Question

please answer please please please​

Answer 1

Question:

$\sf{If\:A=\left[\begin{array}{cc}3&2\\-1&1\\\end{array}\right]\:and \:B=\left[\begin{array}{cc}14&3\\2&4\\\end{array}\right] ,find\:a\:matrix\:C\:such\:that\:AC=B}$

Answer:

$\sf{C=\left[\begin{array}{cc}2&-1\\4&3\\\end{array}\right] }$

Step-by-step explanation:

Given:

$\sf{A=\left[\begin{array}{cc}3&2\\-1&1\\\end{array}\right] }$

$\sf{B=\left[\begin{array}{cc}14&3\\2&4\\\end{array}\right]}$

AC = B

To Find:

Matrix C

Solution:

Given that,

AC = B

Hence,

AC - B = O

where O is the zero matrix.

Let us assume the matrix C is,

$\sf{C=\left[\begin{array}{cc}a&b\\c&d\\\end{array}\right] }$

Substituting the values in the equation we get,

$\left[\begin{array}{cc}3&2\\-1&1\\\end{array}\right] \times \left[\begin{array}{cc}a&b\\c&d\\\end{array}\right] -\left[\begin{array}{cc}14&3\\2&4\\\end{array}\right]=\left[\begin{array}{ccc}0&0\\0&0\\\end{array}\right]$

Multiplying the matrices,

$\left[\begin{array}{cc}3a+2c&3b+2d\\-a+c&-b+d\\\end{array}\right]-\left[\begin{array}{cc}14&3\\2&4\\\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\\\end{array}\right]$

Now subtracting the matrices,

$\left[\begin{array}{cc}3a+2c-14&3b+2d-3\\-a+c-2&-b+d-4\\\end{array}\right] =\left[\begin{array}{cc}0&0\\0&0\\\end{array}\right]$

Equating it we get,

3a + 2c - 14 = 0

3a + 2c = 14 ------(1)

3b + 2d - 3 = 0

3b + 2d = 3-------(2)

-a + c - 2 = 0

-a + c = 2

c = 2 + a-------(3)

-b + d - 4 = 0

-b + d = 4

d = 4 + b-------(4)

Solving equation 1 and 3 by substitution method

3a + 2c = 14

Substitute value of c

3a + 2 (2 + a) = 14

3a + 4 + 2a = 14

5a = 14 - 4

5a = 10

a = 10/5

a = 2

Solving equation 2 and 4 by substitution method,

3b + 2d = 3

Substitute value of d

3b + 2 (4 + b) = 3

3b + 8 + 2b = 3

5b = 3 - 8

5b = -5

b = -5/5

b = -1

Now substitute the value of a in equation 3

c = 2 + 2

c = 4

Substitute the value of b in equation 4

d = 4 + -1

d = 3

Now we know that,

$\sf{C=\left[\begin{array}{cc}a&b\\c&d\\\end{array}\right] }$

Substituting the values we get,

$\sf{C=\left[\begin{array}{cc}2&-1\\4&3\\\end{array}\right] }$

Verification:

AC = B

$\left[\begin{array}{cc}3&2\\-1&1\\\end{array}\right]\times \left[\begin{array}{cc}2&-1\\4&3\\\end{array}\right] =\left[\begin{array}{cc}14&3\\2&4\\\end{array}\right]$

$\left[\begin{array}{cc}6+8&-3+6\\-2+4&1+3\\\end{array}\right] =\left[\begin{array}{cc}14&3\\2&4\\\end{array}\right]$

$\left[\begin{array}{cc}14&3\\2&4\\\end{array}\right] =\left[\begin{array}{cc}14&3\\2&4\\\end{array}\right]$

Hence verified.