Question

If A is a square Matrix such that A² = A, then Find the value of (m + n), If(I + A)³ – 7A = m I + nA. (Where m and n are Constants)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:A \: is \: a \: square \: matrix \: such \: that \: {A}^{2} = A$

Further, given that

$\rm :\longmapsto\: {(I + A)}^{3} - 7A = mI + nA$

Now, Consider

$\rm :\longmapsto\: {(I + A)}^{3} - 7A$

can be rewritten as

$\rm \:  =  \: (I + A)(I + A)(I + A) - 7A$

$\rm \:  =  \: ( {I}^{2} + IA + AI + {A}^{2}) (I + A) - 7A$

We know,

$\purple{\rm :\longmapsto\:\boxed{\tt{ {I}^{2} = I \: \: and \: \: AI = IA = A \: }}}$

So, using this, we get

$\rm \:  =  \: (I +A + A + A)(I + A) - 7A$

$\rm \:  =  \: (I +3A)(I + A) - 7A$

$\rm \:  =  \: {I}^{2} + IA + 3AI + {3A}^{2} - 7A$

$\rm \:  =  \: I+ A + 3A + 3A - 7A$

$\rm \:  =  \: I+ 7A - 7A$

$\rm \:  =  \: I$

can be rewritten as

$\rm \:  =  \: I + 0A$

So,

$\rm :\longmapsto\: {(I + A)}^{3} - 7A = I + 0A$

But, it is given that,

$\rm :\longmapsto\: {(I + A)}^{3} - 7A = mI + nA$

So, on comparing, we get

$\purple{\rm :\longmapsto\:m \: = \: 1 \: \: \: and \: \: \: n \: = \: 0 \: }$

Hence,

$\bf\implies \:m + n = 1$

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LEARN MORE

1. Matrix multiplication is defined when number of columns of pre multiplier is equal to the number of rows of post multiplier.

2. Matrix multiplication is Commutative, i.e. AB = BA

3. Matrix multiplication is Associative, i.e. (AB)C = A(BC)

4. Matrix multiplication is Distributive, i.e. A(B + C) = AB + AC