Question

find x and y
for the matrix​

Answer 1

Question:-

If $\begin{bmatrix} \sf 4 & \sf -5 \\\\ \sf 6 & \sf 7 \end{bmatrix}\begin{bmatrix} \sf x\\\\ \sf y\end{bmatrix} = \begin{bmatrix} \sf - 1 \\ \\ \sf 2 \end{bmatrix}$ , find x & y.

Answer:-

Given:-

$\begin{bmatrix} \sf 4 & \sf -5 \\\\ \sf 6 & \sf 7 \end{bmatrix}\begin{bmatrix} \sf x\\\\ \sf y\end{bmatrix} = \begin{bmatrix} \sf - 1 \\ \\ \sf 2 \end{bmatrix}$

Any two matrices can be multiplied only if no.of columns(vertical lines) in first matrix = no. of rows(horizontal lines) in second matrix.

Multiplication of a 2×2 matrix with a 2×1 matrix:-

$\sf \: \begin{bmatrix} \sf a & \sf b \\\\ \sf c & \sf d \end{bmatrix}\begin{bmatrix} \sf e\\\\ \sf f\end{bmatrix} = \begin{bmatrix} \sf \: ae + bf \\ \\ \sf ce + df \end{bmatrix}$

Hence,

$\implies \sf \: \begin{bmatrix} \sf \: (4)(x) + ( - 5)(y) \\ \\ \sf \: (6)(x) + (7)(y) \end{bmatrix} = \begin{bmatrix} \sf - 1 \\ \\ \sf 2 \end{bmatrix} \\ \\ \\ \implies \sf \begin{bmatrix} \sf \: 4x - 5y \\ \\ \sf \: 6x + 7y \end{bmatrix} = \begin{bmatrix} \sf - 1 \\ \\ \sf 2 \end{bmatrix}$

On comparing both sides we get,

• 4x - 5y = - 1 -- equation (1)

• 6x + 7y = 2 -- equation (2).

Multiply equation (1) by 3.

⟹ 3(4x - 5y) = 3(- 1)

⟹ 12x - 15y = - 3 -- equation (3)

Multiply equation (2) by 2.

⟹ 2(6x + 7y) = 2(2)

⟹ 12x + 14y = 4 -- equation (4)

Now,

Subtract equation (4) from equation (3).

⟹ 12x - 15y - (12x + 14y) = - 3 - 4

⟹ 12x - 15y - 12x - 14y = - 7

⟹ - 29y = - 7

⟹ y = - 7/ - 29

⟹ y = 7/29

Substitute y = 7/29 in equation (1).

⟹ 4x - 5(7/29) = - 1

⟹ 4x - 35/29 = - 1

⟹ 4x = - 1 + 35/29

⟹ x = (1/4) ( - 29 + 35)/ 29

⟹ x = (1/4) (6/29)

⟹ x = 3/58

∴ The values of x & y are 3/58 & 7/29.

Answer 2

$\large\bf\underline\red{Question \: :- }$

$\small\sf{If \begin{gathered} \begin{bmatrix} \sf 4 & \sf -5 \\\\ \sf 6 & \sf 7 \end{bmatrix}\begin{bmatrix} \sf x\\\\ \sf y\end{bmatrix} = \begin{bmatrix} \sf - 1 \\ \\ \sf 2 \end{bmatrix} \end{gathered}find \: x \: and \: y}$

Given :-

$\small\sf{ \begin{gathered} \begin{bmatrix} \sf 4 & \sf -5 \\\\ \sf 6 & \sf 7 \end{bmatrix}\begin{bmatrix} \sf x\\\\ \sf y\end{bmatrix} = \begin{bmatrix} \sf - 1 \\ \\ \sf 2 \end{bmatrix} \end{gathered}}$

In that case, any two matrices can be only if no. of (vertical) columns in first = no. of (horizontal) in second.

Let assume,

$★\small{ \begin{gathered} \begin{bmatrix} \sf a & \sf b \\\\ \sf c & \sf d \end{bmatrix}\begin{bmatrix} \sf e\\\\ \sf f\end{bmatrix} = \begin{bmatrix} \sf ae + bf\\ \\ \sf ce + df \end{bmatrix} \end{gathered}}$

Multiplication of a 2×2 with a 2×1 :-

Putting,

$⟼\small{\begin{gathered} \begin{bmatrix} \sf 4(x) + & \sf -5 (y)\\\\ \sf 6(x) & \sf 7(y) \end{bmatrix} = \begin{bmatrix} \sf - 1 \\ \\ \sf 2 \end{bmatrix} \end{gathered}}$

We get,

$⟼\small{\begin{gathered} \begin{bmatrix} \sf 4 x& \sf -5y \\\\ \sf 6 x& \sf 7y \end{bmatrix}= \begin{bmatrix} \sf - 1 \\ \\ \sf 2 \end{bmatrix} \end{gathered}}$

By Comparing we get two equation,

• 4x - 5y = -1 (1)
• 6x + 7y = 2 (2)

We want to equal at least 1 finding the values,

Multiplying equation (1) by 3

We get,

• 12x - 15y = -3 (3)

Multiplying equation (2) by 2

We get,

• 12x + 14y = 4 (4)

Now, for finding values we subtract them

Subtract equation (4) from (3)

$➠ \: \sf{12x - 15 - (12x + 14y) = - 3 - 4} \\ ➠ \: \sf{12x - 15 - 12x - 14y = - 7} \: \: \: \: \: \: \: \: \: \: \: \\ ➠ \: \sf{ - 29y = -7} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ➠ \: \sf{ y = \frac{ - 7}{ - 29} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$➠ \: \sf\pink{y = \frac{7}{29} }$

Now, We substitute y = 7/29 in equation (1)

$➠ \: \sf{4x - 5( \frac{7}{29}) = -1 } \: \: \: \: \: \\ ➠ \: \sf{4x - \frac{35}{29} = -1 } \: \: \: \: \: \: \: \: \: \: \: \: \: \\ ➠ \: \: \sf{4x = -1 + \frac{35}{29} } \: \: \: \: \: \: \: \: \: \: \: \: \\ ➠ \: \sf{x = ( \frac{1}{4} )( \frac{( - 29 + 35)}{29} )} \\ ➠ \: \sf{x = \frac{1}{4} \times \frac{6}{29} = \frac{6}{116} } \: \: \: \:$

$\sf\pink{➠ \: x = \frac{3}{58} }$

Answer :

${\large{\underline{\boxed{\bf{\red{x = \frac{3}{58} \: and \: y = \frac{7}{29} }}}}}}$

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