Question

how??? Matrix Question​

Answer 1

Answer:

a=3,b=-2

Explanation:

Given

$\left[\begin{array}{c c} (a-4b) & 5 \\ 6 & (-a+b) \end{array}\right]$ =$\left[\begin{array}{c c} 11 & 5 \\ 6 & -5 \end{array}\right]$

As both the matrices are given equal

we can directly equate each term of 1st matrix to 2nd matrix

Equating Ist term in 1st row and 1st column of 1st matrix to the 2nd matrix

$a - 4b = 11............(1)$

Equating 2nd term in 2nd row and 2nd column in 1st matrix to the 2nd matrix

$- a + b = - 5..................(2)$

$(1) + (2)$

$= > a - 4b + ( - a + b) = 11 + ( -5)$

$= > a - a - 4b + b = 11 - 5$

$= > - 3b = 6$

$= > b = \dfrac{6}{ - 3} = - 2$

$\boxed{b = - 2}$

Substituting the value of b in (2)

$- a + b = - 5$

$= > - a - 2 = - 5$

$= > - a = - 5 + 2$

$= > - a = - 3$

$= > a = 3$

$\boxed{a = 3}$

Answer 2

Step-by-step explanation:

Answer:

a=3,b=-2

Explanation:

Given

\begin{lgathered}\left[\begin{array}{c c} (a-4b) & 5 \\ 6 & (-a+b) \end{array}\right]\end{lgathered}

$\begin{lgathered}\left[\begin{array}{c c} (a-4b) & 5 \\ 6 & (-a+b) \end{array}\right]\end{lgathered}$

[

(a−4b)

6

5

(−a+b)

]

=\begin{lgathered}\left[\begin{array}{c c} 11 & 5 \\ 6 & -5 \end{array}\right]\end{lgathered}

[

$=\begin{lgathered}\left[\begin{array}{c c} 11 & 5 \\ 6 & -5 \end{array}\right]\end{lgathered} </p><p>[$

11

6

5

−5

]

As both the matrices are given equal

we can directly equate each term of 1st matrix to 2nd matrix

Equating Ist term in 1st row and 1st column of 1st matrix to the 2nd matrix

a - 4b = 11............(1)a−4b=11............(1)

Equating 2nd term in 2nd row and 2nd column in 1st matrix to the 2nd matrix

- a + b = - 5..................(2)−a+b=−5..................(2)

(1) + (2)(1)+(2)

= > a - 4b + ( - a + b) = 11 + ( -5)=>a−4b+(−a+b)=11+(−5)

= > a - a - 4b + b = 11 - 5=>a−a−4b+b=11−5

= > - 3b = 6=>−3b=6

= > b = \dfrac{6}{ - 3} = - 2=>b=

−3

6

=−2

\boxed{b = - 2}

b=−2

Substituting the value of b in (2)

- a + b = - 5−a+b=−5

= > - a - 2 = - 5=>−a−2=−5

= > - a = - 5 + 2=>−a=−5+2

= > - a = - 3=>−a=−3

= > a = 3=>a=3

\boxed{a = 3}

$\boxed{a = 3}$

a=3

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