$A_{1}$ and $A_{2}$ are two matrices of order 3 $\times$ 3 satisfying the matrix equations $3 A_{1}^{T} + I^{3} = 10 A_{1}$ and $4 I^{3} + 3 A_{2} = 7 A_{2}^{T}$ , then

| A_{1} + A_{2} | = \frac{8 I}{7}

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| A_{1} + A_{2} | + | A_{1} A_{2} | = \frac{8^{3}}{7^{3}} + \frac{1}{7^{3}}

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| A_{1} + A_{2} | = ∣ ∣ \frac{8 I}{7} ∣ ∣

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| A_{1} A_{2} | = ∣ ∣ \frac{7 I}{8} ∣ ∣

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Solution

The correct options are
B $| A_{1} + A_{2} | + | A_{1} A_{2} | = \frac{8^{3}}{7^{3}} + \frac{1}{7^{3}}$
C $| A_{1} + A_{2} | = ∣ ∣ \frac{8 I}{7} ∣ ∣$
Given that
$3 A_{1}^{T} + I^{3} = 10 A_{1}$ $\dots (i)$
$3 A_{2} + 4 I^{3} = 7 A_{2}^{T}$ $\dots (i i)$
Taking transpose of equation (i)
$(3 A_{1}^{T} + I^{3})^{T} = 10 A_{1}^{T} 3 A_{1} + I^{3} = 10 A_{1}^{T}$ $\dots (i i i)$
Similarly from equation (ii)
$(3 A_{2} + 4 I^{3})^{T} = 7 A_{2}$
$3 A_{2}^{T} + 4 I^{3} = 7 A_{2}$ $\dots (i v)$
From equation (i) and (iii) $A_{1} = \frac{13 I}{91} = \frac{I}{7}$
From equation (ii) and (iv)
$A_{2} = I$
$| A_{1} + A_{2} | = ∣ ∣ \frac{8 I}{7} ∣ ∣$
$| A_{1} + A_{2} | + | A_{1} A_{2} | = \frac{8^{3}}{7^{3}} + \frac{1}{7^{3}} = \frac{8^{3} + 1}{7^{3}}$

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