Question

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Answer 1

Answer:

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Step-by-step explanation:

hope u understand this sol

Answer 2

Answer:

1)  Order of 'C' = 1 × 2

$2)Matrix\:'C'=\begin{bmatrix}-2& -1\end{bmatrix}$

Step-by-step explanation:

Given ,

$A=\begin{bmatrix}2 & -1 \\-4 & 5 \end{bmatrix}$

$B=\begin{bmatrix}0& -3\end{bmatrix}$

CA = B

To Find :-

1) order of 'C'

2) matrix of 'C'

Solution :-

CA = B

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

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

$AdjA=\begin{bmatrix}5& 1\\4& 2\end{bmatrix}$

$\left[\therefore If \:'A'\begin{bmatrix}a & b \\c & d \end{bmatrix}, Adj'A'=\begin{bmatrix}d & -b \\-c & a \end{bmatrix} \right]$

det A = (5 × 2) - (1 × 4)

= 10 - 4

= 6

∴ detA = 6

$\displaystyle \sf{ \implies \: C =\begin{bmatrix} 0 & - 3 \end{bmatrix} \times \frac{1}{6} \begin{bmatrix} 5 & 1\\ 4 & 2 \end{bmatrix} }$

[ ∴ A⁻¹ = Adj A/det A ]

$\displaystyle \sf{ \implies \: C = \frac{1}{6} \times \begin{bmatrix} 0 & - 3 \end{bmatrix} \begin{bmatrix} 5 & 1\\ 4 & 2 \end{bmatrix} }$

Multiplying 1st row in 'B' with 1st column in 'A' :-

= (0 × 5) + (-3 × 4)

= 0 - 12

= - 12

[ ∴ '-12' will be in the first row first column in matrix 'C' ]

Multiplying 1st row in 'B' with 2nd column in 'A' :-

= (0 × 1) + ((-3) × 2)

= 0 - 6

= -6

[ ∴ '-6' will be in the 1st row , 2nd column in matrix 'C' ]

$\implies C=\dfrac{1}{6}\times \begin{bmatrix}-12& -6\end{bmatrix}$

$C=\begin{bmatrix}\dfrac{-12}{6}&\dfrac{-6}{6}\end{bmatrix}$

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

Order of 'C' = Number of columns × Number of rows in 'C'

= 1 × 2

∴ Order of 'C' = 1 × 2

$Matrix\:'C'=\begin{bmatrix}-2& -1\end{bmatrix}$