Matrices — Exercise 8.1

Classify the following matrices :

$\text{(i) }\begin{bmatrix*}[r] 2 & -1 \\ 5 & 1 \end{bmatrix*} \\[0.5em]$

$\text{(ii) } \begin{bmatrix} 2 & 3 & -7 \end{bmatrix} \\[0.5em]$

$\text{(iii) } \begin{bmatrix*}[r] 3 \\ 0 \\ -1 \end{bmatrix*} \\[0.5em]$

$\text{(iv) } \begin{bmatrix} 2 & -4 \\ 0 & 0 \\ 1 & 7 \end{bmatrix} \\[0.5em]$

$\text{(v) } \begin{bmatrix*}[r] 2 & 7 & 8 \\ -1 & \sqrt{2} & 0 \end{bmatrix*} \\[0.5em]$

$\text{(vi) } \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \\[0.5em]$

Answer

(i) It is a square matrix of order 2.

(ii) It is a row matrix of order 1 x 3.

(iii) It is a column matrix of order 3 x 1.

(iv) It is a 3 x 2 matrix.

(v) It is a 2 x 3 matrix.

(vi) It is a zero matrix of order 2 x 3.

If a matrix has 4 elements, what are the possible orders it can have ?

Answer

If a matrix has 4 elements the possible orders are,

1 x 4, 2 x 2, 4 x 1.

If a matrix has 8 elements, what are the possible orders it can have ?

Answer

If a matrix has 8 elements the possible orders are,

1 x 8, 4 x 2, 2 x 4, 8 x 1.

Construct a 2 $\times$ 2 matrix whose elements a_ij are given by

(i) a_ij = 2i - j

(ii) a_ij = i.j

Answer

(i) Given a_ij = 2i - j,

∴ a₁₁ = 2(1) - 1 = 1, a₁₂ = 2(1) - 2 = 0,
    a₂₁ = 2(2) - 1 = 3, a₂₂ = 2(2) - 2 = 2.

Hence, the required matrix =$\begin{bmatrix} 1 & 0 \\ 3 & 2 \end{bmatrix}$.

(ii) Given a_ij = i.j,

∴ a₁₁ = 1.1 = 1, a₁₂ = 1.2 = 2,
    a₂₁ = 2.1 = 2, a₂₂ = 2.2 = 4.

Hence, the required matrix =$\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$.

Find the values of x and y if $\begin{bmatrix*}[r] 2x + y \\ 3x - 2y \end{bmatrix*} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$.

Answer

Given $\begin{bmatrix*}[r] 2x + y \\ 3x - 2y \end{bmatrix*} = \begin{bmatrix} 5 \\ 4 \end{bmatrix}$.

By definition of equality of matrices, we get

2x + y = 5    [....Eq 1]
3x - 2y = 4    [....Eq 2]

2x + y = 5
⇒ y = 5 - 2x

Putting value of y in Eq 2,

3x - 2(5 - 2x) = 4
⇒ 3x - 10 + 4x = 4
⇒ 7x = 14
⇒ x = 2.

∴ x = 2, y = 5 - 2x = 5 - 4 = 1.

Hence, the value of x = 2 and y = 1.

Find the values of x if $\begin{bmatrix*}[r] 3x + y & -y \\ 2y - x & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ -5 & 3 \end{bmatrix*}$.

Answer

Given $\begin{bmatrix*}[r] 3x + y & -y \\ 2y - x & 3 \end{bmatrix*} = \begin{bmatrix*}[r] 1 & 2 \\ -5 & 3 \end{bmatrix*}$

By definition of equality of matrices, we get

-y = 2 or y = -2 and
3x + y = 1

Putting value of y = -2 in above equation

⇒ 3x - 2 = 1
⇒ 3x = 3
⇒ x = 1

Hence, the value of x = 1.

If $\begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix*}[r] 5 & 4 \\ 3 & 9 \end{bmatrix*}$, find the values of x and y.

Answer

Given $\begin{bmatrix*}[r] x + 3 & 4 \\ y - 4 & x + y \end{bmatrix*} = \begin{bmatrix} 5 & 4 \\ 3 & 9 \end{bmatrix}$.

By definition of equality of matrices, we get

    x + 3 = 5, y - 4 = 3 and x + y = 9
⇒ x = 2, y = 7 and x + y = 9.

Note that x = 2 and y = 7 satisfies the equation x + y = 9.

Hence, the value of x = 2 and y = 7.

Find the values of x, y and z if $\begin{bmatrix*}[r] x + 2 & 6 \\ 3 & 5z \end{bmatrix*} = \begin{bmatrix*}[r] -5 & y^2 + y \\ 3 & -20 \end{bmatrix*} .$

Answer

Given $\begin{bmatrix*}[r] x + 2 & 6 \\ 3 & 5z \end{bmatrix*} = \begin{bmatrix*}[r] -5 & y^2 + y \\ 3 & -20 \end{bmatrix*} .$

By definition of equality of matrices, we get

x + 2 = -5 or x = -7,
y² + y = 6       [....Eq 1],
5z = -20 or z = -4.

From Eq 1 we get,

$\Rightarrow y^2 + y - 6 = 0 \\[0.5em] \Rightarrow y^2 + 3y - 2y - 6 = 0 \\[0.5em] \Rightarrow y(y + 3) - 2(y + 3) = 0 \\[0.5em] \Rightarrow (y - 2)(y + 3) = 0 \\[0.5em] \Rightarrow y = 2 \text{ or } y = -3. \\[0.5em]$

∴ x = -7, y = 2 or -3 and z = -4.

Hence, the values are:
x = -7, y = 2 and z = -4
OR
x = -7, y = -3 and z = -4.

Find the values of x, y, a and b if $\begin{bmatrix*}[r] x - 2 & y \\ a + 2b & 3a - b \end{bmatrix*} = \begin{bmatrix*}[r] 3 & 1 \\ 5 & 1 \end{bmatrix*}$.

Answer

Given, $\begin{bmatrix*}[r] x - 2 & y \\ a + 2b & 3a - b \end{bmatrix*} = \begin{bmatrix*}[r] 3 & 1 \\ 5 & 1 \end{bmatrix*}$.

By definition of equality of matrices, we get

x - 2 = 3 or x = 5,
y = 1,
a + 2b = 5       [...Eq 1],
3a - b = 1        [...Eq 2].

From Eq 1 we get,

⇒ a + 2b = 5
⇒ a = 5 - 2b

Putting this value of a in Eq 2,

⇒ 3(5 - 2b) - b = 1
⇒ 15 - 6b - b = 1
⇒ 7b = 14
⇒ b = 2.

Using value of b to find value of a,

⇒ a = 5 - 2b
⇒ a = 5 - 2(2)
⇒ a = 5 - 4
⇒ a = 1.

∴ x = 5, y = 1, a = 1 and b = 2.

Hence, the values are:
x = 5, y = 1, a = 1 and b = 2.

Find the values of a, b, c and d if $\begin{bmatrix*}[r] a + b & 3 \\ 5 + c & ab \end{bmatrix*} = \begin{bmatrix*}[r] 6 & d \\ -1 & 8 \end{bmatrix*} .$

Answer

Given $\begin{bmatrix*}[r] a + b & 3 \\ 5 + c & ab \end{bmatrix*} = \begin{bmatrix*}[r] 6 & d \\ -1 & 8 \end{bmatrix*} .$

By definition of equality of matrices, we get

d = 3,
5 + c = -1 or c = -6,
a + b = 6       [...Eq 1],
ab = 8            [...Eq 2].

Putting value of b from Eq 1 in Eq 2,

$a + b = 6 \text{ or } b = 6 - a \\[1em] \Rightarrow a(6 - a) = 8 \\[0.5em] \Rightarrow 6a - a^2 = 8 \\[0.5em] \Rightarrow a^2 - 6a + 8 = 0 \\[0.5em] \Rightarrow a^2 - 4a - 2a + 8 = 0 \\[0.5em] \Rightarrow a(a - 4) - 2(a - 4) = 0 \\[0.5em] \Rightarrow (a - 2)(a - 4) = 0 \\[0.5em] \Rightarrow a = 2 \text{ or } a = 4.$

Now, finding value of b = 6 - a,

if,  a = 2, b = 6 - 2 = 4.

or, a = 4, b = 6 - 4 = 2.

∴ a = 2 or 4, b = 4 or 2, c = -6 and d = 3.

Hence, the values are
a = 2, b = 4, c = -6, d = 3
OR
a = 4, b = 2, c = -6, d = 3.