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Matrices — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsMatrices2 Marks
Question
If A = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]$ and B = $\left[\begin{array}{rrr} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$, then verify (A – B)′ = A′ – B′
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Answer

We will first calculate L.H.S i.e. (A+B)’ and then consecutively we will calculate R.H.S and verify that both are equal.
So, A - B = $\left[\begin{array}{ccc} {-1} & {2} & {3} \\ {5} & {7} & {9} \\ {-2} & {1} & {1} \end{array}\right]-\left[\begin{array}{ccc} {-4} & {1} & {-5} \\ {1} & {2} & {0} \\ {1} & {3} & {1} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {-1-(-4)} & {2-1} & {3-(-5)} \\ {5-1} & {7-2} & {9-0} \\ {-2-1} & {1-3} & {1-1} \end{array}\right]$
$\Rightarrow A-B=\left[\begin{array}{ccc} {3} & {1} & {8} \\ {4} & {5} & {9} \\ {-3} & {-2} & {0} \end{array}\right]$
Therefore, $(A-B)^\prime$ = $\left[\begin{array}{ccc} {3} & {4} & {-3} \\ {1} & {5} & {-2} \\ {8} & {9} & {0} \end{array}\right]$ ...(1)
Now, $A^{\prime}=\left[\begin{array}{ccc} {-1} & {5} & {-2} \\ {2} & {7} & {1} \\ {3} & {9} & {1} \end{array}\right] \text { and } B^{\prime}=\left[\begin{array}{ccc} {-4} & {1} & {1} \\ {1} & {2} & {3} \\ {-5} & {0} & {1} \end{array}\right]$
So, A' - B' = $\left[\begin{array}{ccc} {-1} & {5} & {-2} \\ {2} & {7} & {1} \\ {3} & {9} & {1} \end{array}\right]-\left[\begin{array}{ccc} {-4} & {1} & {1} \\ {1} & {2} & {3} \\ {-5} & {0} & {1} \end{array}\right]$
$\Rightarrow A^{\prime}-B^{\prime}\left[\begin{array}{ccc} {-1-(-4)} & {5-1} & {-2-1} \\ {2-1} & {7-2} & {1-3} \\ {3-(-5)} & {9-0} & {1-1} \end{array}\right]$
$\Rightarrow \mathrm{A}^{\prime}-\mathrm{B}^{\prime}=\left[\begin{array}{ccc} {3} & {4} & {-3} \\ {1} & {5} & {-2} \\ {8} & {9} & {0} \end{array}\right] $...(2)
From equation (1) & (2) we verify that
$(A-B)^\prime=A^\prime-B^\prime$.
Hence verified.

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