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Standard XII

Mathematics

Cramer's Rule

1 0 -1 2

Question

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Solution

The correct option is B
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Similar questions

Draw the graph for the following matrices.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Q. If A + B =

⎡ ⎢ ⎣ \begin{matrix} 1 & 0 & 2 2 & 2 & 2 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

and A - B =

⎡ ⎢ ⎣ \begin{matrix} 1 & 4 & 4 4 & 2 & 0 - 1 & - 1 & 2 \end{matrix} ⎤ ⎥ ⎦

then prove that
A =

⎡ ⎢ ⎣ \begin{matrix} 1 & 2 & 3 3 & 2 & 1 0 & 0 & 2 \end{matrix} ⎤ ⎥ ⎦

and B =

⎡ ⎢ ⎣ \begin{matrix} 0 & - 2 & - 1 - 1 & 0 & 1 1 & 1 & 0 \end{matrix} ⎤ ⎥ ⎦

Evaluate the determinant
$(i)$ $∣ ∣ ∣ ∣ \begin{matrix} 3 & - 1 & - 2 0 & 0 & - 1 3 & - 5 & 0 \end{matrix} ∣ ∣ ∣ ∣$
$(i i)$ $∣ ∣ ∣ ∣ \begin{matrix} 3 & - 4 & 5 1 & 1 & - 2 2 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣$
$(i i i)$ $∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 2 - 1 & 0 & - 3 - 2 & 3 & 0 \end{matrix} ∣ ∣ ∣ ∣$
$(i v)$ $∣ ∣ ∣ ∣ \begin{matrix} 2 & - 1 & - 2 0 & 2 & - 1 3 & - 5 & 0 \end{matrix} ∣ ∣ ∣ ∣$

Q. Evaluate the determinants
(i)

∣ ∣ ∣ ∣ \begin{matrix} 3 & - 1 & - 2 0 & 0 & - 1 3 & - 5 & - 0 \end{matrix} ∣ ∣ ∣ ∣

(ii)

∣ ∣ ∣ ∣ \begin{matrix} 3 & - 4 & 5 1 & 1 & - 2 2 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣

(iii)

∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 2 - 1 & 0 & - 3 - 2 & 3 & 0 \end{matrix} ∣ ∣ ∣ ∣

(iv)

∣ ∣ ∣ ∣ \begin{matrix} 2 & - 1 & - 2 0 & 2 & - 1 3 & - 5 & 0 \end{matrix} ∣ ∣ ∣ ∣

Q. The determinant of the matrix given below is

∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 1 & 0 & 2 - 1 & 1 & 1 & 3 0 & 0 & 0 & 1 1 & - 2 & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣

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The correct option is B 0

The correct option is B
0