Q.
The value of the determinant Δ=∣∣​13​+3​15​+26​3+65​​25​515​​5​10​5​∣∣​ is equal to
2213
166
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Solution:
Taking 5​ common from C2​ and C3​ , we get Δ=(5​)2∣∣​13​+3​15​+26​3+65​​25​3​​12​5​​∣∣​
Applying C1​→C1​−13​C3​−3​C2​, we get Δ=(5)∣∣​−3​00​25​3​​12​5​​∣∣​=5(−3​)(5−6​) =5(18​)−253​=152​−253​