{4} & \frac{1}{8} \ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}$$

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VSAQ-16 : Find the inverse of A = [1 2 3 -5]

$$|A| = 1(-5) – 3(2)$$

$$= -5 – 6$$

$$= -11$$

$$\text{Adj } A = \begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix}$$

$$A^{-1} = \frac{1}{|A|} (\text{Adj } A)$$

$$= -\frac{1}{11} \begin{bmatrix} -5 & -2 \ -3 & 1 \end{bmatrix}$$

$$= \begin{bmatrix} \frac{5}{11} & \frac{2}{11} \ \frac{3}{11} & -\frac{1}{11} \end{bmatrix}$$

$$A^{-1} = \begin{bmatrix} \frac{5}{11} & \frac{2}{11} \ \frac{3}{11} & -\frac{1}{11} \end{bmatrix}$$

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VSAQ-17 : Find the rank of [1 1 1 1 1 1 1 1 1]

$$\text{We take }
A = \begin{bmatrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{bmatrix}$$

$$|A| = \left| \begin{matrix} 1 & 1 & 1 \ 1 & 1 & 1 \ 1 & 1 & 1 \end{matrix} \right| = 0$$

$$\text{Take a } 2 \times 2 \text{ minor, } \left| \begin{matrix} 1 & 1 \ 1 & 1 \end{matrix} \right| = 0$$

$$\text{Take a } 1 \times 1 \text{ minor, } \left| 1 \right| \neq 0$$

$$\text{A = 1}$$

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VSAQ-18 : Find the rank of [1 2 3 2 3 4 0 1 2]

$$\text{We take }
A = \begin{bmatrix} 1 & 2 & 3 \ 2 & 3 & 4 \ 0 & 1 & 2 \end{bmatrix}$$

$$|A| = \left| \begin{matrix} 1 & 2 & 3 \ 2 & 3 & 4 \ 0 & 1 & 2 \end{matrix} \right| = 1(6 – 4) – 2(4 – 0) + 3(2 – 0)$$

$$= 2 – 8 + 6$$

$$= 0$$

$$|A| = 0$$

$$\text{Take a } 2 \times 2 \text{ minor, } \left| \begin{matrix} 1 & 2 \ 2 & 3 \end{matrix} \right| = 3 – 4 = -1 \neq 0$$

$$\text{Rank}(A) = 2$$